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Math 275: Calculus IIILecture Notes by Angel V. Kumchev
Contents
1. Three-dimensional coordinate systems . . . . . . . . . . . . . . .. . . . . . . . . . 12. Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83. Lines and planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 174. Vector functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 235. Space curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 276. Multivariable functions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 367. Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .438. Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 499. Directional derivatives and gradients . . . . . . . . . . . . . . .. . . . . . . . . . . 5810. Tangent planes and normal vectors . . . . . . . . . . . . . . . . . . .. . . . . . . . 6111. Extremal values of multivariable functions . . . . . . . . . .. . . . . . . . . . . . 6812. Double integrals over rectangles . . . . . . . . . . . . . . . . . . .. . . . . . . . . 7513. Double integrals over general regions . . . . . . . . . . . . . . .. . . . . . . . . . 8014. Double integrals in polar coordinates . . . . . . . . . . . . . . .. . . . . . . . . . 8815. Triple integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 9316. Triple integrals in cylindrical and spherical coordinates . . . . . . . . . . . . . . . . 9717. Applications of double and triple integrals . . . . . . . . . .. . . . . . . . . . . . . 10218. Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11019. Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 11820. The fundamental theorem for line integrals . . . . . . . . . . .. . . . . . . . . . . 12621. Green’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 13122. Surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 13823. Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 14524. Gauss’ divergence theorem . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 150Appendix A. Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 156Appendix B. Definite integrals of single-variable functions. . . . . . . . . . . . . . . . 157Appendix C. Polar coordinates in the plane . . . . . . . . . . . . . . . .. . . . . . . . 159Appendix D. Answers to selected exercises . . . . . . . . . . . . . . .. . . . . . . . . 162
1. Three-dimensional coordinate systems
1.1. Cartesian coordinates in space.The three-dimensional Cartesian coordinate systemcon-sists of a pointO in space, called theorigin, and of three mutually perpendicular oriented linesthroughO, called thecoordinate axesand labeled thex-axis, they-axis, and thez-axis. The la-beling of the axes is usually done according to theright-hand rule: if one places one’s right handin space so that the origin lies on one’s palm, they-axis points toward one’s arm, and thex-axispoints toward one’s curled fingers, then thez-axis points toward one’s thumb.
There are several standards for drawing two-dimensional models of three-dimensional objects.In particular, there are three somewhat standard ways to draw a three-dimensional coordinate sys-tem. These are shown on Figure 1.1. Model (c) is the one commonly used in manual sketches.Model (a) is a version of model (c) that is commonly used in textbooks and will be the one weshall use in these notes; unlike model (c), it is based on realmathematics that transforms three-dimensional coordinates to two-dimensional ones. Model (b) is the default model preferred bycomputer algebra systems such asMathematica.
z
yx
(a)
z
y
x
(b)
z
yx
(c)
Figure 1.1. Three-dimensional Cartesian coordinate systems
The coordinate axes determine several notable objects in space. First, we have thecoordinateplanes: these are the unique planes determined by the three different pairs of coordinate axes. Thecoordinate planes determined by thex- andy-axes, by thex- andz-axes, and by they- andz-axesare called thexy-plane, thexz-plane, and theyz-plane, respectively. These three coordinate planedivide space into eight “equal” parts, calledoctants; the octants are similar to the four quadrantsthat the two coordinate axes divide thexy-plane into. The octants are labeled I through VIII, sothat octants I through IV lie above the respective quadrantsin the xy-plane and octants V throughVIII lie below those quadrants.
z
yx
bb
b
b
P
Px
Pz
Py
c
b a
Figure 1.2. The Cartesian coordinates of a point in space
Next, we define the Cartesian coordinates of a pointP in space. We start by finding its orthogonalprojections onto theyz-, the xz-, and thexy-planes. Let us denote those points byPx, Py, andPz,
1
respectively (see Figure 1.2). Leta be the directed distance betweenPx andP (that is,a is the said
distance or its negative according as the arrow−−→PxP points in the direction of thex-axis or opposite
to it), let b be the directed distance betweenPy andP, and letc be the directed distance betweenPz
andP. Then thethree-dimensional Cartesian coordinatesof P are (a,b, c) and we writeP(a,b, c).
Example 1.1.Plot the pointsA(4,3,0), B(2,−2,2), andC(2,2,−2).
Example 1.2. Determine the sets of points described by the equations:z = 1; x = −2; andy = 3, x = −1.
Answer.The equationz = 1 represents a plane parallel to thexy-plane. The equationx = −2represents a plane parallel to theyz-plane. The equationsy = 3, x = −1 represent a vertical lineperpendicular to thexy-plane and passing through the point (3,−1,0). �
1.2. Distance in space.Let P1 = (x1, y1, z1) andP2 = (x2, y2, z2) be two points in space. Then thedistance betweenP1 andP2 is given by
|P1P2| =√
(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2. (1.1)
Example 1.3.The distance between the pointsA(4,3,0) andB(2,−2,2) is
|AB| =√
(4− 2)2 + (3+ 2)2 + (0− 2)2 =√
33.
�
Example 1.4.Find the equation of the sphere of radius 3 centered atC(1,−2,0).
Solution.The sphere of radius 3 centered atC(1,−2,0) is the set of pointsP(x, y, z) such that
|PC| =√
(x− 1)2 + (y+ 2)2 + (z− 0)2 = 3,
that is, the pointsP(x, y, z), whose coordinates satisfy the equation
(x− 1)2 + (y+ 2)2 + z2 = 9. �
More generally, we have the following
Fact. The sphere with center C(a,b, c) and radius r> 0 has an equation
(x− a)2 + (y− b)2 + (z− c)2 = r2. (1.2)
Example 1.5.The equationx2 + y2 + z2 − 4x+ 6y+ z= 7
describes a sphere. Find its center and radius.
Solution.We can rewrite the above equation as
x2 − 4x+ 4+ y2 + 6y+ 9+ z2 + z+ 14 = 7+ 4+ 9+ 1
4
(x− 2)2 + (y+ 3)2 + (z+ 12)2 = 81
4 .
The last equation is of the form (1.2) witha = 2, b = −3, c = −12, andr = 9
2. Hence, the givenequation describes a sphere with center (2,−3,−1
2) and radius92. �
2
1.3. Cylindrical coordinates. You should be familiar from single-variable calculus with the po-lar coordinate system in the plane (see also Appendix C). In the remainder of this lecture, wedefine two sets of coordinates in space that extend the polar coordinates in the plane to the three-dimensional setting. We start with the so-called “cylindrical coordinates”. The basic idea of thecylindrical coordinate system is to replace the Cartesian coordinatesx, y in the xy-plane by thecorresponding polar coordinates and to keep the Cartesianz-coordinate.
z
yx
b
b
P(x, y, z)
P0(x, y)
z
θ r
Figure 1.3. The cylindrical coordinates of a point in space
Let P be a point in space with Cartesian coordinates (x, y, z). Thecylindrical coordinatesof Pare the triple (r, θ, z), where (r, θ) are the polar coordinates of the pointP0(x, y) in the xy-plane(see Figure 1.3). In other words, we pass between Cartesian and cylindrical coordinates in spaceby passing between Cartesian and polar coordinates in thexy-plane and keeping thez-coordinateunchanged. The formulas for conversion from cylindrical toCartesian coordinates are
x = r cosθ, y = r sinθ, z= z, (1.3)
and those for conversion from Cartesian to cylindrical coordinates are
r =√
x2 + y2, θ = arctan(y/x), z= z. (1.4)
The formulaθ = arctan(y/x) comes with some strings attached (see§C.2 in Appendix C).
Example 1.6.Find the Cartesian coordinates of the pointsP(2, π,e) andQ(4,arctan(−2),2.7).
Solution.The Cartesian coordinates ofP are (x, y,e), where
x = 2 cosπ = −2, y = 2 sinπ = 0.
That is,P(−2,0,e).The Cartesian coordinates ofQ are (x, y,2.7), where
x = 4 cos(arctan(−2)) = 4 cos(arctan 2)=4√5,
y = 4 sin(arctan(−2)) = −4 sin(arctan 2)= − 8√5.
That is,Q( 4√5,− 8√
5,2.7). �
Example 1.7.Find the cylindrical coordinates of the pointsP(3,4,5) andQ(−2,2,−1).
Solution.The cylindrical coordinates ofP are (r, θ,5), where
r =√
32 + 42 = 5, tanθ =43,
3
andθ lies in the first quadrant. Hence,θ = arctan(43) and the cylindrical coordinates ofP are(5,arctan(43),5).
The cylindrical coordinates ofQ are (r, θ,−1), where
r =√
22 + (−2)2 = 2√
2, tanθ = −1,
and θ lies in the second quadrant. Hence,θ = 3π4 and the cylindrical coordinates ofP are
(2√
2, 3π4 ,−1). �
Example 1.8.Find the Cartesian equation of the surfacer = cosθ. What kind of surface is this?
Solution.We have
r = cosθ ⇐⇒ r2 = r cosθ ⇐⇒ x2 + y2 = x.
The last equation is the equation of a circle of radius12 centered at the point (1
2,0) in thexy-plane:
x2 + y2 = x ⇐⇒ x2 − x+ 14 + y2 = 1
4 ⇐⇒ (x− 12)2 + y2 = 1
4.
Here, however, we use the equationx2 + y2 = x to describe a surface in space, so we view it as anequation inx, y, z. Sincez does not appear in the equation, if thex- andy-coordinates of a pointsatisfy the equationx2+y2 = x, then the point is on the surface. In other words, ifa2+b2 = a, thenall points (a,b, z), whosex-coordinate isa andy-coordinate isb, are on the surface. We also knowthat all the points (x, y,0), with x2 + y2 = x, are on the surface: these are the points on the circlein the xy-plane with the same equation. Putting these two observations together, we conclude thatthe given surface is the cylinder that consists of all the vertical lines passing through the points ofthe circlex2 + y2 = x in thexy-plane (see Figure 1.4). �
yx
z
Figure 1.4. The cylinderx2 + y2 = x
1.4. Spherical coordinates.We now introduce the so-called “spherical coordinates” in space. Inorder to define this set of coordinates, we choose a special point O, theorigin, a plane that containsO, and a pair of mutually perpendicular axes passing throughO—one that lies in the chosen planeand another that is orthogonal to the chosen plane. Since we want to relate the spherical coordinatesof a point to its Cartesian coordinates, it is common to chooseO to be the origin ofO(0,0,0) ofthe Cartesian coordinate system, the special plane to be thexy-plane, and the two axes to be thex-axis and thez-axis. With these choices, thespherical coordinatesof a pointP(x, y, z) in space are(ρ, θ,φ), whereρ = |OP|, θ is the angular polar coordinate of the pointP0(x, y) in thexy-plane, andφ is the angle betweenOP and the positive direction of thez-axis (see Figure 1.5). It is commonto restrictφ to the range 0≤ φ ≤ π.
4
z
yx
O
b
b
P(x, y, z)
P0(x, y)
z= ρ cosφ
θ
φ
ρ
r = ρ sinφ
Figure 1.5. The spherical coordinates of a point in space
Consider a pointP(x, y, z) in space and letP0(x, y,0). Also, let (ρ, θ,φ) be the spherical coordi-nates ofP. From the right triangle△OPP0, we find that
z= ±|P0P| = |OP| cosφ = ρ cosφ, |OP0| = |OP| sinφ = ρ sinφ.
On the other hand, using polar coordinates in thexy-plane, we find that
x = |OP0| cosθ = ρ sinφ cosθ, y = |OP0| sinθ = ρ sinφ sinθ.
Thus, the formulas for conversion from spherical to Cartesian coordinates are
x = ρ sinφ cosθ, y = ρ sinφ sinθ, z= ρ cosφ. (1.5)
The formulas for conversion from Cartesian to spherical coordinates are
ρ =√
x2 + y2 + z2, θ = arctan(y/x), φ = arccos(z/ρ), (1.6)
where the formula forθ comes with some strings attached (the same as in the cases of polar andcylindrical coordinates).
Example 1.9.Find the Cartesian coordinates of the pointP(2, π/3,3π/4).
Solution.The Cartesian coordinates ofP are
x = 2 sin(3π/4) cos(π/3) =1√2, y = 2 sin(3π/4) sin(π/3) =
√3√2, z= 2 cos(3π/4) = −
√2. �
Example 1.10.Find the spherical coordinates of the pointsP(1,−1,−√
6) andQ(−2,2,−1).
Solution.The spherical coordinates ofP are
ρ =√
1+ 1+ 6 =√
8, φ = arccos(
−√
6/√
8)
= arccos(
−√
3/2)
=5π6,
and the solutionθ oftanθ = −1, 3π/2 < θ < 2π.
Hence,θ = 7π/4 and the spherical coordinates ofP are (√
8,7π/4,5π/6).The spherical coordinates ofQ are
ρ =√
4+ 4+ 1 = 3, φ = arccos(−1/3),
and the solutionθ oftanθ = −1, π/2 < θ < π.
Hence,θ = 3π/4 and the spherical coordinates ofQ are (3,3π/4,arccos(−1/3)). �
5
Example 1.11.Find the Cartesian equation of the surfaceρ2(sin2φ − 2 cos2φ) = 1.
Solution.Sincex2 + y2 = ρ2 sin2φ cos2 θ + ρ2 sin2φ sin2 θ = ρ2 sin2φ,
we haveρ2 sin2φ − 2ρ2 cos2φ = 1 ⇐⇒ x2 + y2 − 2z2 = 1.
The last equation represents a surface known as “one-sheet hyperboloid”; it is shown on Figure 1.6.�
yx
z
Figure 1.6. The hyperboloidx2 + y2 − 2z2 = 1
yx
z
Figure 1.7. The region 2≤ ρ ≤ 3, π/2 ≤ φ ≤ π
Example 1.12.Describe the solid defined by the inequalities
2 ≤ ρ ≤ 3, π/2 ≤ φ ≤ π.
Solution.The second inequality restricts the points of the solid to the lower half-space. The in-equalityρ ≤ 3 describes the points in space that are within distance of 3 from the origin, that is,the interior of the spherex2 + y2 + z2 = 9. Similarly the inequalityρ ≥ 2 describes the exterior ofthe spherex2 + y2 + z2 = 4. Thus, the given solid is obtained from the lower half of theinterior ofthe spherex2 + y2 + z2 = 9 by removing from it the interior of the smaller spherex2 + y2 + z2 = 4(see Figure 1.7). �
Example 1.13.Find the cylindrical and spherical equations of the spherex2 + y2 + z2 + 4z= 0.
Solution.By (1.4),x2 + y2 = r2, so the cylindrical form of the given equation is
x2 + y2 + z2 + 4z= 0 ⇐⇒ r2 + z2 + 4z= 0.
By (1.6),x2 + y2 + z2 = ρ2, so the spherical form of the given equation is
ρ2 + 4ρ cosφ = 0 ⇐⇒ ρ = −4 cosφ. �
Exercises
1. Which of the pointsP(1,3,2), Q(0,−2,−1), andR(5,3,−3):(a) is closest to thexy-plane; (b) lies in thexz-plane; (c) lies in theyz-plane?
2. Describe the set inR3 represented by the equationx+ 2y = 2.
3. Find the side lengths of the triangle with verticesA(0,6,2), B(3,4,1), C(1,3,4). Is this a special kind of atriangle: equilateral, isosceles, right?
6
4. Find an equation of the sphere that passes throughP(4,2,1) and has centerC(1,1,1).
5. Show that the equationx2 + y2 + z2 = x+ 2y+ 4z represents a sphere and find its center and radius.
6. Describe verbally the set inR3 represented by the given equation or inequality:(a) x = −3(d) (x− 1)2 + y2 + z2 ≥ 4
(b) z≤ 2(e) 4< x2 + y2 + z2 < 9
(c) −1 ≤ y ≤ 2(f) xy= 0
7. Write inequalities to describe the region:(a) the half-space consisting of all points to the right of the planey = −2;(b) the interior of the upper hemisphere of radius 2 centeredat the origin.
8. Find the Cartesian coordinates of the point whose cylindrical coordinates are:(a) (2,0,3) (b) (3, π/4,−3) (c) (3,7π/6,2)
9. Find the cylindrical coordinates of the point whose Cartesian coordinates are:(a) (1,−1,4) (b) (−2,2,2) (c) (
√3,1,−2)
10. Find the Cartesian coordinates of the point whose spherical coordinates are:(a) (1,0,0) (b) (3, π/4,3π/4) (c) (3,7π/6, π/2)
11. Find the spherical coordinates of the point whose Cartesian coordinates are:(a) (1,1,2) (b) (−1,1,−
√2) (c) (
√3,1,0)
12. Find the cylindrical and spherical equations of the surface represented by the (Cartesian) equation:(a) x2 + y2 = 2z (b) x2 + 2x+ y2 + z2 = 1 (c)z=
√
x2 + y2
13. Find the Cartesian equation of the surface represented by the given cylindrical or spherical equation. Ifpossible, describe the surface verbally.
(a)ρ cosφ = −2(d) ρ = 2 sinφ sinθ
(b) r = 2 sinθ(e)ρ sin2φ − 2 cosφ = 0
(c) r2 + z2 = 4(f) r cosθ − 2r sinθ = 3
14. Describe verbally the set inR3 represented by the given conditions:(a)θ = π/4(c) r = 1, 0 ≤ z≤ 1(e)ρ ≤ 1, −π/2 ≤ θ ≤ π/2
(b) ρ = 2(d) 2≤ ρ ≤ 4, 0 ≤ φ ≤ π/2(f) ρ sinφ ≤ 1, π/2 ≤ φ ≤ π
7
2. Vectors
2.1. Geometric vectors. A vector is a quantity (such as velocity, acceleration, force, impulse,etc.) that has bothmagnitudeanddirection. In mathematics, we consider two types of vectors:geometric and algebraic. Ageometric vectoris a directed line segment (also called anarrow): thelength of the arrow represents the magnitude of the vector and the arrow points in the direction ofthe vector. We will denote a vector byv or ~v; if v is the vector with initial pointA and terminal
point B, we also writev =−−→AB.
Two vectorsu andv areequal, u = v, if they have equal magnitudes (lengths) and point in thesame direction. The vector of zero length is called thezero vectorand is denoted0; it is the onlyvector with no specific direction.
2.2. Vector addition. If u =−−→ABandv =
−−→BC are two vectors (so that the terminal point of the first
vector is the initial point of the second), we definetheir sumu + v by (see Figure 2.1a)
u + v =−−→AB+
−−→BC =
−−→AC.
If u =−−→ABandv =
−−→CD, with C , B, we first representv by an equal arrow
−−−→BD′ and then apply the
above definition:u + v =−−−→AD′ (see Figure 2.1b).
A
B
C
u
vu +
v
(a)A
BC
DD′
u
vv u +
v
(b)
Figure 2.1. Vector addition
2.3. Multiplication by scalars. Let v be a vector and letc be a real number (called also ascalar).Thescalar multiple cv is the vector whose length is|c| times the length ofv and whose direction isthe same as the direction ofv if c > 0 or opposite to the direction ofv if c < 0. If c = 0 or v = 0,thencv = 0.
The vector (−1)v has the same length asv but the opposite direction. We call this vector thenegative ofv and denote it by−v. For two vectorsu andv, we definetheir differenceu − v by
u − v = u + (−v).
Often, we visualize the sum and the difference of two vectorsu andv by drawing them so that theyshare the same initial point and considering the parallelogram they define. Then the sumu + v isrepresented by the diagonal of the parallelogram that passes through the common initial point ofuandv and the differenceu − v is represented by the other diagonal (see Figure 2.2).
8
u
vu +
v
(a)
u
v
u − v
(b)
Figure 2.2. The sum and the difference ofu andv
2.4. Coordinate representation of vectors. Algebraic vectors.Often, it is convenient to intro-duce a coordinate system and to work with vectors algebraically instead of geometrically. If we
represent a (geometric) vectora by the arrow−−→OA, whereO is the origin of a Cartesian coordinate
system, the terminal pointA of a has coordinates (a1,a2) or (a1,a2,a3), depending on the dimen-sion of the coordinate system. We call these coordinates thecoordinatesor thecomponentsof thealgebraic vectora and write
a = 〈a1,a2〉 or a = 〈a1,a2,a3〉.
We will use the same algebraic vector to represent all the geometric vectors equal to−−→OA. For
example, ifA(1,−1), B(−1,3), andC(0,2), then−−→BC =
−−→OA= 〈1,−1〉.
Thus, for any two pointsA(a1,a2,a3) andB(b1,b2,b3), the algebraic vectorv representing the arrow−−→AB is
v = 〈b1 − a1,b2 − a2,b3 − a3〉.From now on, we will state all the formulas for three-dimensional vectors. One usually obtainsthe respective two-dimensional definition or result simplyby dropping the last component of thevectors; any exceptions to this rule will be mentioned explicitly when they occur.
2.5. Operations with algebraic vectors. Note that when we write a vectora algebraically, thedistance formula provides a convenient tool for computing the length (also called thenorm) ‖a‖ ofthat vector: ifa = 〈a1,a2,a3〉, then
‖a‖ =(
a21 + a2
2 + a23
)1/2. (2.1)
Furthemore, ifa = 〈a1,a2,a3〉 and b = 〈b1,b2,b3〉 are vectors andc is a scalar, we have thefollowing coordinate formulas fora+ b, a− b, andca:
a+ b = 〈a1 + b1,a2 + b2,a3 + b3〉, (2.2)
a− b = 〈a1 − b1,a2 − b2,a3 − b3〉, (2.3)
ca = 〈ca1, ca2, ca3〉. (2.4)
In other words, we add vectors, subtract vectors, and multiply vectors by scalars by simply per-forming the respective operations componentwise.
Example 2.1.Consider the vectorsa = 〈1,−2,3〉 andb = 〈0,1,−2〉. Then
2a+ b = 〈2,−4,6〉 + 〈0,1,−2〉 = 〈2,−3,4〉9
and
‖2a+ b‖ =√
22 + (−3)2 + 42 =√
29.
�
The algebraic operations with vectors have several properties that resemble some of the proper-ties of the algebraic operations with numbers known to you since grade school.
Theorem 2.1.Leta, b, andc denote vectors and let c, c1, c2 denote scalars. Then:
1. a+ b = b + a2. a+ (b + c) = (a+ b) + c3. a+ 0 = a4. a+ (−a) = 05. c(a+ b) = ca+ cb6. (c1 + c2)a = c1a+ c2a7. (c1c2)a = c1(c2a)8. 1a = a9. ‖a‖ ≥ 0, with ‖a‖ = 0 if and only ifa = 0
10. ‖ca‖ = |c| · ‖a‖11. ‖a+ b‖ ≤ ‖a‖ + ‖b‖ (the triangle inequality)
Proof. The proofs of 1)–8) are all similar: they use (2.2)–(2.4) to deduce 1)–8) from the respectiveproperties of numbers. For example, to prove the distributive law 6), we writea = 〈a1,a2,a3〉 andargue as follows:
(c1 + c2)a = 〈(c1 + c2)a1, (c1 + c2)a2, (c1 + c2)a3〉 by (2.4)
= 〈c1a1 + c2a1, c1a2 + c2a2, c1a3 + c2a3〉 by the distributive law for numbers
= 〈c1a1, c1a2, c1a3〉 + 〈c2a1, c2a2, c2a3〉 by (2.2)
= c1a+ c2a by (2.4).
The proofs of 9) and 10) are also easy. For instance,
‖ca‖ =(
(ca1)2 + (ca2)
2 + (ca3)2)1/2
=(
c2(a21 + a2
2 + a23))1/2=√
c2(
a21 + a2
2 + a23
)1/2= |c| · ‖a‖.
The proof of 11) is slightly more involved, so we won’t discuss it. �
2.6. Unit vectors and the standard basis.The simplest nonzero vectors are:
e1 = 〈1,0,0〉 , e2 = 〈0,1,0〉 , e3 = 〈0,0,1〉 .
These are called thestandard basis vectors. Sometimes, especially in physics and engineering,the vectorse1, e2, ande3 are denoted byi, j , andk, respectively, but we shall stick with the abovenotation, because it is more common in mathematics (it is also more consistent with generalizationsto higher dimensions).
Because of (2.2)–(2.4), it is always possible to express a given vector as a combination of thethree standard basis vectors. For example,
〈1,2,4〉 = 〈1,0,0〉 + 〈0,2,0〉 + 〈0,0,4〉 = 〈1,0,0〉 + 2 〈0,1,0〉 + 4 〈0,0,1〉 = e1 + 2e2 + 4e3.10
More generally, ifa = 〈a1,a2,a3〉, we have
a = 〈a1,0,0〉 + 〈0,a2,0〉 + 〈0,0,a3〉 = a1e1 + a2e2 + a3e3. (2.5)
Notice thate1, e2, ande3 have length 1. In general, vectors of length 1 are calledunit vectors.We remark that ifa is any nonzero vector (i.e.,a , 0), then there is a unique unit vector that pointsin the same direction asa—namely, the vector
u =1‖a‖
a =a‖a‖.
Indeed, ifc = ‖a‖−1, thena andu = ca point in the same direction (sincec > 0) and
‖u‖ = ‖ca‖ = |c| · ‖a‖ = ‖a‖−1 · ‖a‖ = 1.
Example 2.2.Consider the vectora = 〈2,1,2〉. Its length is
‖a‖ =√
22 + 12 + 22 = 3,
so a unit vector in the direction ofa is13a =
⟨
23,
13,
23
⟩
.
�
2.7. The dot product. We can now add two vectors and we can multiply a vector by a scalar. Isthere a way to multiply two vectors? The first way that comes tomind is to multiply the vectorscomponentwise:
ab = 〈a1,a2,a3〉 〈b1,b2,b3〉 = 〈a1b1,a2b2,a3b3〉 ,but this way of multiplying two vectors turns out to have no practical meaning. Instead, throughthe remainder of this lecture, we will talk about two meaningful ways to multiply vectors: the dotproduct and the cross product.
Definition. Given three-dimensional vectorsa = 〈a1,a2,a3〉 andb = 〈b1,b2,b3〉, we define theirdot producta · b to be
a · b = a1b1 + a2b2 + a3b3. (2.6)
Notice that the dot product of two vectors is a scalar, not a vector. Sometimes, the dot productis called also theinner or thescalar product ofa and b. Because of this, you should avoid thetemptation to refer to the scalar multipleca as the “scalar product ofc anda”.
Example 2.3.The dot product ofa = 〈4,2,−1〉 andb = 〈−2,2,1〉 is
a · b = (4)(−2)+ (2)(2)+ (−1)(1)= −5.
The dot product ofu = 2e1 + 3e2 + e3 andv = e1 + 4e2 is
u · v = (2)(1)+ (3)(4)+ (1)(0)= 14.
�
The next theorem summarizes some important algebraic properties of the dot product.
Theorem 2.2.Leta, b, andc denote vectors and let c1, c2 denote scalars. Then:
1. a · a = ‖a‖22. a · 0 = 03. a · b = b · a
11
4. (c1a) · (c2b) = c1c2(a · b)5. (a+ b) · c = a · c+ b · c
Proof. These follow from the definition of the dot product and from the properties of the realnumbers. For example, ifa = 〈a1,a2,a3〉 andb = 〈b1,b2,b3〉, we have
a · b = a1b1 + a2b2 + a3b3 by (2.6)
= b1a1 + b2a2 + b3a3 by the commutative law for real numbers
= b · a by (2.6).
That is, 3) holds. �
2.8. Geometric meaning of the dot product. We said earlier that our definition of the dot productis motivated by applications. The next theorem relates the dot product of two vectors to theirgeometrical representations.
Theorem 2.3. If a andb are two vectors, then
a · b = ‖a‖ ‖b‖ cosθ,
whereθ is the angle betweena andb.
Corollary. If a andb are two nonzero vectors andθ is the angle between them, then
cosθ =a · b‖a‖ ‖b‖
.
Example 2.4.Find the angle betweena = 〈1,2,−1〉 andb = 〈0,2,1〉.
Solution.The angleθ betweena andb satisfies
cosθ =(1)(0)+ (2)(2)+ (−1)(1)
√
12 + 22 + (−1)2√
02 + 22 + 12=
3√6√
5=
3√30,
soθ = arccos(
3√30
)
= 0.9911. . . . �
We say that two vectorsa andb areorthogonal(or perpendicular) if a · b = 0. By the corollary,a · b = 0 implies that the angle betweena andb is 90◦, so this definition makes perfect sense.
Note that ifa = 〈a1,a2,a3〉, then we can use the dot products ofa and the standard basis vectorsej to extract the components ofa:
a1 = a · e1, a2 = a · e2, a3 = a · e3.
More generally, suppose thatu is a unit vector. Thena · u is a scalar and
projua = (a · u)u
is a vector that is parallel tou. This vector is called the(orthogonal) projection ofa ontou. It hasthe property that the difference between it and the original vectora is perpendicular tou:
(
a− projua)
· u = 0.
The two-dimensional case of this is illustrated on Figure 2.3. Note that the size of the dot producta · u equals the length of the projection.
12
ua
projua
a− projua
Figure 2.3. The projection of a vector onto a unit vector
2.9. The cross product. The dot product of two vectors is a scalar. In this section, wedefineanother product of two vectors: their “cross product”, which is a vector. Our discussion requiresbasic familiarity with 2× 2 and 3× 3 determinants, so you may want to give a quick look toAppendix A before proceeding further.
Definition. Let a = 〈a1,a2,a3〉 andb = 〈b1,b2,b3〉 be two three-dimensional vectors. Then theircross producta× b is the vector
a× b =
∣
∣
∣
∣
a2 a3
b2 b3
∣
∣
∣
∣
e1 −∣
∣
∣
∣
a1 a3
b1 b3
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
a1 a2
b1 b2
∣
∣
∣
∣
e3.
Remarks. There are a couple of things we should point out:
1. The cross product is defined only for three-dimensional vectors. There is no analog of thisoperation for two-dimensional vectors.
2. The defining formula may be confusing at first. On the other hand, following the exactlabelling is important because of the properties of the cross product (see the Theorembelow). Another representation of the cross product that ismuch easier to remember is
a× b =
∣
∣
∣
∣
∣
∣
e1 e2 e3
a1 a2 a3
b1 b2 b3
∣
∣
∣
∣
∣
∣
. (2.7)
This expression is not exactly a determinant in the sense described in Appendix A (sinceits first row consists of vectors instead of numbers), but if we expand this “determinant” asif we were dealing with the determinant of a regular 3× 3 matrix, we get exactly the crossproduct ofa andb.
The following theorem summarizes the algebraic propertiesof the cross product.
Theorem 2.4. Let a, b, and c denote three-dimensional vectors and let c, c1, c2 denote scalars.Then
1. b × a = −(a× b)2. (c1a) × (c2b) = c1c2(a× b)3. a× (b + c) = a× b + a× c4. (a+ b) × c = a× c+ b × c5. a× a = 0 anda× 0 = 06. a× (b × c) = (a× b) × c may fail.
13
Example 2.5.We have
e1 × e2 =
∣
∣
∣
∣
∣
∣
e1 e2 e3
1 0 00 1 0
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
0 01 0
∣
∣
∣
∣
e1 −∣
∣
∣
∣
1 00 0
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
1 00 1
∣
∣
∣
∣
e3 = e3,
and similarly,e2 × e3 = e1, e3 × e1 = e2.
By property 1), we also have
e2 × e1 = −e3, e1 × e3 = −e2, e3 × e2 = −e1.
Note thate1 × (e2 × e1) = e1 × (−e3) = −(e1 × e3) = e2
and(e1 × e2) × e1 = e3 × e1 = e2,
soe1 × (e2 × e1) = (e1 × e2) × e1. On the other hand,
e1 × (e1 × e2) = e1 × e3 = −e2,
whereas by property 5),(e1 × e1) × e2 = 0× e2 = 0.
Thus,e1 × (e1 × e2) , (e1 × e1) × e2; this illustrates property 6) above. �
Example 2.6.Compute the cross product ofa = 〈3,−2,1〉 andb = 〈1,−1,1〉.
Solution.By (2.7),
a× b =
∣
∣
∣
∣
∣
∣
e1 e2 e3
3 −2 11 −1 1
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
−2 1−1 1
∣
∣
∣
∣
e1 −∣
∣
∣
∣
3 11 1
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
3 −21 −1
∣
∣
∣
∣
e3
= −e1 − 2e2 − e3 = 〈−1,−2,−1〉 . �
Second solution.We can also argue by using the algebraic properties in the theorem and the valuesof the cross products ofe1,e2, ande3:
(3e1 − 2e2 + e3) × (e1 − e2 + e3)
= 3e1 × e1 − 2e2 × e1 + e3 × e1 − 3e1 × e2 + 2e2 × e2 − e3 × e2 + 3e1 × e3 − 2e2 × e3 + e3 × e3
= 0+ 2e3 + e2 − 3e3 + 0+ e1 − 3e2 − 2e1 + 0 = −e1 − 2e2 − e3. �
2.10. Geometric meaning of the cross product.
Theorem 2.5.Leta = 〈a1,a2,a3〉 andb = 〈b1,b2,b3〉 be three-dimensional vectors. Then:
i) a×b is perpendicular to botha andb. The direction ofa×b is determined by the right-handrule.
ii) ‖a× b‖ = ‖a‖ ‖b‖ | sinθ|, whereθ is the angle betweena andb. That is, the length ofa× bis equal to the area of the parallelogram determined bya andb.
iii) a× b = 0 if and only ifa andb are parallel.14
iv) The volume V of the parallelepiped determined by three vectors a,b, andc which are notcoplanar(that is, they do not lie in one plane) is given by the formula
V =∣
∣a · (b × c)∣
∣.
Furthermore, three vectorsa,b, andc are coplanar if and only ifa · (b × c) = 0.
Example 2.7.Find two unit vectors perpendicular toa = 〈1,2,3〉 andb = 〈2,1,1〉.Solution.We know from part 1) of the theorem thata× b is orthogonal to botha andb. We findthat
a× b =
∣
∣
∣
∣
∣
∣
e1 e2 e3
1 2 32 1 1
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
2 31 1
∣
∣
∣
∣
e1 −∣
∣
∣
∣
1 32 1
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
1 22 1
∣
∣
∣
∣
e3 = −e1 + 5e2 − 3e3.
Hence,‖a× b‖ =
√
(−1)2 + 52 + (−3)2 =√
35,and two (in fact the only two) unit vectors orthogonal toa andb are
u =a× b‖a× b‖
=
⟨
−1√35,
5√35,−3√35
⟩
and − u =⟨
1√35,−5√35,
3√35
⟩
. �
Example 2.8.Find the area of the triangle with verticesP(1,2,3), Q(−1,3,2), andR(3,−1,2).
Solution.The area of△PQRis one half of the area of the parallelogram determined by thevectors−−→PQ= 〈−2,1,−1〉 and
−−→PR= 〈2,−3,−1〉. We know from part ii) of Theorem 2.5 that the area of the
parallelogram is the length of
−−→PQ× −−→PR=
∣
∣
∣
∣
∣
∣
e1 e2 e3
−2 1 −12 −3 −1
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
1 −1−3 −1
∣
∣
∣
∣
e1 −∣
∣
∣
∣
−2 −12 −1
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
−2 12 −3
∣
∣
∣
∣
e3 = −4e1 − 4e2 + 4e3.
Therefore, the area of the triangle is12
√16+ 16+ 16= 2
√3. �
Exercises
1. LetO(0,0), A(1,2), B(3,0), andM(−2,1) be points inR2. Draw the geometric vectorsv =−−→ABandw =
−−→MO
and use them to draw the following vectors:(a)v + 2w (b) w − v (c) w + 1
2v
2. Find‖b‖, a− 3b, ‖a+ 12b‖, and 2a+ b − 3e2, wherea andb are the following vectors:
(a)a = 〈1,−2〉 , b = 〈2,0〉(c) a = 〈2,0,−1〉 , b = 〈−1,1,1〉
(b) a = 3e1 + 2e2, b = −e1 + e2
(d) a = e2 +12e3, b = −2e1 + e2 + e3
3. Find a vectorb that has the same direction asa = 〈1,−1〉 and length‖b‖ = 3.
4. Finda · b, wherea andb are the following vectors:(a)a = 〈3,−2,1〉 , b = 〈1,0,2〉(c) a = 3e1 + 2e2 + 4e3, b = −e1 + 2e2 +
12e3
(b) a = 〈2,0,−1〉 , b = 〈2,1,4〉(d) ‖a‖ = 2, ‖b‖ = 3, ∢(a,b) = 60◦
5. Find the angle between the given vectors:(a)a = 〈2,0,−1〉 , b = 〈2,1,4〉(c) a = 3e1 + 2e2, b = −e1 + e3
(b) a = 〈2,0〉 , b =⟨
1,√
3⟩
6. Find the three angles of the triangle with verticesA(1,0,0), B(2,1,0), C(0,1,2).
7. Determine whether the given vectors are parallel, perpendicular, or neither:15
(a)a = 〈2,1,−1〉 , b =⟨
12 ,1,2
⟩
(c) a = 〈1,−2,2〉 , b = 〈−2,4,−4〉(b) a = 〈1,2,3〉 , b =
⟨
1,√
2,√
3⟩
8. (a) Find the projection ofa = 〈1,−1,4〉 ontou =⟨
13 ,
23 ,−
23
⟩
.(b) Find the projection ofa = 〈2,−1,2〉 onto the direction ofb = 〈2,−1,0〉.
9. Finda× b, wherea andb are the following vectors:(a)a = 〈2,−3,1〉 , b = 〈1,1,2〉 (b) a = 3e1 + 2e2 + 2e3, b = −e1 + 2e2 + e3
10. Leta = 〈−1,2,−4〉, b = 〈7,3,−4〉, andc = 〈−2,1,0〉. Evaluate the expressions or explain why they areundefined:
(a)a · (2b + c)(d) c · (−2a× b)
(b) ‖b × (2a)‖(e) ‖2c‖ (a× (a+ 3b))
(c) b × (c× a)(f) a× (c · b)
11. Find a unit vectoru orthogonal to bothx = 〈−2,3,−6〉 andy = 〈2,2,−1〉:(a) by using the cross product; (b) without using the cross product.
12. Find the area of the triangle with verticesA(3,2,1), B(2,4,5), andC(3,1,4).
13. Find the volume of the parallelepiped with adjacent edges OA, OB, andOC, whereA(2,3,0), B(3,−1,2),andC(1,0,−4).
14. Use the method from the second solution of Example 2.6 to find r × s, where
r = cosφ cosθe1 + cosφ sinθe2 − sinφe3, s= − sinφ sinθe1 + sinφ cosθe2.
15. Use the method from the second solution of Example 2.6 to findu × v, where
u = e1 + acosφe2 + asinφe3, v = −bsinφe2 + bcosφe3.
16. Letr = 〈x, y, z〉 anda = 〈a,b, c〉. Describe the set of pointsP(x, y, z) such that‖r − a‖ = R.
17. Prove theCauchy–Schwarz inequality: if a,b are three-dimensional vectors, then
|a · b| ≤ ‖a‖ ‖b‖.
18. Prove thetriangle inequality: if a,b are three-dimensional vectors, then
‖a+ b‖ ≤ ‖a‖ + ‖b‖.
19. Are the following statements true or false?(a) If a , 0 anda · b = a · c, then it follows thatb = c.(b) If a , 0 anda× b = a× c, then it follows thatb = c.(c) If a , 0, a · b = a · c, anda× b = a× c, then it follows thatb = c.
16
3. Lines and planes
3.1. Equations of a straight line in space.In our discussion of lines and planes, we shall use thenotion of a position vector. Aposition vectoris a vector whose initial point is at the origin. Thus,the components of a position vector are the same as the coordinates of its terminal point.
We shall specify a line in space using a point on the line and the direction of the line. Ifr0 is theposition vector of a pointP0(x0, y0, z0) on the line and ifv is a vector parallel to the line, then theterminal point of a position vectorr is on the line if and only if
r = r0 + tv for somet ∈ R. (3.1)
We call (3.1) aparametric vector equationof the line ℓ throughP0 and parallel tov. In thisequation, we construct a generic pointP on ℓ in two steps: first we get onℓ by going to the givenpoint P0; then we slide alongℓ by a multiple of the given vectorv until we reachP (see Figure3.1). The value of the parametert indicates the distance alongℓ (measured in units equal to‖v‖)that we need to move to reachP: for example,t = 0 corresponds toP = P0 andt = −2 correspondsto a point that is at a distance 2‖v‖ from P0 in the direction opposite tov.
r0
r
tv
ℓ
O
P0
P
Figure 3.1. The parametric vector equation of a line
Writing v andr in component form,v = 〈a,b, c〉 andr = 〈x, y, z〉, we can express (3.1) in termsof the components of the vectors involved:
〈x, y, z〉 = 〈x0, y0, z0〉 + t 〈a,b, c〉 = 〈x0 + ta, y0 + tb, z0 + tc〉 ,or equivalently:
x = x0 + ta, y = y0 + tb, z= z0 + tc (t ∈ R). (3.2)These are theparametric scalar equationsof ℓ. In general, the vectorv = 〈a,b, c〉 is determinedup to multiplication by a nonzero scalar, that is, we can use any nonzero scalar multiple ofv inplace ofv. This will change the coefficients in equations (3.1) and (3.2), but not the underlying setof points. We shall refer to the coordinatesa,b, c of any vectorv parallel toℓ asdirection numbersof ℓ.
Another common way to specify a lineℓ in space are its symmetric equations. Given the scalarequations (3.2), withabc, 0, we can eliminatet to obtain the equations
x− x0
a=
y− y0
b=
z− z0
c. (3.3)
These are thesymmetric equationsof ℓ. If one of a,b, or c is zero—say,b = 0, we obtain thesymmetric equations ofℓ by keeping that scalar equation and eliminatingt from the other two:
y = y0,x− x0
a=
z− z0
c.
17
If two amonga,b, c are equal to zero, the corresponding scalar equations represent the symmetricequations ofℓ after discarding the third parametric equation: ifa = b = 0, the symmetric equationsof ℓ are
x = x0, y = y0.
Finally, we can describe a line in space by the coordinates ofany two distinct pointsP1(x1, y1, z1)
and P2(x2, y2, z2) on the line. IfP1 and P2 are such points,−−−→P1P2 = 〈x2 − x1, y2 − y1, z2 − z1〉 is
parallel to the line, so we can write the symmetric equations(3.3) in the formx− x1
x2 − x1=
y− y1
y2 − y1=
z− z1
z2 − z1, (3.4)
(or something like it, for other types of symmetric equations).
Example 3.1.Find the equations of the lineℓ through the pointsP(1,0,3) andQ(2,−1,4).
Solution.To obtain the symmetric equations, we note that−−→PQ= 〈1,−1,1〉 is a vector parallel toℓ,
so (3.3) can be written asx− 1
1=
y− 0−1=
z− 31
⇐⇒ x− 1 = −y = z− 3.
The parametric scalar equations (based again on−−→PQ) are
x = 1+ t, y = 0+ t(−1) = −t, z= 3+ t (t ∈ R),
and the parametric vector equation is
r (t) = 〈1,0,3〉 + t 〈1,−1,1〉 = 〈1+ t,−t,3+ t〉 (t ∈ R). �
Example 3.2. Find a parametric vector equation of the lineℓ passing throughP(−1,2,−3) andparallel to the line 2(x+ 1) = 4(y− 3) = z.
Solution.We can write the equations of the second line as
x+ 112
=y− 3
14
= z,
sov =⟨
12,
14,1
⟩
is a vector parallel to the second line. Therefore,v is also parallel toℓ. That is,ℓpasses throughP and is parallel tov. Its vector equation then is
r (t) = 〈−1,2,−3〉 + t⟨
12,
14,1
⟩
=⟨
−1+ 12t,2+ 1
4t,−3+ t⟩
(t ∈ R). �
Example 3.3. Find the point of intersection and the angle between the lines with the vectorparametrizations
r1(t) = 〈1+ t,−1− t,−4+ 2t〉 and r2(u) = 〈1− u,1+ 3u,2u〉 .
Solution. If it exists, the point of intersection of the given lines will have coordinates (x, y, z) suchthat
x = 1+ t = 1− u, y = −1− t = 1+ 3u, z= −4+ 2t = 2u,
for somet,u. From the equations forx, we find thatt = −u. Hence, the other two equationsbetweent andu become
−1+ u = 1+ 3u, −4− 2u = 2u =⇒ u = −1, t = −u = 1.
Therefore, the point of intersection is (2,−2,−2).18
To find the angle between the lines, we find the angleθ between their parallel vectors:v =〈1,−1,2〉 for the first line andu = 〈−1,3,2〉 for the second. Thus, by the corollary to Theorem 2.3,
cosθ =u · v‖u‖ ‖v‖
=−1− 3+ 4√
6√
14= 0 =⇒ θ = 90◦. �
3.2. Equations of a plane. We shall specify a plane in space using a point on the plane andadirection perpendicular to the plane. Ifr0 is the position vector of a pointP0(x0, y0, z0) on the planeandn = 〈a,b, c〉 is a vector perpendicular (also callednormal) to the plane, then a generic point
P(x, y, z) is on the plane if and only if−−→P0P is perpendicular ton (see Fig. 3.2). Introducing the
position vectorr of P, we can express this condition as−−→P0P · n = 0 ⇐⇒ (r − r0) · n = 0, (3.5)
or equivalently,n · r = n · r0. (3.6)
These are thevector equationsof the given plane.
z
yx
n
O
P0
Pr − r0
Figure 3.2. The vector equation of a plane
If we express the dot product in (3.5) in terms of the components of the vectors, we get
a(x− x0) + b(y− y0) + c(z− z0) = 0, (3.7)
or thescalar equationof the given plane. Given the last equation, we can easily obtain the plane’slinear equation:
ax+ by+ cz+ d = 0, (3.8)simply by writingd = −ax0 − by0 − cz0.
Example 3.4. Find the equation of the plane that containsP(1,−2,3) and is perpendicular toe2 + 2e3.
Solution.The scalar equation of this plane is
0(x− 1)+ 1(y+ 2)+ 2(z− 3) = 0 ⇐⇒ y+ 2z− 4 = 0. �
Example 3.5.Find the equation of the planeπ that containsP(3,−1,5) and is parallel to the planewith equation 4x+ 2y− 7z+ 5 = 0.
Solution.A normal vector for the given plane isn = 〈4,2,−7〉. Hence, the scalar equation ofπ,which also hasn as its normal vector, is
4(x− 3)+ 2(y+ 1)− 7(z− 5) = 0 ⇐⇒ 4x+ 2y− 7z+ 25= 0. �
19
Example 3.6. Find the equation of the plane that contains the pointsP(1,2,3), Q(−1,3,2), andR(3,−1,2).
Solution.The cross product−−→PQ× −−→PR is a normal vector for this plane. By Example 2.8,
−−→PQ× −−→PR= −4e1 − 4e2 + 4e3,
so we can use the normal vectorn = 〈1,1,−1〉. We obtain the scalar equation
(x− 1)+ (y− 2)− (z− 3) = 0 ⇐⇒ x+ y− z= 0. �
Example 3.7.Find the angle between the planes 2x− y+ 3z= 5 and 5x+ 5y− z= 1.
Solution.The angle between the two planes equals the angle between their normal vectors:n1 =
〈2,−1,3〉 andn2 = 〈5,5,−1〉, respectively. Ifθ is the angle betweenn1 andn2, we use the dotproduct to find
cosθ =n1 · n2
‖n1‖ ‖n2‖=
10− 5− 3√14√
51=
2√714. �
Example 3.8.Find the line of intersection of the planesx+ y+ z+ 1 = 0 andx− y+ z+ 2 = 0.
Solution.We shall compute the symmetric equations of the given line. First, by adding the twoequations, we obtain the equation
2x+ 2z+ 3 = 0 ⇐⇒ x = −z− 3/2 ⇐⇒ x =z+ 3/2−1
.
On the other hand, subtracting the two given equations, we get
2y− 1 = 0 ⇐⇒ y = 1/2.
Hence, the symmetric equations of the given line are
x =z+ 3/2−1
, y = 1/2.
Note that these symmetric equations lead to the parametric scalar equations
x = t, y = 1/2, z= −3/2− t (t ∈ R). �
Example 3.9.Find the distance fromP(3,−5,2) to the planex− 2y+ z= 5.
Solution 1.The distance fromP to the given plane is equal to the distance|PQ|, whereQ is theintersection point of the plane and the line throughP that is perpendicular to the plane. The normalvector to the planen = 〈1,−2,1〉 is parallel to the linePQ, so the symmetric equations of that lineare
x− 31=
y+ 5−2=
z− 21.
Combining these equations with the equation of the plane, we obtain a linear system
x− 3 =y+ 5−2, z− 2 =
y+ 5−2, x− 2y+ z= 5,
whose solution (43,−53,
13) is the pointQ. Therefore,
|PQ| =(
(−53)2 + (10
3 )2 + (−53)2
)1/2= 5
3
√6. �
20
Solution 2.We now give a second solution, which is somewhat trickier, but has the advantage thatcan be easily generalized (see the next example). LetQ(u, v,w) be the same point as before. In this
solution, we avoid the explicit computation ofu, v,w. We have−−→PQ= 〈u− 3, v+ 5,w− 2〉.
By the definition ofQ, we know that−−→PQ is normal to the given plane, so
−−→PQ is parallel to the
normal vectorn = 〈1,−2,1〉. Hence, by the properties of the dot product,∣
∣n · −−→PQ∣
∣ = ‖n‖ |PQ|∣
∣ cos∢(n,−−→PQ)
∣
∣ =√
6 |PQ|,since the angle between the two vectors is 0◦ or 180◦. On the other hand, using the definition ofthe dot product, we have
∣
∣n · −−→PQ∣
∣ = |(u− 3)− 2(v+ 5)+ (w− 2)| = |u− 2v+ w− 15| = |5− 15| = 10.
Here, we have used thatu − 2v + w = 5, becauseQ lies on the given plane. Comparing the two
expressions for the dot product∣
∣n · −−→PQ∣
∣, we conclude that|PQ| = 10/√
6. �
Example 3.10.Find the distance fromP(x0, y0, z0) to the planeax+ by+ cz+ d = 0.
Solution*. We essentially repeat the second solution of the previous example. LetQ(u, v,w) bethe intersection point of the plane and the line throughP that is perpendicular to the plane. Then−−→PQ = 〈u− x0, v− y0,w− z0〉. Since
−−→PQ is normal to the given plane, it is parallel to its normal
vectorn = 〈a,b, c〉. Hence, by the properties of the dot product,∣
∣n · −−→PQ∣
∣ = ‖n‖ |PQ|∣
∣ cos∢(n,−−→PQ)
∣
∣ =√
a2 + b2 + c2 |PQ|,since the angle between the two vectors is 0◦ or 180◦. On the other hand, using the definition ofthe dot product, we have
∣
∣n · −−→PQ∣
∣ = |a(u− x0) + b(v− y0) + c(w− z0)|= |au+ bv+ cw− ax0 − by0 − cz0|= | − d − ax0 − by0 − cz0| = |ax0 + by0 + cz0 + d|.
Here, we have used thatau+ bv+ cw= −d, becauseQ lies on the given plane. Comparing the two
expressions for the dot product∣
∣n · −−→PQ∣
∣, we conclude that the desired distance is
|PQ| = |ax0 + by0 + cz0 + d|√a2 + b2 + c2
. �
Exercises
1. Find the equations of the line that passes through the point (1,−2,3) and is perpendicular to the plane withequation−2x+ y− z+ 4 = 0.
2. Find the equations of the line that passes through the points (0,−2,3) and (1,1,5).
3. Let ℓ1 be the line with equationsx = −1+ 2t, y = 3t, z = 2 and letℓ2 be the line through the points (4,3,2)and (6,−1,4). Determine whether these two lines are parallel, intersecting, or skew. If they intersect, findtheir intersection point.
4. Find the intersection points of the line with equationsx− 3
2=
y+ 15= z and the coordinate planes.
5. Find an equation of the plane that passes through the point(1,−2,3) and is perpendicular to the vector〈3,0,−2〉.
21
6. Find an equation of the plane that contains the line with equationsx− 1
2=
y+ 23=
z− 24
and is parallel to
the plane 3x− 2y = 5.
7. Find an equation of the plane through the points (1,−1,0), (2,0,3), and (−2,2,1).
8. Find the equations of the intersection line of the planesx+ y− z= 2 and 2x− 3y+ z= 6.
9. Find an equation of the plane that contains both the lineℓ1 with equationsx = 1+ 2t, y = −1+ 3t, z = −2tand the lineℓ2 with equationsx = −2− 3t, y = t, z= 4+ 4t.
10. Find the angle between the given planes:(a) 2x− y+ 3z= 0, x+ 2y+ z= 4; (b) x+ y+ z= 4, x+ 2y− 3z= 6.
11. The planes 2x− 2y− z= 1 and 6x− 6y− 3z= −5 are parallel. Find the distance between them.
22
4. Vector functions
4.1. Definition. A vector-valued function(or avector function) is a function whose domain is aset of real numbers and whose range is a set of vectors. In general, a vector function has the form
f (t) = 〈 f1(t), f2(t), f3(t)〉 = f1(t)e1 + f2(t)e2 + f3(t)e3.
where f1, f2, and f3 are functions in the regular sense (functions for which boththe inputs and out-puts are real numbers). The functionsf1, f2, and f3 are sometimes called thecomponent functionsof the vector functionf . When we want to distinguish that the values of a functionf are numbers(as opposed to vectors), we will refer tof as ascalar function.
Example 4.1. As we know from Lecture #3, the parametric equation of the line in space passingthrough (1,0,−5) and parallel to the vector〈1,2,1〉 is, in fact, a vector function:
r (t) = 〈1+ t,2t,−5+ t〉 (t ∈ R).
The line is the set of values of this function,not the graph (the graph is four-dimensional).Here is another example of a vector function:
f (t) =⟨
cost, (t − 5)−1, ln t⟩
;
its domain is (0,5)∪ (5,∞). �
4.2. Calculus of vector functions: limits. Vector functions can have limits, derivatives, an-tiderivatives, and definite integrals—just like scalar functions. As we will see, all of these canbe computed componentwise. Because of that, once one is proficient in the calculus of scalar-valued functions, one can easily master the calculus of vector functions. We begin with limits andcontinuity. The limit of a vector function is a vector.
Definition. If f (t) = 〈 f1(t), f2(t), f3(t)〉 is defined neara, then thelimit of f (t) as t→ a is defined by
limt→a
f (t) =⟨
limt→a
f1(t), limt→a
f2(t), limt→a
f3(t)⟩
,
provided that the limits of all three component functions exist.
Definition. A vector functionf (t) = 〈 f1(t), f2(t), f3(t)〉, defined neara, is continuous at aif
limt→a
f (t) = f (a).
Example 4.2.We have
limt→7
⟨
cost, (t − 5)−1, ln t⟩
=
⟨
limt→7
cost, limt→7
(t − 5)−1, limt→7
ln t⟩
=⟨
cos 7, 12, ln 7
⟩
.
�
4.3. Calculus of vector functions: derivatives.
Definition. If f (t) = 〈 f1(t), f2(t), f3(t)〉 is continuous ata, then itsderivativef ′(a) at a is defined by
f ′(a) = limt→a
f (t) − f (a)t − a
= limh→0
f (a+ h) − f (a)h
,
provided that the limit on the right exists. Iff ′(a) exists, we say thatf is differentiable at a.23
Theorem 4.1. A vector functionf (t) = 〈 f1(t), f2(t), f3(t)〉 is differentiable at a if and only if allthree component functions f1, f2, f3 are. Whenf ′(a) exists, we have
f ′(a) =⟨
f ′1(a), f ′2(a), f ′3(a)⟩
.
In other words, the derivative of a vector function can be computed using componentwise differen-tiation.
Proof. We have
f ′(a) = limt→a
f (t) − f (a)t − a
= limt→a
〈 f1(t) − f1(a), f2(t) − f2(a), f3(t) − f3(a)〉t − a
= limt→a
⟨
f1(t) − f1(a)t − a
,f2(t) − f2(a)
t − a,
f3(t) − f3(a)t − a
⟩
=⟨
f ′1(a), f ′2(a), f ′3(a)⟩
. �
Example 4.3. If f (t) =⟨
cost, (t − 5)−1, ln t⟩
, then
f ′(t) =⟨
− sint,−(t − 5)−2, t−1⟩
.
�
Theorem 4.2. Let f andg be differentiable vector functions, let u be a differentiable scalar func-tion, and let c be a scalar. Then the following formulas hold:
1.[
f (t) + g(t)]′= f ′(t) + g′(t);
2.[
cf (t)]′= cf ′(t);
3.[
u(t)f (t)]′= u(t)f ′(t) + u′(t)f (t);
4.[
f (t) · g(t)]′= f (t) · g′(t) + f ′(t) · g(t);
5.[
f (t) × g(t)]′= f (t) × g′(t) + f ′(t) × g(t);
6.[
f (u(t))]′= f ′(u(t))u′(t).
Proof. We shall prove 4). Iff (t) = 〈 f1(t), f2(t), f3(t)〉 andg(t) = 〈g1(t),g2(t),g3(t)〉, then
f (t) · g(t) = f1(t)g1(t) + f2(t)g2(t) + f3(t)g3(t).
Hence, the left side of 4) equals[
f (t) · g(t)]′=(
f1(t)g1(t) + f2(t)g2(t) + f3(t)g3(t))′
= f ′1(t)g1(t) + f1(t)g′1(t) + f ′2(t)g2(t) + f2(t)g
′2(t) + f ′3(t)g3(t) + f3(t)g
′3(t).
On the other hand, we have
f ′(t) · g(t) =⟨
f ′1(t), f ′2(t), f ′3(t)⟩
· 〈g1(t),g2(t),g3(t)〉 = f ′1(t)g1(t) + f ′2(t)g2(t) + f ′3(t)g3(t),
f (t) · g′(t) = 〈 f1(t), f2(t), f3(t)〉 ·⟨
g′1(t),g′2(t),g
′3(t)
⟩
= f1(t)g′1(t) + f2(t)g
′2(t) + f3(t)g
′3(t),
so the right side of 4) equals
f ′1(t)g1(t) + f ′2(t)g2(t) + f ′3(t)g3(t) + f1(t)g′1(t) + f2(t)g
′2(t) + f3(t)g
′3(t).
This establishes 4). �
24
4.4. Calculus of vector functions: integrals. The definition of the definite integral of a continu-ous function via Riemann sums (see Appendix B) can also be extended to vector functions. Here,we will adopt an equivalent definition, which is a little lesstransparent but considerably easier touse.
Definition. If f (t) = 〈 f1(t), f2(t), f3(t)〉 is continuous in [a,b], then itsdefinite integral∫ b
a f (t) dt isdefined by
∫ b
af (t) dt =
⟨∫ b
af1(t) dt,
∫ b
af2(t) dt,
∫ b
af3(t) dt
⟩
.
Remark. Note that∫ b
a f (t) dt is avector (not a scalar and not a vector function).
Example 4.4. If f (t) =⟨
cost, (t − 5)−1, ln t⟩
, then∫ 2
1f (t) dt =
⟨∫ 2
1cost dt,
∫ 2
1(t − 5)−1 dt,
∫ 2
1ln t dt
⟩
=
⟨
[
sint]2
1,[
ln |t − 5|]2
1,[
t ln t − t]2
1
⟩
=⟨
sin 2− sin 1, ln(34),2 ln 2− 1
⟩
.
�
Theorem 4.3. Let f andg be continuous vector functions, letu be a vector, and let c be a scalar.Then the following formulas hold:
1.∫ b
a
[
f (t) + g(t)]
dt =∫ b
a f (t) dt+∫ b
a g(t) dt;
2.∫ b
a
[
cf (t)]
dt = c∫ b
a f (t) dt;
3.∫ b
a
[
u · f (t)]
dt = u ·∫ b
a f (t) dt;
4.∫ b
a
[
u × f (t)]
dt = u ×∫ b
a f (t) dt;
5.∫ b
a f ′(t) dt = f (b) − f (a).
Proof. We shall prove 4). Both sides of the identity are three-dimensional vectors, so it suffices toshow that their respective components are equal. Ifu = 〈u1,u2,u3〉 andf (t) = 〈 f1(t), f2(t), f3(t)〉,then
u × f (t) = 〈u2 f3(t) − u3 f2(t),u3 f1(t) − u1 f3(t),u1 f2(t) − u2 f1(t)〉 .Thus, the first component of the left side of 4) is
∫ b
a
(
u2 f3(t) − u3 f2(t))
dt = u2
∫ b
af3(t) dt− u3
∫ b
af2(t) dt.
Since the right side of this identity is also the first component of the right side of 4), this shows thatthe first components of the two sides of 4) match. The proof that the remaining two componentsof the two sides of 4) match is similar. �
Exercises
1. Evaluate limt→a
⟨
ln(t + 1)t,sin 2t
t,et
⟩
, where: (a)a = 1; (b)a = 0.
25
2. Find the derivative of the given vector function:(a) f (t) =
⟨
ln(
t2 + 1)
,arcsin 2t,et⟩
(c) f (t) = t3e1 − 3e2te3
(b) f (t) =⟨
5t3 + 2t,et2 sint,et2 cost⟩
(d) f (t) =√
t2 + 1e1 − cos 2te2 + sin 2te3
3. Evaluate the given integral:(a)
∫ 10
((
t2 + 4)−1
e1 + t(
t2 − 2)−1
e3)
dt (b)∫ 1−1
⟨
5t3 + 2t,e2t, sint cos4 t⟩
dt
4. Findf (t), if f ′(t) = 2te1 + ete2 + (ln t) e3 andf (1) = e1 + e2.
26
5. Space curves
You are familiar with parametric plane curves from single-variable calculus. Vector functionsare a great tool for dealing with curves, both in the plane andin space. In this lecture, we usevector functions to give a unified treatment of plane curves and space curves.
5.1. Review of plane curves.Recall that a curveγ in the plane is the set
γ ={
(x, y) : x = x(t), y = y(t), t ∈ I}
,
whereI is some interval inR (finite or infinite; open, closed, or semiopen) andx(t) andy(t) arecontinuous functions oft. We refer to the pair of equations
x = x(t), y = y(t) (t ∈ I ) (5.1)
asparametric equationsor aparametrizationof γ. A curveγ has many parametrizations, and some-times replacing one parametrization by another can change aparticular property of the curve, evenif the set of points stays the same. Thus, when we want to distinguish a particular parametrization,we talk about theparametric curveγ with equations (5.1).
5.2. Definition of space curve.Next, we extend the above definitions to three dimensions.
Definition. If x(t), y(t), z(t) are continuous functions defined on an intervalI , then the set
γ ={
(x, y, z) : x = x(t), y = y(t), z= z(t), t ∈ I}
is called aspace curve. The equations
x = x(t), y = y(t), z= z(t) (t ∈ I ) (5.2)
are calledparametric equationsor aparametrizationof γ; t is theparameter.
Alternatively, we can think of the curveγ with equations (5.2) as the range of the vector function
r (t) = x(t)e1 + y(t)e2 + z(t)e3 (t ∈ I ). (5.3)
Indeed, if we draw the vectorsr (t) so that they all start at the origin, then the resulting geometricvectors will be exactly the position vectors of the points onthe curve (see Figure 5.1). We shallrefer to (5.3) as theparametric (vector) equationof γ. Note that the plane curve (5.1) can be treated
z
yx
O
r (t)
γ
Figure 5.1. Vector functions and curves in space
as a special case of a space curve:
r (t) = x(t)e1 + y(t)e2 + 0e3 (t ∈ I ).27
Example 5.1.Consider the space curveγ with parametric equation
r (t) = (cost)e1 + (sint)e2 + te3 (0 ≤ t < ∞).
The point (cost, sint,0) traces the unit circlex2 + y2 = 1 in thexy-plane in the counterclockwisedirection, so the point (cost, sint, t) onγmoves in the positive direction of thez-axis along a spirallying above that circle. This curve is known as thehelix; it is shown on Figure 5.2. �
z
yx
Figure 5.2. The helix
z
yx
Figure 5.3. r (t) = 〈t cost, t sint, t〉
Example 5.2.Consider the space curveγ with parametric equation
r (t) = (t cost)e1 + (t sint)e2 + te3 (0 ≤ t < ∞).
The initial point of the curve is the origin. For any givent, we have
(t cost)2 + (t sint)2 = t2,
so thex, y, andz-coordinates of a point onγ always satisfy the equationx2 + y2 = z2. Thus,γ is aninfinite spiral that starts at the origin and unwinds upward and outward; it is shown on Fig. 5.3.�
Example 5.3.Given two pointsP1(x1, y1, z1) andP2(x2, y2, z2), the space curve
r (t) = (1− t) 〈x1, y1, z1〉 + t 〈x2, y2, z2〉 (0 ≤ t ≤ 1)
is the line segment with endpointsP1 andP2. Indeed, we can expressr (t) in the form
r (t) = 〈x1 + t(x2 − x1), y1 + t(y2 − y1), z1 + t(z2 − z1)〉 ,
which is the parametric equation of the line throughP1 that is parallel to−−−→P1P2. Thus, if the domain
for t was the whole real lineR, the curve would be the straight line throughP1 andP2. Restrictingthe domain fort to the interval [0,1], we obtain the piece of the line betweenr (0) = 〈x1, y1, z1〉 = P1
andr (1) = 〈x2, y2, z2〉 = P2, that is, the line segmentP1P2. �
5.3. Smooth curves.At first glance, continuous curves appear to be a good model for the geo-metric objects that we think of as “curves”: that is, two- or three-dimensional objects that areessentially one-dimensional. However, it turns out that there are some rather peculiar continuousparametric curves. For example, in 1890 the Italian mathematician Giuseppe Peano stunned themathematical community with his construction of a continuous plane curve that fills the entireunit square (see§5.7); there are also examples of continuous space curves that fill an entire cube.In order to rule out such pathological curves, we shall restrict our attention to a special class ofparametric curves called “piecewise smooth”.
28
Definition. A curveγ given by a vector functionr (t), t ∈ I , is calledsmoothif r ′(t) is continuousandr ′(t) , 0 (except possibly at the endpoints ofI ). A curveγ consisting of a finite number ofsmooth pieces is calledpiecewise smooth.
Example 5.4.For the helixr (t) = (cost)e1 + (sint)e2 + te3, t ≥ 0, we have
r ′(t) = (− sint)e1 + (cost)e2 + e3 , 0.
Hence, the helix is a smooth curve. �
Example 5.5.Determine whether the curveγ defined by the vector function
r (t) = t2e1 + t3e2 + (cost)e3 (t ∈ R)
is smooth, piecewise smooth, or neither.
Solution.The derivativer ′(t) = 2te1 + 3t2e2 − sinte3
vanishes whent = 0. Thus,γ is not smooth. On the other hand, the curves
γ1 : r (t) = t2e1 + t3e2 + (cost)e3 (t ≥ 0)
andγ2 : r (t) = t2e1 + t3e2 + (cost)e3 (t ≤ 0)
are smooth, since they fail the conditionr ′(t) , 0 only at an endpoint of the intervals on whichthey are defined. Sinceγ is the union ofγ1 andγ2, it follows thatγ is piecewise smooth. �
5.4. Tangent lines. The requirement thatr ′(t) , 0 for a smooth curve is not arbitrary. To explainwhere it comes from, we now consider the following problem.
Problem. Letγ be a smooth parametric curve with parametric equation
r (t) = x(t)e1 + y(t)e2 + z(t)e3 (t ∈ I ).
Find an equation of the tangent line toγ at the pointr (a) = (x0, y0, z0).
Idea of solution.The tangent line will have the parametric equations
x = x0 + at, y = y0 + bt, z= z0 + ct (t ∈ R),
whereu = 〈a,b, c〉 is any vector parallel to the tangent line. In other words, weneed to determinethe direction of the tangent line.
Recall that the tangent line throughP(x0, y0, z0) is the limit of the secant lines throughP and apoint Q(x, y, z) on the curve asQ→ P. If Q = r (t), then a vector parallel to the secant line throughP andQ is
−−→PQ= 〈x− x0, y− y0, z− z0〉 = r (t) − r (a).
Thus,
direction of tangent line atP = limt→a
[
direction of secant line throughP,Q]
= limt→a
[
direction ofr (t) − r (a)]
= limt→a
[
direction of normalizedr (t) − r (a)]
= limt→a
r (t) − r (a)‖r (t) − r (a)‖
= limt→a
(
r (t) − r (a))
/(t − a)
‖r (t) − r (a)‖/(t − a).
29
By the definition of derivative, the limit of the numerator isr ′(a). By a slightly more complicatedargument (which we omit), the limit of the denominator is‖r ′(a)‖. Thus, a vector parallel to thetangent line isr ′(a)/‖r ′(a)‖. �
z
yx
O
r ′(t)
P
Figure 5.4. The tangent vectorr ′(t) at P = r (t)
We now see the reason for the requirement of non-vanishing ofthe derivative in the definition ofsmoothness: with that requirement, a smooth curve has a tangent line at every point. Furthermore,the vectorr ′(t) is parallel to the tangent line to the curve at the endpointP of r (t). For this reason,r ′(t) is called also thetangent vectorto the curver (t) (see Figure 5.4). Theunit tangent vectortor (t) is given by the formula
T(t) =r ′(t)‖r ′(t)‖
.
It points in the same direction as the tangent vectorr ′(t) but has length 1.
Example 5.6.For the helix, we have
r ′(t) = (− sint)e1 + (cost)e2 + e3,
so the unit tangent vector is
T(t) =r ′(t)‖r ′(t)‖
=〈− sint, cost,1〉
√
(− sint)2 + (cost)2 + 12=
⟨
− sint√2,cost√
2,
1√2
⟩
.
�
Example 5.7.Find the equation of the tangent line to the helix at the pointP corresponding to thevalue of the parametert = π.
Solution.We haveP(−1,0, π). The tangent line passes throughP and is parallel to the tangentvectorr ′(π) = 〈0,−1,1〉 (see Example 5.6). Hence, its parametric equation is
r (t) = 〈−1,0, π〉 + t 〈0,−1,1〉 = 〈−1+ t,−t, π + t〉 (t ∈ R);
the symmetric equations of the tangent line arex = −1,y−1=
z− π1
. �
Example 5.8.Find the angle between the helix and the curveγ defined by the vector function
r2(t) = −te1 + (t − 1)2e2 + πte3 (t ≥ 0)
at P(−1,0, π).30
Solution.Let σ denote the helix. IfP lies on bothγ andσ, the angle betweenγ andσ at Pis the angle between their tangent vectors. It is easy to see that both curves do pass throughP:r1(π) = r2(1) = 〈−1,0, π〉. We know from Example 5.7 that the tangent vector to the helixat P isv1 = 〈0,−1,1〉. The tangent vector toγ at P is
v2 = r ′2(1) = 〈−1,2(t − 1), π〉t=1 = 〈−1,0, π〉 .
Hence, ifα is the angle betweenγ andσ at P, we have
cosα =v1 · v2
‖v1‖ ‖v2‖=
0 · (−1)+ (−1) · 0+ 1 · π√
02 + (−1)2 + 12√
(−1)2 + 02 + π2=
π√2+ 2π2
,
that is,
α = arccos
(
π√2+ 2π2
)
≈ 0.8315. �
5.5. Normal vectors*. We now want to find vectors that are perpendicular to the curver (t), thatis, we want to find vectors orthogonal to the tangent vectorr ′(t), or equivalently, to the unit tangentvectorT(t). One such vector is theunit normal vector
N(t) =T′(t)‖T′(t)‖
.
In order to prove thatN(t) andT(t) are orthogonal, we differentiate both sides of the identity
T(t) · T(t) = ‖T(t)‖2 = 1.
We obtain
T′(t) · T(t) + T(t) · T′(t) = 0 =⇒ 2T(t) · T′(t) = 0 =⇒ T(t) · T′(t) = 0.
That is,T′(t) is orthogonal toT(t).Another vector orthogonal toT(t) is theunit binormal vector
B(t) = T(t) × N(t).
By the properties of the cross product,B(t) is perpendicular to bothT(t) andN(t) and has lengthequal to the area of the parallelogram with sidesT(t) andN(t). Since this parallelogram is a squareof side length 1,B(t) is a unit vector.
Remark. Note that it is important to useT(t) and notr ′(t) when we compute the normal vector.In fact, r ′′(t) need not be orthogonal tor ′(t) and in most cases won’t be. For example, ifr (t) =⟨
1, t, t2⟩
, we haver ′(t) = 〈0,1,2t〉 andr ′′(t) = 〈0,0,2〉, so
r ′(t) · r ′′(t) = 4t , 0 unlesst = 0.
On the other hand,T(t) andT′(t) are orthogonal:
T(t) =⟨
0, (1+ 4t2)−1/2,2t(1+ 4t2)−1/2⟩
, T′(t) =⟨
0,−4t(1+ 4t2)−3/2,2(1+ 4t2)−3/2⟩
.
31
5.6. Arc length. You are probably familiar with the formula for arc length of aparametric planecurve: if γ is a parametric curve in thexy-plane with parametric equations (5.1), whereI = [a,b],then its lengthL(γ) is given by the formula
L(γ) =∫ b
a
√
x′(t)2 + y′(t)2 dt. (5.4)
In this section, we generalize this formula to space curves.We argue from scratch, so even if (5.4)is new to you, this should not be an obstacle.
First, we must say what we mean by “length” of a curve. We shallfocus on piecewise smoothcurves given by a parametric equation of the form (5.3). The intuitive idea of length is that wecan take the curve, “straighten” it into a line segment, and then measure the length of that linesegment. Now, imagine that we live in a “rigid” world, where we cannot straighten our curve (orbend a thread so it would lay on top of the curve). We could try to cut a number of short straightline segments and piece them together so that the resulting polygon lies close to our original curve;we can then measure the total length of the polygon and argue that the length of the curve is aboutthe same. This is the basic idea behind our definition of arc length.
z
yx
Figure 5.5. Polygonal approximation to a curve by a 13-gon
Let γ be a parametric curve given by (5.3) withI = [a,b]. For eachn ≥ 1, set∆n = (b − a)/nand define the numbers that partition [a,b] into n subintervals of equal lengths:
t0 = a, t1 = a+ ∆n, t2 = a+ 2∆n, . . . , tn = a+ n∆n = b.
We writexi = x(ti), yi = y(ti), zi = z(ti) for the coordinates of the pointPi(xi , yi , zi) that correspondsto the valuet = ti of the parameter. The pointsP0,P1,P2, . . . ,Pn partition γ into n arcs (seeFigure 5.5). LetLn be the length of the polygon with vertices at those points, that is,
Ln =
n∑
i=1
|Pi−1Pi | =n
∑
i=1
√
(xi − xi−1)2 + (yi − yi−1)2 + (zi − zi−1)2.
If the sequence{Ln}∞n=1 converges and limn→∞
Ln = L, we callL thearc lengthof the given curve.
Using the above definition and the properties of smooth functions and of definite integrals, wecan prove the following formula for the length of a piecewisesmooth curve.
Theorem 5.1.Letγ be a piecewise smooth parametric curve with equation
r (t) = x(t)e1 + y(t)e2 + z(t)e3 (t ∈ [a,b]).32
Then the arc length ofγ is
L(γ) =∫ b
a
√
x′(t)2 + y′(t)2 + z′(t)2 dt =∫ b
a‖r ′(t)‖dt. (5.5)
Remarks. 1. Note that (5.4) is a special case of (5.5): namely, the casewhenz(t) = 0.
2. Note that (5.5) gives the length of the parametric curve and not the length of the graph. Forexample, the parametric equation
r (t) = coste1 + sinte2 + 0e3 (0 ≤ t ≤ 4π)
represents the unit circle in thexy-plane. If we apply (5.5) to this curve, we will find that its lengthis 4π (and not 2π). The explanation for this is that the above curve traces theunit circle twice.
Example 5.9. If f ′ is continuous on [a,b], the length of the curvey = f (x) is
L =∫ b
a
√
1+ f ′(x)2 dx.
This formula follows from the theorem by representing the given graph as a parametric curve withparametrization
r (t) = te1 + f (t)e2 + 0e3 (t ∈ [a,b]).�
Example 5.10.Find the length of the curveγ given byr (t) =⟨
2t3/2, cos 2t, sin 2t⟩
, where 0≤ t ≤ 1.
Solution.We haver ′(t) = 3t1/2e1 − 2 sin 2te2 + 2 cos 2te3,
so (5.5) yields
L(γ) =∫ 1
0‖r ′(t)‖dt =
∫ 1
0
√
(
3t1/2)2+ (−2 sin 2t)2 + (2 cos 2t)2 dt
=
∫ 1
0
√
9t + 4 sin2 2t + 4 cos2 2t dt =∫ 1
0
√9t + 4dt
=19
∫ 13
4u1/2 du=
2(
13√
13− 8)
27≈ 2.879. �
5.7. Space-filling curves*. In this section, we describe the general idea of the construction ofPeano’s space-filling curve. We start with a piecewise linear curve contained in the unit square.This curve is then shrunk to a fraction of its original size and several copies of it are placed through-out the unit square in such a way that they do not intersect andcan be easily connected one toanother with straight lines. This yields a more complicatedpolygon that also lies within the unitsquare. The whole process (shrinking, copying, connecting) is then repeated iteratively. The firsttwo iterations of Peano’s original constructions are displayed on Figure 5.6; Figure 5.7 shows thefirst five iterations of an alternative construction by Hilbert. The result of the iterative process isa sequence of curves. It can be proved that the sequence of continuous functions underlying thissequence of curves has a limit, which is also a continuous function. It should be intuitively clear(and this can be proved rigorously) that the limit curve fillsthe entire square.
33
Figure 5.6. The first two iterations in Peano’s construction
Figure 5.7. The first five iterations in Hilbert’s construction
Exercises
1. Does the pointP(0, π,1) lie on the curve parametrized by the given vector functionr (t)? If so, for whatvalue(s) oft?
(a)r (t) = 〈sint,2 arccost, t + 1〉 (b) r (t) = 〈sint, t, cos 2t〉 (c) r (t) = te2 + tante3
2. Do the curvesr1(t) =⟨
t, t2, t3⟩
andr2(t) =⟨
t − 2, t2 − 3t + 2, t2 + t − 6⟩
intersect?
3. Let two objects move in space along trajectories described by the vector functionsr1(t) =⟨
t, t2, t3⟩
andr2(t) =
⟨
t − 2, t2− 3t + 2, t2+ t − 6⟩
. Will the objects ever collide? That is, will they ever be at the same pointin space at the same time?
4. Find the unit tangent vectorT(t) to the curver (t) = 〈sin 3t, cos 3t, ln(t + 1)〉 at the point witht = 0.
5. Find the equations of the tangent line to the curver (t) =⟨
t3 − 2, t2 + 1,3t + 1⟩
at the point (−2,1,1).
6. Find the equations of the tangent line to the curver (t) = 〈5 cost,5 sint,2 cos 2t〉 at the point (−5,0,2).34
7. Find the points on the curver (t) = 〈5 cost,5 sint,2 cos 2t〉 where the tangent line is parallel to thex-axis.
8. Is the given curve smooth? If not, then is it piecewise smooth?(a)r (t) =
⟨
t3 + t, sin 2t,3t2 − 1⟩
(b) r (t) =⟨
cost, t2, cos 2t⟩
(c) r (t) = 〈cost, cos 2t, cos 3t〉
9. Find the intersection point of the curvesr1(t) =⟨
2t, t2 + t, t3⟩
andr2(t) =⟨
t, t2 − 3t + 4, t2 + t − 5⟩
. Find theangle between the curves at that point, that is, find the anglebetween the tangent vectors at the point.
10. Find the arc length of the curver (t):(a) r (t) =
⟨
1, t2, t3⟩
, 0 ≤ t ≤ 1 (b) r (t) =⟨
t, ln t,2√
2t⟩
, 1 ≤ t ≤ 2
11. Let r (t), a ≤ t ≤ b, be a parametrization of a smooth curveγ. Let L =∫ b
a ‖r′(t)‖dt be the length ofγ and
define the new function
s(t) =∫ t
a‖r ′(u)‖du (a ≤ t ≤ b).
(a) Evaluates(a) ands(b).
(b) Use the Fundamental Theorem of Calculus to show thats′(t) > 0 for all t with a < t < b.
(c) Let t = t(s), 0 ≤ s ≤ L, be the inverse function of the functions(t), and define the vector functionf (s) = r (t(s)), 0 ≤ s≤ L. Show that‖f ′(s)‖ = 1 for all s with 0 < s< L.
(d) Convince yourself that the vector functionf (s) from part (c) is another parametrization of the curveγ.This parametrization has a special property:f (s) is the position vector of the point onγ that iss units alongthe curve after the initial point. This parametrization is known as theparametrization with respect to arclengthor thenatural parametrizationand plays an important role in the study of space curves.
(e) Parametrize with respect to arc length the curver (t) =⟨
e3t sin 4t,e3t cos 4t,5⟩
, 0 ≤ t ≤ 13 ln 4.
35
6. Multivariable functions
6.1. Functions of two variables. In this lecture we introduce multivariable functions. These arefunctions that depend on multiple inputs (usually 2, 3, or 4), but produce only a single output.We start our discussion with the simplest example: functions of two variables. For example, thetemperatureT at any point on the surface of the Earth at this very moment depends on two piecesof information: the longitudex and the latitudey of the point. Thus, we can think ofT as a functionT(x, y) of two variablesx andy or as a function of the ordered pair (x, y). We now give a formaldefinition.
Definition. A function f of two variablesis a rule that assigns to each ordered pair of real numbers(x, y) in a setD a unique real number denotedf (x, y). The setD is thedomainof f and the set ofall values thatf takes on is itsrange, that is, the range off is { f (x, y) : (x, y) ∈ D}.
6.2. Functions of three and more variables.The example that we used above to motivate theintroduction of functions of two variables is somewhat artificial. A more natural (and also moreuseful) quantity is the temperatureT at a point (x, y, z) in space at this very moment. ThenT isa functionT(x, y, z) of three variables—the coordinates (x, y, z) of the point. Note that with thisfunction in hand we don’t have to assume that the temperaturein this room and the temperature onthe wing of an airplane flying 39,000 feet above us are the same. An even more realistic exampleis that of the temperatureT at the point (x, y, z) in space at timet. This quantity is a functionT(x, y, z, t) of four variables. Note that when we fix the value oft to the timet0 at the “verymoment” a couple of minutes ago at which we defined the function T(x, y, z), T(x, y, z, t0) turnsinto T(x, y, z). We now give a formal definition.
Definition. A function f of n variables x1, x2, . . . , xn is a rule that assigns to each orderedn-tupleof real numbers (x1, x2, . . . , xn) in a setD a unique real number denotedf (x1, x2, . . . , xn). The setD is thedomainof f and the set of all values thatf takes on is itsrange. In particular, afunction fof three variablesis a rule that assigns to each ordered triple of real numbers (x, y, z) in a setD (inR
3) a unique real number denotedf (x, y, z).
6.3. Examples of domains and ranges of multivariable functions.
Example 6.1.The functionf (x, y) =√
1− xy is defined whenever
1− xy≥ 0 ⇐⇒ xy≤ 1.
If x > 0, the last inequality is satisfied fory ≤ x−1; if x < 0, the inequality is satisfied fory ≥ x−1;and if x = 0, it is satisfied for ally. That is, the domain off contains the points in I quadrant onand below the graph ofy = x−1, the points in III quadrant on and above the graph ofy = x−1, andall the points in II and IV quadrants, including thex- andy-axes (see Figure 6.1). The range offis [0,∞). Clearly, f cannot attain negative values. On the other hand, ifk ≥ 0,
√
1− xy= k ⇐⇒ 1− xy= k2 ⇐⇒ xy= 1− k2.
Hence, ifk ≥ 0, k , 1, f (x, y) = k for all points (x, y) on the hyperbolaxy= 1−k2; also, f (x, y) = 1for all points (x, y) on the coordinate axes. That is, every numberk ∈ [0,∞) does belong to therange off . �
36
x
y
Figure 6.1. The domain ofz=√
1− xy
yx
z
Figure 6.2. z= 3− x2 − y2
Example 6.2. The domain of the functionf (x, y, z) = ln(16− x2 − y2 − 9z2) consists of the points(x, y, z) in space whose coordinates satisfy
16− x2 − y2 − 9z2 > 0 ⇐⇒ x2
16+
y2
16+
z2
16/9< 1.
Since lnx is an increasing function, the range off consists of the logarithms of the positive num-bers in the range of the functionw = 16− x2 − y2 − 9z2. Becausex2 + y2 + 9z2 can attain anynon-negative value and cannot attain any negative value, itfollows thatw takes on the numbers inthe interval (−∞,16]. Thus, the range off consists of the logarithms of the numbers in (0,16], thatis, the range off is (−∞, ln 16]. �
6.4. Graphs.
Definition. If f is a function ofn variablesx1, x2, . . . , xn with domainD, then thegraphof f is theset of all points (x1, x2, . . . , xn, z) in Rn+1 with (x1, x2, . . . , xn) ∈ D andz= f (x1, x2, . . . , xn), that is,the graph off is the set
{
(x1, x2, . . . , xn, f (x1, x2, . . . , xn)) : (x1, x2, . . . , xn) ∈ D}
.
In particular, the graph of a functionf (x, y) defined on a setD in R2 is the surface given by{
(x, y, f (x, y)) : (x, y) ∈ D}
.
Example 6.3.The graph of the functionf (x, y) = 3− x2 − y2 is the surface
z= 3− x2 − y2.
This surface is displayed on Figure 6.2. �
6.5. Level curves and surfaces.While the graph of a function of two variables provides an excel-lent visualization of the function, it might be quite difficult to sketch without advanced technology.What can we do using only manual (and mental) labor? One answeris well-known to cartogra-phers. When drawing topographic or termal maps, cartographers often produce two-dimensionaldrawings marked with curves of constant elevation or constant temperature. We can do the samefor any function f (x, y) of two variables: we can graphf in the xy-plane by marking its “levelcurves”. The precise definition is: thek-level curveof the function f (x, y) is the set of all points(x, y) in the domainD of f which satisfy the equationf (x, y) = k.
37
This idea becomes even more important in the case of functions of three variables. Note thatthe graphw = f (x, y, z) of such a function is a four-dimensional object, which we can studyanalytically but cannot represent graphically in our three-dimensional universe. On the other hand,we can draw thelevel surfacesof f : these are the sets of points in the domain off for whichf (x, y, z) = k for some fixed numberk.
Example 6.4.Describe the level curves of the function
f (x, y) =x
x2 + y2.
Solution.We want to draw the curvex
x2 + y2= k
for all k. If we exclude the origin from consideration (since the curve is undefined there), we canrepresent this curve by the equation
kx2 + ky2 − x = 0.
Whenk = 0, this equation represents they-axis x = 0, so the level curvef (x, y) = 0 is they-axiswith the origin removed from it. Whenk , 0, we can transform the equation of the level curvefurther to
x2 − xk+ y2 = 0 ⇐⇒
(
x− 12k
)2
+ y2 =1
4k2,
which is the equation of a circle centered at (12k,0) of radius 1
2|k| (k can be negative). Since the origindoes lie on this circle, the level curvef (x, y) = k is the circle with the origin removed from it. Acontour map off showing the level curves withk = 0,±1
3,±12,±1,±2 is shown on Figure 6.3. �
13
12
12−13 −1
2−1 −2
0
x
y
Figure 6.3. The level curves off (x, y) = x/(x2 + y2)
38
6.6. Limits of multivariable functions. Since all the important concepts in the single-variablecalculus were defined in terms of limits, if we are to generalize those concepts to functions of twoand more variables, we must first define the notion of a limit ofa multivariable function. Thefollowing definition generalizes the formal definition of limit to functions of two variables.
Definition. Let f be a function of two variables whose domainD includes points arbitrarily closeto (a,b). We say that thelimit of f (x, y) as(x, y) approaches(a,b) is L and write
lim(x,y)→(a,b)
f (x, y) = L or f (x, y)→ L as (x, y)→ (a,b),
if for any given numberǫ > 0, there is a corresponding numberδ = δ(ǫ) > 0 such that
0 <√
(x− a)2 + (y− b)2 < δ =⇒ | f (x, y) − L| < ǫ.
This definition isvery formal. It is also very useful in proofs (something we will not be con-cerned with) and very imposing at first sight. A more intuitive definition reads as follows.
Definition. Let f be a function of two variables whose domainD includes points arbitrarily closeto (a,b). We say that thelimit of f (x, y) as(x, y) approaches(a,b) is L and write
lim(x,y)→(a,b)
f (x, y) = L or f (x, y)→ L as (x, y)→ (a,b),
if f (x, y) approachesL as (x, y) approaches (a,b), independent of how (x, y) approaches (a,b).
We can define limits of functions of three or more variables ina similar fashion.
Definition. Let f be a function of three variables whose domainD includes points arbitrarily closeto (a,b, c). We say that thelimit of f (x, y, z) as(x, y, z) approaches(a,b, c) is L and write
lim(x,y,z)→(a,b,c)
f (x, y, z) = L or f (x, y, z)→ L as (x, y, z)→ (a,b, c),
if f (x, y, z) approachesL as (x, y, z) approaches (a,b, c), independent of how (x, y, z) approaches(a,b, c). More precisely, the limit off (x, y, z) as (x, y, z) approaches (a,b, c) is L, if for any givennumberǫ > 0, there is a corresponding numberδ = δ(ǫ) > 0 such that
0 <√
(x− a)2 + (y− b)2 + (z− c)2 < δ =⇒ | f (x, y, z) − L| < ǫ.
The limits of multivariable functions can be much trickier than those of a single-variable func-tions. The primary reason for that is the infinitude of ways inwhich the arguments (x, y) (or (x, y, z))can approach the point (a,b) (or (a,b, c)). This point is best illustrated by examples of limits thatdo not exist.
Example 6.5.The limit
lim(x,y)→(0,0)
xx2 + y2
does not exist. Indeed, we know from Example 6.4 (see also Fig. 6.3) that the level curves of thisfunction all want to pass through (0,0) (but since the point is not in the domain off , they can’t).Thus, if we approach (0,0) along two distinct level curves (say, alongf (x, y) = 1 andf (x, y) = −1),the function values will approach distinct values (1 and−1, respectively). However, if the limitexisted, f (x, y) would have to approach the same value independent of the way(x, y) approaches(0,0). Therefore, the limit does not exist. �
39
Example 6.6.Show that the limit lim(x,y)→(0,0)
xy4
x2 + y8does not exist.
Solution.First, consider the behavior of the function as (x, y)→ (0,0) along the straight liney = x.We have
limx→0
f (x, x) = limx→0
x5
x2 + x8= lim
x→0
x3
1+ x6= 0.
On the other hand, if we let (x, y)→ (0,0) along the curvex = y4, we get
limy→0
f (y4, y) = limy→0
y8
y8 + y8=
12.
Since different approach paths lead to different limiting values, the limit does not exist. �
Example 6.7.For any numbersa,b, c, we have
lim(x,y)→(a,b)
x = a, lim(x,y)→(a,b)
y = b, lim(x,y)→(a,b)
c = c.
Using these facts and the two-variable versions of the limitlaws:
lim(x,y)→(a,b)
[ f (x, y) ± g(x, y)] = lim(x,y)→(a,b)
f (x, y) ± lim(x,y)→(a,b)
g(x, y), etc.,
we can compute the limit of any polynomial inx andy at any point (a,b) in the plane. For example,
lim(x,y)→(2,3)
xy=(
lim(x,y)→(2,3)
x)(
lim(x,y)→(2,3)
y)
= 2 · 3 = 6.
�
6.7. Continuity of multivariable functions. We can use the notion of limit of multivariable func-tions to define the notion of continuity.
Definition. Let f (x, y) be a function of two variables whose domainD includes (a,b) and pointsarbitrarily close to (a,b). We say thatf (x, y) is continuous at(a,b), if
lim(x,y)→(a,b)
f (x, y) = f (a,b).
We say thatf (x, y) is continuous in D, if it is continuous at every point (a,b) ∈ D.
Definition. Let f (x, y, z) be a function of three variables whose domainD includes (a,b, c) andpoints arbitrarily close to (a,b, c). We say thatf (x, y, z) is continuous at(a,b, c), if
lim(x,y,z)→(a,b,c)
f (x, y, z) = f (a,b, c).
We say thatf (x, y, z) is continuous in D, if it is continuous at every point (a,b, c) ∈ D.
We want also to extend the notion of piecewise continuity to multivariable functions, and inparticular, to functions of two variables. Recall thatf (x) is called piecewise continuous on [a,b]if it is continuous everywhere in [a,b] except for a finite number of points, where it can have onlyjump or removable discontinuities. The two-dimensional version of this definition reads as follows.
Definition. Let f (x, y) with domainD. We say thatf (x, y) is piecewise continuous in D, if it iscontinuous at every point (a,b) ∈ D, except possibly for the points on a finite number of smoothcurves inD, sayγ1, γ2, . . . , γn, where f (x, y) may have jump discontinuities.
40
Remark. As in the case of single-variable functions, polynomials inseveral variables are continu-ous. Likewise, rational functions in several variables arecontinuous except where the denominatoris 0. Finally, the composition of a continuous function in several variables with a continuousfunction is continuous.
Example 6.8.The function
f (x, y) = cos
(
x2 − y2
x2 + y2
)
is continuous everywhere in its domain, which is all the points inR2 except (0,0). Indeed,f (x, y)is the composition of the continuous functionw = cost and the rational function
t =x2 − y2
x2 + y2.
�
Example 6.9.Show that the function
f (x, y) =
sin(x2 + y2)x2 + y2
if ( x, y) , (0,0),
1 if (x, y) = (0,0),
is continuous everywhere.
Solution. Indeed, recalling that
limx→0
sinxx= 1,
we find that the function
g(x) =
sinxx
if x , 0,
1 if x = 0,
is continuous everywhere. Furthermore,x2 + y2 is also continuous everywhere. Therefore, thefunction f (x, y) = g(x2 + y2) is continuous everywhere. �
Exercises
1. Find the domain and the range of the given function:(a) f (x, y) = y2e−x2/2 (b) f (x, y) = ln
(
x+ y2)
(c) f (x, y, z) =z
x2 − y2 − 1
2. Sketch the domain of the given function:(a) f (x, y) = ln
(
x+ y2)
(b) f (x, y) =
√x+ 2y
x2 + y2 − 1
3. For each function, draw the level curves for levels−1,0,1,2, and 3; then sketch the graphz= f (x, y):(a) f (x, y) = 1+ y2 (b) f (x, y) = 2x+ 3y− 5 (c) f (x, y) = x2 + y2
4. Find the given limit or show that it does not exist:(a) lim
(x,y)→(1,3)
(
x4 − 4xy2 + 2x3y)
(d) lim(x,y)→(0,0)
3x2yx4 + y2
(b) lim(x,y)→(0,0)
xyx2 + y2
(e) lim(x,y)→(0,0)
xy√
x2 + y2
(c) lim(x,y,z)→(1,0,2)
sin(2xy)1− zy2
(f) lim(x,y)→(0,0)
xy√1+ xy− 1
41
5. Find the set of points inR2 orR3 where the given function is continuous:(a) f (x, y) = sin
(
x4 + x2y2)
(c) f (x, y) =
xyx2 + y2
if ( x, y) , (0,0),
1 if (x, y) = (0,0).
(b) f (x, y, z) = ln(
4− x2 − y2 − 4z2)
(d) f (x, y) =
xy√1+ xy− 1
if ( x, y) , (0,0),
1 if (x, y) = (0,0).
42
7. Surfaces
7.1. Introduction. In single-variable calculus, the interplay between curvesand functions is es-sential: we use geometrical questions about curves to motivate the study of calculus and then applyour knowledge of calculus to answer questions about plane curves. A similar symbiosis occurs inmultivariable calculus between functions and surfaces. However, while we are familiar with a fairnumber of plane curves, our supply of known surfaces is limited essentially to planes, spheres, andgraphsz = f (x, y). We introduced also the level surfaces of functions of three variables: these arethe sets
Σ ={
(x, y, z) ∈ R3 : F(x, y, z) = k}
.
However, there are hardly any level surfaces that we recognize. The main goal of this lecture is tointroduce some other surfaces that are commonly used in multivariable calculus.
7.2. Cylinders.
Definition. A cylinder is a surface that consists of all lines—calledrullings—that are parallel to agiven line and pass through a given plane curve.
Example 7.1.We have seen already one example of a cylinder: in Example 1.8, we explained thatthe equationx2 + y2 = x describes the surfaceΣ in space that consists of all the points on verticallines that pass through a circle in thexy-plane. Note that by a similar argument, any equation ofone of the forms
f (x, y) = 0, f (x, z) = 0, or f (y, z) = 0represents a cylinder in space. For example, the equation 2y2 + z2 = 1 represents the cylinder thatpasses through the ellipse 2y2 + z2 = 1 in theyz-plane and has rullings parallel to thex-axis. Notethat the three-dimensional equations of the ellipse are
2y2 + z2 = 1, x = 0.
Similarly, in space, the equationx = y2 represents the cylinder that passes through the parabolax = y2 in thexy-plane and has rullings parallel to thez-axis. �
7.3. Quadric surfaces. Quadric surfaces are generalizations of the conic sectionsto three dimen-sions.
Definition. A quadric surfaceis the set of all points (x, y, z) ∈ R3 that satisfy a second-degreeequation in the variablesx, y, andz:
A11x2 + A12xy+ A13xz+ A22y
2 + A23yz+ A33z2 + B1x+ B2y+ B3z+ B0 = 0,
whereA11, . . . ,A33, B0, . . . , B3 are constants.
In other words, a quadric surface is a level surface of a quadratic function of three variables. Inthe next couple of examples, we visualize some quadric surfaces by sketching their traces in planesparallel to the coordinate planes.
Example 7.2.Consider the quadric surface
x2
4+ y2 + z2 = 1.
The trace of this surface in the horizontal planez= k has equations
x2
4+ y2 = 1− k2, z= k,
43
which represent an ellipse, provided that
1− k2 > 0 ⇐⇒ −1 < k < 1.
Similarly, the traces of the surface in the planesy = k, −1 < k < 1, andx = k, −2 < k < 2, are alsoellipses. Thus, the given surface is called anellipsoid. �
Example 7.3.Consider the quadric surface
z= x2 − 2y2.
The trace of this surface in a planex = k is
z= k2 − 2y2, x = k,
which is a parabola open downward. Likewise, the trace in a planey = k is a parabola openupward,
z= x2 − 2k2, y = k,
and the trace in a vertical planez= k is the hyperbola
x2 − 2y2 = k, z= k.
This surface is called anhyperbolic paraboloid. �
Remark. At first, one might think that there is no neat classification of all quadric surfaces, but infact there is. First of all, by a change of the coordinates, every quadric surface can be brought intothe form
Ax2 + By2 +Cz2 + Dz+ E = 0,
where eitherD = 0 or C = E = 0. If one of the variables is missing—say, ifA = 0 or B = 0,then the quadric surface is a cylinder parallel to the respective coordinate axis. This leaves onlysix possibilities, which lead to the following six surfaces: cone, ellipsoid, elliptic paraboloid,hyperbolic paraboloid, one-sheet hyperboloid, two-sheet hyperboloid. These surfaces in standardpositions are displayed on Figure 7.1 and some basic facts about them are summarized in Table 7.1.
Example 7.4.Classify the quadric surfaceΣ with equation
4y2 + z2 − x+ 16y− 4z+ 19= 0.
Solution.We can rewrite this equation in the form
4(y+ 2)2 + (z− 2)2 − (x+ 1) = 0.
If we now change the coordinates to
X = y+ 2, Y = z− 2, Z = x+ 1,
we find that a point inP(x, y, z) in the xyz-space is onΣ if and only if P(X,Y,Z) is on the surfaceZ = 4X2 + Y2 in the XYZ-space. The latter surface is an elliptic paraboloid witha = 1
4, b = 1,andc = 1. Note that in thexyz-space the vertex of the paraboloid is at (−1,2,−2) (these are thexyz-coordinates of the origin of theXYZ-coordinate system) and its axis is the liney = −2, z = 2(these are equations of theZ-axis inxyz-coordinates). �
44
Surface Equation Traces Axis
xy yz xz
EllipticParaboloid
zc=
x2
a2+
y2
b2Ellipses Parabolas Parabolasz-axis
HyperbolicParaboloid
zc=
x2
a2− y2
b2Hyperbolas Parabolas Parabolasz-axis
Ellipsoidx2
a2+
y2
b2+
z2
c2= 1 Ellipses Ellipses Ellipses
One-sheetHyperboloid
x2
a2+
y2
b2− z2
c2= 1 Ellipses Hyperbolas Hyperbolasz-axis
Conex2
a2+
y2
b2− z2
c2= 0 Ellipses Hyperbolas Hyperbolasz-axis
Two-sheetHyperboloid
x2
a2+
y2
b2− z2
c2= −1 Ellipses Hyperbolas Hyperbolasz-axis
Table 7.1. The non-degenerate quadric surfaces
z
yxElliptic Paraboloid
z
y
x
Hyperbolic Paraboloid
z
y
x
Ellipsoidz
yx
One-sheet Hyperboloid
z
yx
Cone
z
yx
Two-sheet Hyperboloid
Figure 7.1. The non-degenerate quadric surfaces
7.4. Parametric surfaces. By now, you have probably come to the realization that a “surface”is a three-dimensional geometric object that is essentially two-dimensional. In the case of graphsz = f (x, y) this is quite obvious: the points (x, y, z) on the graph are in one-to-one correspondencewith the points (x, y) in the domain off (x, y). For example, the points on the elliptic paraboloidz= x2 + 2y2 (see Fig. 7.1(a)) are in one-to-one correspondence with thepoints in thexy-plane. Onthe other hand, the two-dimensionality of a level surface, such as the ellipsoidx2 + 2y2 + z2 = 4,
45
is less obvious from its equation (though it is quite intuitive from Fig. 7.1(c)). Level surfacesare also hard to plot, because there is no easy way to generatepoints on a level surface. On theother hand, level surfaces are the natural analytic representation for some basic examples: the conex2 + y2 − z2 = 0, the cylinderx2 + y2 = 1, and the spherex2 + y2 + z2 = 1 are all level surfacesthat cannot be represented as graphs of functions of two variables. In this section, we introduce thenotion of a “parametric surface”.
Definition. If x(u, v), y(u, v), z(u, v) are three continuous functions defined on a setD in R2, thenthe setΣ of the points (x, y, z) satisfying theparametric equations
x = x(u, v), y = y(u, v), z= z(u, v) ((u, v) ∈ D), (7.1)
is called aparametric surface; the variablesu, v are called theparameters. Equations (7.1) and thevector functionr : R2→ R3, given by
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D), (7.2)
are referred to asparametrizationsof Σ.
Remark. Like level surfaces, parametric surfaces generalize of thegraphsz= f (x, y) of functionsof two variables. Indeed, suppose thatΣ is the surfacez = f (x, y), (x, y) ∈ D. Then we canparametrizeΣ by the equations
x = u, y = v, z= f (u, v) ((u, v) ∈ D).
On the other hand, a parametric surface is almost as easy to plot as the graph of a function: to plotz = f (x, y), we let the point (x, y) roam through the domainD and plot the points (x, y, f (x, y)); toplot the surfaceΣ parametrized by (7.2), we let (u, v) roam throughD and plot the endpoints of theposition vectorsr (u, v).
Example 7.5. Identify the surfaceΣ, parametrized by the vector function
r (u, v) = ue1 + ucosve2 + usinve3,
where−∞ < u < ∞ and 0≤ v ≤ 2π.
Solution.For a point (x, y, z) onΣ, we have
y2 + z2 = (ucosv)2 + (usinv)2 = u2 = x2.
This is the equation of a cone whose axis of symmetry is thex-axis. �
Example 7.6. Find a parametrization of the planeΠ through the pointsA(1,2,3), B(1,−1,−2),andC(−2,0,1).
Solution.Let us first find an equation of the plane. Two vectors paralleltoΠ are−−→AB= 〈0,−3,−5〉
and−−→BC = 〈−3,1,3〉. Hence, the vectorn =
−−→AB× −−→BC is normal toΠ. We have
n = (−3e2 − 5e3) × (−3e1 + e2 + 3e3)
= 9(e2 × e1) − 9(e2 × e3) + 15(e3 × e1) − 5(e3 × e2) = −4e1 + 15e2 − 9e3,
so an equation forΠ is−4(x− 1)+ 15(y− 2)− 9(z− 3) = 0.
46
Let us solve this equation for one of the variables, say forx:
−4(x− 1)+ 15(y− 2)− 9(z− 3) = 0 ⇐⇒ x− 1 = 154 (y− 2)− 9
4(z− 3)
⇐⇒ x = 154 y− 9
4z+ 14.
We see thatΠ is the graph of the function
x = 154 y− 9
4z+ 14,
so it is parametrized by the vector function
r (u, v) =(
154 u− 9
4v+ 14
)
e1 + ue2 + ve3,
where−∞ < u, v < ∞. �
Example 7.7. Let a = 〈a1,a2,a3〉 andb = 〈b1,b2,b3〉 be two vectors that are not parallel to eachother. Find a parametrization of the planeΠ through the origin that is parallel toa andb.
Solution.We can try to argue similarly to the previous problem. However, the equation ofΠ is(
a2b3 − a3b2
)
x+(
a3b1 − a1b3
)
y+(
a1b2 − a2b1
)
z= 0,
and it is not clear which variable to solve for: any of the coefficients can be zero. Thus, we will usea different approach. LetP(x, y, z) be an arbitrary point inΠ. We consider a parallelogramOAPB
O
P
A
B
a
rb
Figure 7.2. The parallelogramOAPB
in Π, whose sidesOAandOBare parallel toa andb, respectively (see Figure 7.2). Then−−→OA= ua
and−−→OB= vb for some numbersu, v. If we denote the position vector ofP by r , we conclude that
r = ua+ vb. By varyingu, v we get the entire planeΠ, so one possible parametrization is
r (u, v) = (a1u+ b1v)e1 + (a2u+ b2v)e2 + (a3u+ b3v)e3 (−∞ < u, v < ∞). �
Example 7.8.Find a parametrization of the elliptic cylinder
x2 + 4z2 = 1, y ≥ 5.
Solution.A parametrization of the ellipsex2 + 4z2 = 1 in thexz-plane is given by the functions
x = cosu, z= 12 sinu (0 ≤ u ≤ 2π).
The points on the given cylinder must havex- andz-coordinates that satisfy these two equationsand ay-coordinate withy ≥ 5. Thus, we can parametrize the cylinder by
r (u, v) = cosue1 + ve2 +12 sinue3,
where 0≤ u ≤ 2π andv ≥ 5. �
47
Example 7.9. Find a parametrization of the part of the plane 3x + 2y + z = 4 that lies within theparaboloidz= x2 + 2y2 + 1.
Solution.By solving the equation of the plane forz, z = 4 − 3x − 2y, we can easily obtain aparametrization of the whole plane:
r (u, v) = ue1 + ve2 + (4− 3u− 2v)e3 (−∞ < u, v < ∞).
To parametrize just the given part of the plane, we restrictu, v to its projection onto thexy-plane.The plane and the paraboloid intersect along the curve that contains the simultaneous solutions ofthe equations
3x+ 2y+ z= 4, z= x2 + 2y2 + 1.Eliminatingz, we find that thex- andy-coordinates of the points on this curve satisfy
4− 3x− 2y = x2 + 2y2 + 1 =⇒(
x+ 32
)2+ 2
(
y+ 12
)2= 23
4 .
Thus, the intersection curve of the plane and the ellipsoid consists of all points on the plane whosex- andy-coordinates satisfy the last equation. Consequently, the intersection curve projects onto anellipse in thexy-plane, and the part of the plane 3x+2y+z= 4 that lies within the given paraboloidprojects onto the interior of that ellipse. We obtain the parametrization
r (u, v) = ue1 + ve2 + (4− 3u− 2v)e3,(
u+ 32
)2+ 2
(
v+ 12
)2 ≤ 234 . �
Exercises
1. Find the traces of the given quadric surface in the planesx = k, y = k, andz = k. Use that information andTable 7.1 to classify the surface.
(a)z= x2 − y2 (b) 4y2 − x2 − z2 = 1 (c)y2 + 4z2 = 5− x2
2. Rewrite the equation of the given quadric surface in a standard form and use the result and Table 7.1 toclassify the surface.
(a) 3x+ 6y2 − 3z2 = 9 (b) x2 + 4y2 + z2 − 4x+ 2z= 4
3. Parametrize the curve of intersection of the cylinderx2 + y2 = 9 and the planez= x+ y+ 2.
4. Identify the given parametric surface:(a) r (u, v) = ucosve1 + usinve2 + u2e3 (b) r (θ, z) = 2 cosθe1 + sinθe2 + ze3
5. Parametrize the plane that contains the linesx− 1
2=
y+ 13=
z−2
andx+ 2−3
= y =z− 4
4.
6. Parametrize the half of the hyperboloidx2 − 4y2 − 4z2 = 16 that lies in the half-spacex ≤ 0.
7. Parametrize the part of the hemispherex2 + y2 + z2 = 5, z≥ 0 that lies inside the cylinderx2 + y2 = 1.
8. Parametrize the part of the paraboloidz= 4− x2 − y2 that lies above thexy-plane by using:(a) x andy as parameters; (b) the polar coordinates in the plane as parameters.
9. Let 0< a < b. A torus of radii a and bis the surface of revolution obtained by rotating aboutz-axis the circle(x− b)2 + z2 = a2 in thexz-plane. Parametrize the torus of radiia andb.
48
8. Partial derivatives
8.1. First-order partials. If f is a function of two variables, itspartial derivatives fx and fy at(a,b) are defined by
fx(a,b) =∂ f∂x
(a,b) = ∂x f (a,b) = limh→0
f (a+ h,b) − f (a,b)h
and
fy(a,b) =∂ f∂y
(a,b) = ∂y f (a,b) = limh→0
f (a,b+ h) − f (a,b)h
.
If f is a function ofn variablesx1, x2, . . . , xn, itspartial derivatives fx1, fx2, . . . , fxn at (a1,a2, . . . ,an)are defined by
fx j (a1,a2, . . . ,an) =∂ f∂xj
(a1,a2, . . . ,an) = ∂x j f (a1,a2, . . . ,an)
= limh→0
f (a1, . . . ,aj + h, . . . ,an) − f (a1, . . . ,aj , . . . ,an)h
( j = 1,2, . . . ,n).
Remark. Note that if we define the (single-variable) functions
F1(x) = f (x,b) and F2(y) = f (a, y),
then
fx(a,b) = limh→0
f (a+ h,b) − f (a,b)h
= limh→0
F1(a+ h) − F1(a)h
= F′1(a)
and
fy(a,b) = limh→0
f (a,b+ h) − f (a,b)h
= limh→0
F2(b+ h) − F2(b)h
= F′2(b).
That is, the partial derivativesfx and fy of f are the (usual) derivatives of the functionsF1(x) andF2(y), which we obtain by fixing, respectively, the variablesy andx in f (x, y). This observationmeans that we can compute partial derivatives by treating asconstants all variables except the onewith respect to which we are differentiating.
Example 8.1.Find the first-order partials of the functions
f (x, y) =x
x2 + y2, g(x, y, z) = exy sinxz.
Solution.The partial derivatives off are
fx(x, y) =1 · (x2 + y2) − x · (2x)
(x2 + y2)2=
y2 − x2
(x2 + y2)2
and
fy(x, y) = x · −(2y)(x2 + y2)2
=−2xy
(x2 + y2)2.
The partials ofg are
gx(x, y, z) = yexy sinxz+ exy(zcosxz),
gy(x, y, z) = xexy sinxz, gz(x, y, z) = exy(xcosxz). �
49
8.2. Higher-order partials. Just as we can differentiate the derivativef ′(x) of a single-variablefunction to obtain its second derivative, we can differentiate the partial derivatives of a multivari-able function to obtain its higher-order partials. For example, a functionf of two variables hasfour second-order partial derivatives fxx, fxy, fyx, and fyy:
fxx =∂2 f∂x2=∂
∂x
(
∂ f∂x
)
, fxy =∂2 f∂y∂x
=∂
∂y
(
∂ f∂x
)
,
fyx =∂2 f∂x∂y
=∂
∂x
(
∂ f∂y
)
, fyy =∂2 f∂y2=∂
∂y
(
∂ f∂y
)
.
Example 8.2.Find the second-order partials of the functionf (x, y) = x2 cosy+ y2 sinx.
Solution.We have
fx(x, y) = 2xcosy+ y2 cosx, fy(x, y) = −x2 siny+ 2ysinx.
The second-order partials off then are:
fxx = 2 cosy− y2 sinx, fxy = −2xsiny+ 2ycosx,
fyx = −2xsiny+ 2ycosx, fyy = −x2 cosy+ 2 sinx. �
Remark. Notice that the second-order partials of the function in theexample satisfyfxy = fyx. Inother words, for this function it isn’t important whether wedifferentiate first with respect tox andthen with respect toy or vice versa. One might ask whether this is always true. The answer is:“only for ‘nice’ functions”. The following theorem gives a necessary condition for this to hold.
Theorem 8.1.Suppose that f is a function of two variables defined near(a,b) and having second-order partials fxy and fyx that are continuous at(a,b). Then
fxy(a,b) = fyx(a,b).
The following example illustrates the usefulness of the above theorem.
Example 8.3.Find the second-order partials of
f (x, y, z) = sin(x2 + 2y2 − z).
Solution.The first-order partials off are:
fx(x, y, z) = 2xcos(x2 + 2y2 − z),
fy(x, y, z) = 4ycos(x2 + 2y2 − z),
fz(x, y, z) = − cos(x2 + 2y2 − z).
There are nine second-order partials, but we really have to compute only six of them. We have
fxx(x, y, z) = 2 cos(x2 + 2y2 − z) + 2x(
− 2xsin(x2 + 2y2 − z))
,
fyy(x, y, z) = 4 cos(x2 + 2y2 − z) + 4y(
− 4ysin(x2 + 2y2 − z))
,
fzz(x, y, z) = − sin(x2 + 2y2 − z).
For the mixed partials, we have
fxy(x, y, z) = 2x(
− 4ysin(x2 + 2y2 − z))
= −8xysin(x2 + 2y2 − z),
fzy(x, y, z) = −(
− 4ysin(x2 + 2y2 − z))
= 4ysin(x2 + 2y2 − z),
fzx(x, y, z) = −(
− 2xsin(x2 + 2y2 − z))
= 2xsin(x2 + 2y2 − z).50
Note that the last three functions are continuous (they are products of polynomials inx, y, z andcompositions of sine functions and polynomials inx, y, z). Furthermore, it is clear that the partialsfyx, fxz, and fyz will also be continuous. But then the theorem implies that, infact, we have
fyx = fxy, fxz = fzx, fyz = fzy. �
Example 8.4*. Find fxy(0,0) and fyx(0,0) for
f (x, y) =
xy(y2 − x2)x2 + y2
if ( x, y) , (0,0),
0 if (x, y) = (0,0).
Solution.The first order partials off are:
fx(x, y) =
y5 − x4y− 4x2y3
(x2 + y2)2if ( x, y) , (0,0),
0 if (x, y) = (0,0),
and
fy(x, y) =
xy4 + 4x3y2 − x5
(x2 + y2)2if ( x, y) , (0,0),
0 if (x, y) = (0,0).
These formulas are straightforward everywhere except at (x, y) = (0,0), where we use the definitionof partial derivatives:
fx(0,0) = limh→0
f (h,0)− f (0,0)h
= limh→0
0− 0h= 0
and
fy(0,0) = limh→0
f (0,h) − f (0,0)h
= limh→0
0− 0h= 0.
Then
fxy(0,0) = limh→0
fx(0,h) − fx(0,0)h
= limh→0
h− 0h= 1
and
fyx(0,0) = limh→0
fy(h,0)− fy(0,0)h
= limh→0
−h− 0h= −1,
that is, f fails to satisfyfxy(0,0) = fyx(0,0). On the other hand, when (x, y) , (0,0), we have
fxy(x, y) = fyx(x, y) =y6 + 9x2y4 − 9x4y2 − x6
(x2 + y2)3.
This is not surprising, considering that the function
y6 + 9x2y4 − 9x4y2 − x6
(x2 + y2)3
is continuous everywhere except at (0,0) (since it is a rational function inx andy, whose denomi-nator vanishes only at (0,0)). �
51
8.3. Tangent planes.Recall that the equation of the tangent line to the curvey = f (x) at a pointP(x0, y0) on the curve is given by
y− y0 = f ′(x0)(x− x0).
The following theorem is the analog of this formula for functions of two variables. It describesthe equation of the tangent plane to the surfacez= f (x, y) at a pointP(x0, y0, z0) on that surface interms of the partials off . Strictly speaking, we should firstdefinewhat a tangent plane is, but wewill leave that for Lecture #10. The same applies to the proofof the theorem.
Theorem 8.2. Suppose that f has continuous partial derivatives at(x0, y0). Then the equation ofthe tangent plane to the surface z= f (x, y) at the point P(x0, y0, z0) is
z− z0 = fx(x0, y0)(x− x0) + fy(x0, y0)(y− y0).
Example 8.5.Find the tangent plane to the surfacez= ex2−y2at (−1,0,e).
Solution.First, observe that the point is indeed on the surface:
e= e(−1)2−02.
The partials of the given function are
∂z∂x= 2xex2−y2
,∂z∂y= −2yex2−y2
,
whence∂z∂x
(−1,0) = −2e,∂z∂y
(−1,0) = 0.
Applying the theorem, we find that the equation of the tangentplane is
z− e= −2e(x− (−1))+ 0(y− 0) ⇐⇒ 2ex+ z+ e= 0. �
8.4. Linear approximations and differentiable functions*. Recall that for a single-variablefunction f which has a derivativef ′(a) at x = a (we called such functions “differentiable”), thetangent line at (a, f (a)):
y = f (a) + f ′(a)(x− a),
provides a very good approximation tof (x) for x neara:
f (x) ≈ f (a) + f ′(a)(x− a).
When we deal with functions of two variables, we may encounterfunctions f (x, y) for which bothpartial derivativesfx(a,b) and fy(a,b) exist, but the plane
z= f (a,b) + fx(a,b)(x− a) + fy(a,b)(y− b)
is not tangent to the surfacez = f (x, y) and does not provide a good approximation tof (x, y) for(x, y) near (a,b). Thus, in the case of functions of two or more variables, we reserve the term“differentiable function” for functions for which this type of approximation (known as thelinearapproximation to f at(a,b)) does hold.
Definition. A function z = f (x, y) of two variables is calleddifferentiable at(a,b), if near (a,b)we have
f (x, y) = f (a,b) + fx(a,b)(x− a) + fy(a,b)(y− b) + ǫ1(x, y)(x− a) + ǫ2(x, y)(y− b),52
where
lim(x,y)→(a,b)
ǫ1(x, y) = lim(x,y)→(a,b)
ǫ2(x, y) = 0.
Similarly, a functionw = f (x, y, z) of three variables is calleddifferentiable at(a,b, c), if near(a,b, c) we have
f (x, y, z) = f (a,b, c) + fx(a,b, c)(x− a) + fy(a,b, c)(y− b) + fz(a,b, c)(z− c)
+ ǫ1(x, y, z)(x− a) + ǫ2(x, y, z)(y− b) + ǫ3(x, y, z)(z− c),
where
lim(x,y,z)→(a,b,c)
ǫ1(x, y, z) = lim(x,y,z)→(a,b,c)
ǫ2(x, y, z) = lim(x,y,z)→(a,b,c)
ǫ3(x, y, z) = 0.
So, if the existence of partial derivatives does not guarantee the differentiability of f , whatcondition does? It turns out that the existence ofcontinuouspartials suffices.
Theorem 8.3. If the partial derivatives fx and fy exist near(a,b) and are continuous at(a,b), thenf is differentiable at(a,b).
8.5. The chain rule(s). Recall that the chain rule in single-variable calculus givesa formula forthe derivative of a composite function: ify = f (x(t)), then
dydt=
d fdx
dxdt.
In this section, we discuss the generalizations of the chainrule to functions of two and morevariables. Our first result concerns composite functions ofone variable obtained from functions oftwo variables via composition.
Theorem 8.4.Let z= f (x, y) be a differentiable function of two variables and let x= x(t), y = y(t)be two differentiable functions of the same variable t. Then z= f (x(t), y(t)) is a differentiablefunction of t and
dzdt=∂ f∂x
dxdt+∂ f∂y
dydt.
Remark. Take a note that we use “partial d’s”∂∂x and ∂
∂y to denote derivatives of functions of two
variables, but we use “straight d’s”ddt to denote derivatives of functions of a single variable. Itis important to follow this convention when dealing with a mixture of multivariable and single-variable functions.
Example 8.6.Let z= f (x, y) = x ln(x+ 2y), x = cost, andy = sint. Findz′(0).
Solution.The partials off are
fx(x, y) = ln(x+ 2y) +x
x+ 2y, fy(x, y) =
2xx+ 2y
,
and the derivatives ofx(t) andy(t) are
x′(t) = − sint, y′(t) = cost.53
Hence, by the above theorem, the derivativez′(0) of the composite functionz= f (x(t), y(t)) at t = 0is:
dzdt
(0) =∂z∂x
(1,0)dxdt
(0)+∂z∂y
(1,0)dydt
(0)
=
(
ln(1+ 0)+1
1+ 0
)
(−0)+
(
21+ 0
)
(1) = 2.
In general, we have
dzdt=
(
ln(cos(t) + 2 sint) +cost
cost + 2 sint
)
(− sint) +
(
2 costcost + 2 sint
)
(cost). �
Next, we give formulas for the derivatives of a composite functionz= f (x(s, t), y(s, t)).
Theorem 8.5. Let z= f (x, y), x = x(s, t), and y= y(s, t) be three differentiable functions of twovariables. Then z= f (x(s, t), y(s, t)) is a differentiable function of(s, t) with partial derivatives
∂z∂s=∂ f∂x∂x∂s+∂ f∂y∂y∂s,
∂z∂t=∂ f∂x∂x∂t+∂ f∂y∂y∂t.
Example 8.7.Let z= x2 − xy, x = ssint, andy = t sins. Find∂z/∂s and∂z/∂t.
Solution.We havezx(x, y) = 2x− y, zy(x, y) = −x,
andxs(s, t) = sint, xt(s, t) = scost, ys(s, t) = t coss, yt(s, t) = sins.
Hence, the partials of the composite functionz(x(s, t), y(s, t)) are
zs(s, t) = zx(x(s, t), y(s, t))xs(s, t) + zy(x(s, t), y(s, t))ys(s, t)
=(
2ssint − t sins)
sins+(
− ssint)
(t coss)
and
zt(s, t) = zx(x(s, t), y(s, t))xt(s, t) + zy(x(s, t), y(s, t))ts(s, t)
=(
2ssint − t sins)
(scost) +(
− ssint)
sins. �
Remark. Often, we do not substitutex(s, t) andy(s, t) in the formulas for the partials ofz(s, t). Insuch cases, the answer in the above example would take the form
zs(s, t) = (2x− y) sins− xtcoss, zt(s, t) = (2x− y)scost − xsins.
We now state the most general version of the chain rule for multivariable functions.
Theorem 8.6. Let u = f (x1, . . . , xn) be a differentiable function of n variables and let each xj =
xj(t1, . . . , tm) be a differentiable function of (the same) m variables. Then
u = f (x1(t1, . . . , tm), . . . , xn(t1, . . . , tm))
is a differentiable function of t1, . . . , tm with partial derivatives
∂u∂ti=∂ f∂x1
∂x1
∂ti+ · · · + ∂ f
∂xn
∂xn
∂ti(i = 1, . . . ,m).
Example 8.8.Find the partial derivatives ofw(x, y, t), where
w =√
u2 + v2, u = y+ xcost, v = x+ ysint.54
Solution.The partial derivatives of these functions are:
∂w∂u=
u√u2 + v2
,∂w∂v=
v√u2 + v2
,
∂u∂x= cost,
∂u∂y= 1,
∂u∂t= −xsint,
∂v∂x= 1,
∂v∂y= sint,
∂v∂t= xcost.
Hence, the partial derivatives of the composite functionw(x, y, t) are:
∂w∂x=∂w∂u∂u∂x+∂w∂v∂v∂x=
ucost√u2 + v2
+v√
u2 + v2=
ucost + v√u2 + v2
,
∂w∂y=∂w∂u∂u∂y+∂w∂v∂v∂y=
u√u2 + v2
+vsint√u2 + v2
=u+ vsint√
u2 + v2,
∂w∂t=∂w∂u∂u∂t+∂w∂v∂v∂t=−uxsint√
u2 + v2+
vxcost√u2 + v2
=−uxsint + vxcost√
u2 + v2. �
Example 8.9.Find the partial derivativewyx of w(x, y, t), where
w =√
u2 + v2, u = y+ xcost, v = x+ ysint.
Solution.To find a higher-order partial derivative of a composite function, we “mix” the chain rulewith the other rules for partial differentiation. We have
∂2w∂x∂y
=∂
∂x
(
∂w∂y
)
=∂
∂x
(
u+ vsint√u2 + v2
)
=∂
∂u
(
u+ vsint√u2 + v2
)
∂u∂x+∂
∂v
(
u+ vsint√u2 + v2
)
∂v∂x.
We now compute the partials with respect tou andv:
∂
∂u
(
u+ vsint√u2 + v2
)
=∂
∂u
(
(u+ vsint)(u2 + v2)−1/2)
= (u2 + v2)−1/2 + (u+ vsint)(−u)(u2 + v2)−3/2
= (u2 + v2)−3/2[
(u2 + v2) − u(u+ vsint)]
=v2 − uvsint(u2 + v2)3/2
,
∂
∂v
(
u+ vsint√u2 + v2
)
=∂
∂v
(
(u+ vsint)(u2 + v2)−1/2)
= (sint)(u2 + v2)−1/2 + (u+ vsint)(−v)(u2 + v2)−3/2
= (u2 + v2)−3/2[
(u2 + v2) sint − v(u+ vsint)]
=u2 sint − uv(u2 + v2)3/2
.
55
Substituting these partials into the expression forwyx, we get
∂2w∂x∂y
=∂
∂u
(
u+ vsint√u2 + v2
)
∂u∂x+∂
∂v
(
u+ vsint√u2 + v2
)
∂v∂x
=v2 − uvsint(u2 + v2)3/2
· cost +u2 sint − uv(u2 + v2)3/2
· 1
=u2 sint − uv(1+ sint cost) + v2 cost
(u2 + v2)3/2. �
Exercises
1. Find the first partial derivatives of the given function:
(a) f (x, y) = x4 − 4x2y3/2
(d) f (x, y) =(
x2 + xy+ y2)x
(b) f (x, y, z) = e2xy sinxz(e) f (x, z) =
√x2 + z2 + 1
(c) f (x, y, z) = ln(3x+ 7yz)(f) f (r, θ) = r tan2 θ
2. Find the second partial derivatives of the given function:
(a) f (x, y) = x4 − 4x2y3/2 (b) f (x, y) = cos(
x2 + xy+ y2)
(c) f (x, z) =√
x2 + z2 + 1
3. Find the indicated partial derivative of the given function:
(a) f (x, y) = x4 − 4x2y3/2; fxyx
(c) f (x, y) = cos(
x2 + xy+ y2)
; fxxy
(b) f (x, z) =√
x2 + z2 + 1; ∂3 f /∂x∂2z(d) f (x, y, z) = xz2e2yz; fxyz, fzzy
4. Use the chain rule to calculatedF/dt:(a) F(t) = f (x(t), y(t)), f (x, y) = 2xy+ y3, x(t) = t2 + 3t + 4, y(t) = t2
(b) F(t) = f (x(t), y(t)), f (x, y) = xyex2+4y2, x(t) = 2 cost, y(t) = sint
(c) F(t) = f (x(t), y(t), z(t)), f (x, y, z) = x2 + y2 − 3xyz, x(t) = t sint, y(t) = t cost, z(t) = 12t2
(d) F(t) = f (x(t), y(t), z(t)), f (x, y, z) = ln(xy+ 2yz− 3xz), x(t) = 2et, y(t) = 3e−t, z(t) = e2t
5. Use the chain rule to calculate∂F/∂sand∂F/∂t:(a) F(s, t) = f (x(s, t), y(s, t)), f (x, y) = 2x2y− xy2, x(s, t) = t2 + 3s2, y(s, t) = 3st(b) F(s, t) = f (x(s, t), y(s, t)), f (x, y) = arctan
(
x2 − 4y2)
, x(s, t) = 2set + 2se−t, y(s, t) = tes − te−s
(c) F(s, t) = f (x(s, t), y(s, t), z(s, t)), f (x, y, z) = ln(
x2 + 2y2 + 3z2)
, x(s, t) = scost, y(s, t) = ssint,z(s, t) = t2
6. Use the chain rule to calculate the given partial derivative at the given point:(a)gr (r, θ) at (r, θ) = (2, π), g(x, y) = xy2 − x2y, x = r cosθ, y = r sinθ(b) hu(u, v) at (u, v) = (2,−3), h(x, y, z) = ex2−y2−z2
, x = 2u+ v, y = u− v, z= u+ v+ 1
7. Show that the functionu(x, y) = aey cosx+ bey sinx is a solution of the differential equationuxx + uyy = 0.
8. Show that any functionu(x, t) of the form
u(x, t) = f (x+ at) + g(x− at)
is a solution of the differential equationutt = a2uxx.
9. Find the linear approximation to the given differentiable function near the given point:
(a) f (x, y) = ln(2xy+ x+ y), (2,1) (b) f (x, y, z) = arctan(
xy+ xz+ 2yz)
, P(3,2,1)
10. Show that the functionf (x, y) = xy is differentiable at (1,2) by finding functionsǫ1(x, y) andǫ2(x, y) thatsatisfy the definition of differentiability in§8.4.
56
11. (a) Calculatefx(0,0) and fy(0,0) for the function
f (x, y) =
xyx2 + y2
if ( x, y) , (0,0),
0 if (x, y) = (0,0).
(b) Show thatf (x, y) is not continuous at (0,0).(c) Show thatf (x, y) is not differentiable at (0,0).
57
9. Directional derivatives and gradients
9.1. Directional derivatives. Given a function of two variablesf (x, y), we know how to computeits rate of change in thex-direction and in they-direction: the rate of change in thex-direction isgiven by the partial derivative with respect tox,
fx(x, y) =∂ f∂x
(x, y) = limh→0
f (x+ h, y) − f (x, y)h
,
and the rate of change in they-direction is given by the partial derivative with respect to y,
fy(x, y) =∂ f∂y
(x, y) = limh→0
f (x, y+ h) − f (x, y)h
.
The natural question then is how to compute the rate of changeof f in other directions. Thisquestion leads us to the notion of a “directional derivative”.
Definition. If f is a function of two variables and ifu = 〈a,b〉 is a unit vector, thedirectionalderivative of f in theu-direction is
Du f (x, y) = limh→0
f (x+ ha, y+ hb) − f (x, y)h
, (9.1)
if the limit exists. If f is a function of three variables and ifu = 〈a,b, c〉 is a unit vector, thedirectional derivative of f in theu-direction is
Du f (x, y, z) = limh→0
f (x+ ha, y+ hb, z+ hc) − f (x, y, z)h
. (9.2)
Remark. Note that if we introduce the functiong(t) = f (x+ ta, y+ tb), the limit on the right sideof (9.1) is also
limh→0
g(h) − g(0)h
= g′(0);
that is,Du f (x, y) = g′(0). On the other hand, by the chain rule,
g′(t) = fx(x+ ta, y+ tb) · a+ fy(x+ ta, y+ tb) · b =⇒ g′(0) = fx(x, y)a+ fy(x, y)b.
This is (essentially) how one proves the following result.
Theorem 9.1. If f is a differentiable function of two variables andu = 〈a,b〉 is a unit vector, then
Du f (x, y) = fx(x, y)a+ fy(x, y)b.
If f is a differentiable function of three variables andu = 〈a,b, c〉 is a unit vector, then
Du f (x, y, z) = fx(x, y, z)a+ fy(x, y, z)b+ fz(x, y, z)c.
Example 9.1. Find the directional derivative off (x, y) =2x
x− yat the point (1,0) in the direction
of v = e1 −√
3e2.
Solution.First, notice thatv is not unit and that the unit vector in the direction ofv is
u =v‖v‖=〈1,−
√3〉√
1+ 3=
⟨
12,−√
32
⟩
.
The partials off are
fx(x, y) =2(x− y) − 2x
(x− y)2=−2y
(x− y)2, fy(x, y) =
2x(x− y)2
,
58
so the theorem gives
Du f (x, y) =12−2y
(x− y)2+−√
32
2x(x− y)2
=−y−
√3x
(x− y)2, Du f (1,0) = −
√3. �
Example 9.2. Find the directional derivative off (x, y, z) =x
y+ zat the point (2,1,1) in the direc-
tion of v = 〈1,2,3〉.
Solution.The unit vector in the direction ofv is
u =v‖v‖=〈1,2,3〉√1+ 4+ 9
=
⟨
1√14,
2√14,
3√14
⟩
,
and the partial derivatives off are
fx(x, y, z) =1
y+ z, fy(x, y, z) =
−x(y+ z)2
, fz(x, y, z) =−x
(y+ z)2.
Hence,
Du f (2,1,1) =12
1√14− 1
22√14− 1
23√14=−2√14. �
9.2. The gradient vector.
Definition. If f is a function of two variablesx andy, its gradient, denoted∇ f or gradf , is thevector function
∇ f = 〈 fx(x, y), fy(x, y)〉 = ∂ f∂x
e1 +∂ f∂y
e2.
If f is a function of three variablesx, y, andz, its gradient, ∇ f or gradf , is
∇ f = 〈 fx(x, y, z), fy(x, y, z), fz(x, y, z)〉 =∂ f∂x
e1 +∂ f∂y
e2 +∂ f∂z
e3.
Remark. Note that using the gradient vector, we can restate the formulas in Theorem 9.1 in theform
Du f (x) = ∇ f (x) · u, (9.3)
wherex = (x, y) or x = (x, y, z).
Theorem 9.2. Let f be a differentiable function of two or three variables. The maximum value ofthe directional derivative Du f (x) is ‖∇ f (x)‖; it occurs whenu has the same direction as∇ f (x).
Proof. By (9.3) and the properties of the dot product,
Du f (x) = ∇ f (x) · u = ‖∇ f (x)‖ ‖u‖ cosθ = ‖∇ f (x)‖ cosθ,
whereθ is the angle between∇ f (x) andu. Since cosθ ≤ 1, with equality whenθ = 0 (i.e., when∇ f (x) andu point in the same direction), we find that
Du f (x) ≤ ‖∇ f (x)‖,
with an equality whenu points in the direction of the gradient. �
59
Example 9.3.By Example 9.2, the gradient of the function
f (x, y, z) =x
y+ z
is
∇ f (x, y, z) =
⟨
1y+ z
,−x
(y+ z)2,−x
(y+ z)2
⟩
.
Hence, the largest directional derivative at (2,1,1) occurs in the direction of
∇ f (2,1,1) =
⟨
12,−12,−12
⟩
.
Its value is‖∇ f (2,1,1)‖ =√
32 . �
Exercises
1. Find the directional derivative of the given function in the given direction:(a) f (x, y) =
(
x2 + y2)
e2xy, the direction of〈1,2〉 (b) f (x, y, z) = ln(
xy+z2)
, the direction ofe2−2e3
2. Find the directional derivative ofF(x, y, z) =√
4x2 − y2 − z2 atP(2,−1,3) in the direction ofv = 〈−1,−2,2〉.
3. Find the directional derivative of
G(x, y, z) = arcsin
(
2x4x2 + y2 + z2
)
at the pointP(1,2,0) in the direction of the pointQ(−1,3,1).
4. Find the maximum rate of change off at the given point and the direction in which it occurs:(a) f (x, y) = tan(xy+ x+ y), P(2,−1) (b) f (x, y, z) = xy+ yz2, Q(1,−3,1)
60
10. Tangent planes and normal vectors
10.1. Tangent planes and normal vectors to level surfaces.We now fulfill a promise made inLecture 8. We start by defining the tangent plane to a level surface.
Definition. Let F(x, y, z) be a differentiable function, and letP(x0, y0, z0) be a point on the surfaceΣ with equationF(x, y, z) = k. Thetangent plane toΣ at P is the plane, if it exists, that containsPand the tangent vectors to any smooth curveγ onΣ that passes throughP.
The following theorem establishes a formula for the equation of the tangent plane to a levelsurface.
Theorem 10.1.Let F(x, y, z) be a differentiable function, and let P(x0, y0, z0) be a point on thesurfaceΣ with equation F(x, y, z) = k. If ∇F(x0, y0, z0) , 0, then the tangent plane toΣ at the pointP exists and is perpendicular to∇F(x0, y0, z0). In particular, the tangent plane has equation
Fx(x0, y0, z0)(x− x0) + Fy(x0, y0, z0)(y− y0) + Fz(x0, y0, z0)(z− z0) = 0. (10.1)
Proof. Note that ifγ is a curve onΣ given by the vector functionr (t) = x(t)e1+ y(t)e2+ z(t)e3, thenthe component functions ofr (t) must satisfy the equation ofΣ:
F(x(t), y(t), z(t)) = k.
When we differentiate this equation using the chain rule, we get
∂F∂x
dxdt+∂F∂y
dydt+∂F∂z
dzdt= 0 ⇐⇒ ∇F · r ′(t) = 0.
In other words, the tangent vector to the curve,r ′(t), is perpendicular to the gradient vector∇F.First of all, this proves the existence of the tangent plane—it is the plane throughP that is perpen-dicular to∇F(x0, y0, z0). Second, we get that (10.1) is the equation of the tangent plane. �
Example 10.1. Find the equation of the tangent plane at a pointP(x0, y0, z0) of the one-sheethyperboloid
F(x, y, z) =x2
a2+
y2
b2− z2
c2= 1.
Solution.We have
Fx(x0, y0, z0) =2x0
a2, Fy(x0, y0, z0) =
2y0
b2, Fz(x0, y0, z0) = −
2z0
c2,
so the equation of the tangent plane to the hyperboloid at a point P(x0, y0, z0) is
2x0
a2(x− x0) +
2y0
b2(y− y0) −
2z0
c2(z− z0) = 0,
which is equivalent to2xx0
a2+
2yy0
b2− 2zz0
c2=
2x20
a2+
2y20
b2− 2z2
0
c2.
SinceP lies on the hyperboloid, the right side is equal to 2 and we canrewrite the last equation inthe form
xx0
a2+
yy0
b2− zz0
c2= 1. �
Example 10.2.Find the equation of the tangent plane at a pointP(x0, y0, z0) of the surfacez =f (x, y), where f (x, y) is a differentiable function.
61
Solution.We can rewrite the equation of the surface as
f (x, y) − z= 0,
which is of the form considered above withF(x, y, z) = f (x, y) − z andk = 0. Since
Fx(x0, y0, z0) = fx(x0, y0), Fy(x0, y0, z0) = fy(x0, y0), Fz(x0, y0, z0) = −1,
we deduce that the equation of the tangent plane to the surface at (x0, y0, z0) is
fx(x0, y0)(x− x0) + fy(x0, y0)(y− x0) − (z− z0) = 0,
which is equivalent to the equation stated without proof in§8.3. Therefore, Theorem 8.2 is merelyspecial case of Theorem 10.1. �
Finally, we define the normal line to a surface.
Definition. Let F(x, y, z) be a differentiable function and letP(x0, y0, z0) be a point of the surfaceΣ with equationF(x, y, z) = k. Thenormal line toΣ at P is the line that passes throughP and isorthogonal to the tangent plane atP.
The same reasoning that we used to establish (10.1) shows that the symmetric equations of thenormal line toΣ at P is
x− x0
Fx(x0, y0, z0)=
y− y0
Fy(x0, y0, z0)=
z− z0
Fz(x0, y0, z0),
with the proper modifications, if some of the partials vanishat (x0, y0, z0).
10.2. Tangent planes and normal vectors to parametric surfaces.Suppose thatΣ is a surfacewith parametrization
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D)
andP0(x0, y0, z0) = r (u0, v0) is a point onΣ. We shall focus on curvesγ onΣ that are the images ofparametric curves
u = u(t), v = v(t) (t ∈ I )
that lie inD and pass through the point (u0, v0).
Definition. The tangent plane toΣ at P is the plane, if it exists, that containsP and the tangentvectors to any smooth curveγ onΣ of the above type.
We remark that this definition, at least at first glance, seemsmore restrictive than the definitionwe gave in§10.1 for the tangent plane to a level surface: we look only at some curves onΣ insteadof looking at all such curves. However, it can be shown that our definition and the definitionthat refers to all the curves onΣ yield the same notion of tangent plane. The following theoremestablishes the existence of the tangent plane at a “smooth”point of a parametric surface.
Theorem 10.2.Let r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3, (u, v) ∈ D, be a differentiable vectorfunction with partials
ru = xue1 + yue2 + zue3, r v = xve1 + yve2 + zve3,
and let P(x0, y0, z0) = r (u0, v0) be a point on the parametric surfaceΣ given byr (u, v). If
n = ru(u0, v0) × r v(u0, v0) , 0,
then the tangent plane toΣ at the point P exists and is perpendicular to the vectorn.62
Proof. Let γ be a curve onΣ of the type considered in the above definition. It can be parametrizedby means of the vector function
f (t) = r (u(t), v(t)) = x(u(t), v(t))e1 + y(u(t), v(t))e2 + z(u(t), v(t))e3 (t ∈ I ).
Thus, we can use the chain rule to express its tangent vectorf ′(t) in terms of the derivatives ofr (u, v), u(t), andv(t):
dfdt=
dxdt
e1 +dydt
e2 +dzdt
e3
=
(
∂x∂u
dudt+∂x∂v
dvdt
)
e1 +
(
∂y∂u
dudt+∂y∂v
dvdt
)
e2 +
(
∂z∂u
dudt+∂z∂v
dvdt
)
e3
=
(
∂x∂u
dudt
e1 +∂y∂u
dudt
e2 +∂z∂u
dudt
e3
)
+
(
∂x∂v
dvdt
e1 +∂y∂v
dvdt
e2 +∂z∂v
dvdt
e3
)
=dudt
(
∂x∂u
e1 +∂y∂u
e2 +∂z∂u
e3
)
+dvdt
(
∂x∂v
e1 +∂y∂v
e2 +∂z∂v
e3
)
= u′(t)ru + v′(t)r v.
In particular, the tangent vectorf ′(t0) at the pointP0 of γ is
f ′(t0) = u′(t0)ru(u0, v0) + v′(t0)r v(u0, v0).
In other words, the tangent vector is the sum of a vector parallel to ru(u0, v0) and a vector parallelto r v(u0, v0). It follows that the cross productn, which is perpendicular to bothru(u0, v0) andr v(u0, v0), will be perpendicular to the tangent vectorf ′(t0) to the curveγ. Note that the last vectoris independent of the curve and is completely determined by the parametrization of the surfaceΣ.This establishes the existence of the tangent plane: it is the plane through the pointP0 with normalvectorn. �
Example 10.3.Find an equation of the tangent plane to the surfaceΣ with parametrization
r (u, v) = cosu(4+ 2 cosv)e1 + sinu(4+ 2 cosv)e2 + 2 sinve3 (0 ≤ u, v ≤ 2π)
at the point (4,0,2) = r (0, π/2).
Solution.We have
ru(u, v) = − sinu(4+ 2 cosv)e1 + cosu(4+ 2 cosv)e2,
r v(u, v) = −2 cosusinve1 − 2 sinusinve2 + 2 cosve3,
so the normal vector toΣ at (4,0,2) is
n = (ru × r v)(0, π/2) = 〈0,4,0〉 × 〈−2,0,0〉 = (4e2) × (−2e1) = −8(e2 × e1) = 8e3.
Thus, the equation of the tangent plane is
8(z− 2) = 0 ⇐⇒ z= 2. �
Example 10.4.Find the equation of the tangent plane at a pointP(x0, y0, z0) of the surfacez =f (x, y), where f (x, y) is a differentiable function.
Solution.We can parametrize the given surface as
r (x, y) = xe1 + ye2 + f (x, y)e3 ((x, y) ∈ D),63
so the normal vectorn to it is
n = r x × r y =
∣
∣
∣
∣
∣
∣
e1 e2 e3
1 0 fx
0 1 fy
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
0 fx
1 fy
∣
∣
∣
∣
e1 −∣
∣
∣
∣
1 fx
0 fy
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
1 00 1
∣
∣
∣
∣
e3 = − fxe1 − fye2 + e3.
Thus, the equation of the tangent plane at (x0, y0, z0) is
− fx(x0, y0)(x− x0) − fy(x0, y0)(y− y0) + (z− z0) = 0,
which is equivalent to the equation we gave earlier in Theorem 8.2 and Example 10.2. �
10.3. Smooth and piecewise smooth surfaces.Recall that we called the parametric curveγ givenby the vector functionr (t), t ∈ I , smooth if the tangent vector to the curve,r ′(t), does not vanishfor t ∈ I . We later saw that this condition ensures thatγ has a tangent line at each point. In the nextdefinition, we define a “smooth surface” so that it has a tangent plane at each point.
Definition. The level surfaceΣ with equationF(x, y, z) = k is calledsmooth, if the functionF(x, y, z) is differentiable at every point ofΣ and∇F(x, y, z) , 0, except possibly at the bound-ary ofΣ. The parametric surfaceΣ, given by
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D), (10.2)
is calledsmooth, if the partialsru and r v are defined for all (u, v) ∈ D and ru × r v , 0, exceptpossibly at the boundary ofD.
Note that it is an immediate consequence of this definition and Theorems 10.1 and 10.2 that asmooth surface (either level or parametric) has a tangent plane at each point.
Definition. The level surfaceΣ with equationF(x, y, z) = k is calledpiecewise smooth, if thefunctionF(x, y, z) is continuous and we can representΣ as the union of a finite number of smoothsurfaces, sayΣ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn. The parametric surfaceΣ, given by (10.2) is calledpiecewisesmooth, if r (u, v) is continuous and we can representΣ as the union of a finite number of smoothsurfaces, sayΣ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn.
Example 10.5.Determine whether the surfaceΣ is smooth, piecewise smooth, or neither:(a) Σ is the elliptic paraboloidz= x2 + 4y2.(b) Σ is the conez2 = x2 + y2.
First solution. (a) We viewΣ as a level surface ofF(x, y, z) = z− x2 − 4y2. Since
∇F(x, y, z) = 〈−2x,−8y,1〉 , 0,
the paraboloid is smooth.
(b) We viewΣ as a level surface ofF(x, y, z) = z2 − x2 − y2. Since
∇F(x, y, z) = 〈−2x,−2y,2z〉 = 0 ⇐⇒ x = y = z= 0,
the cone is not smooth. However, we can breakΣ into two surfaces: the level surfaceΣ1 of
F(x, y, z) = z2 − x2 − y2 (z≥ 0),
and the level surfaceΣ2 of
F(x, y, z) = z2 − x2 − y2 (z≤ 0),
each of which is smooth, because the origin is a boundary point for them. Thus, the cone ispiecewise smooth. �
64
Second solution.(b) We viewΣ as a parametric surface given by
r (u, v) = vcosue1 + vsinue2 + ve3,
where 0≤ u ≤ 2π, −∞ < v < ∞. Then
ru = −vsinue1 + vcosue2, r v = cosue1 + sinue2 + e3,
and
ru × r v = (−vsinue1 + vcosue2) × (cosue1 + sinue2 + e3)
= −vsin2 u(e1 × e2) − vsinu(e1 × e3) + vcos2 u(e2 × e1) + vcosu(e2 × e3)
= −vsin2 ue3 + vsinue2 − vcos2 ue3 + vcosue1 = vcosue1 + vsinue2 − ve3.
Hence,ru × r v = 0 whenv = 0 and the cone is not smooth. However, we can breakΣ into twosurfaces: the parametric surfaceΣ1 given by
r (u, v) = vcosue1 + vsinue2 + ve3 (0 ≤ u ≤ 2π, v ≥ 0),
and the parametric surfaceΣ2 given by
r (u, v) = vcosue1 + vsinue2 + ve3 (0 ≤ u ≤ 2π, v ≤ 0),
each of which is smooth, because the origin is a boundary point for them. Thus, the cone ispiecewise smooth. �
10.4. Orientable surfaces. We conclude this lecture with the notion of orientation of a surface,which we shall need in our discussion of surface integrals later in these notes. Intuitively, we thinkof a surface as a two-dimensional object, which is placed in space and thus has an “upper side” anda “lower side.” However, this is not always the case, even fornice, smooth surfaces. We use theconcept of orientation of a surface to separate the intuitive, two-sided surfaces from the one-sidedones. However, before giving a formal definition, we presentan example of a surprisingly simpleone-sided surface.
A
B
D
C
Figure 10.1. The Mobius stripbefore gluing
Figure 10.2. The Mobius stripafter gluing
Example 10.6.Take a long strip of paper in the form of a rectangleABCD, the vertices listedcounterclockwise so thatAB andCD are the two short sides of the strip (see Figure 10.1). Thentwist the paper half a turn and glue the short sides so thatA gets glued toC and B to D. Theresulting surface (see Figure 10.2) has only one side! Indeed, if you take a paintbrush, startedat the cut, and moved along the surface parallel to its edges,you will be able to paint the entirestrip without lifting the paintbrush. This surface is knownas theMobius strip. You may think atfirst that the Mobius strip is an esoteric example and that “normal” surfaces, given by nice, simple
65
formulas, don’t behave this way. Think again! The Mobius stripis given by a nice, simple formula:its parametric equations are
x = 2 cosu+ vcos(u/2), y = 2 sinu+ vcos(u/2), z= vsin(u/2),
where 0≤ u ≤ 2π and−12 ≤ v ≤ 1
2. �
Definition. Let Σ be a smooth surface inR3. For each nonboundary pointP(x, y, z) on the surface,let n1(x, y, z) andn2(x, y, z) be the two unit vectors atP that are normal toΣ. We say thatΣ isorientable, if it is possible to select a unit normal vectorn(x, y, z) at every nonboundary point ofΣ in such a way thatn(x, y, z) varies continuously as (x, y, z) varies overΣ. A choice of the unitnormal vectorn as described above is called anorientationof Σ.
Remark. If a surfaceΣ is orientable, then there are two distinct orientations of that surface. Forexample, let us choose a distinguished pointP0(x0, y0, z0) onΣ. Let us select and fix one of the twounit normal vectors atP0; we denote it byn0. Then for each nonboundary pointP nearP0 onlyone of the unit normal vectors atP will be close ton0, the other will be close to−n0. Thus, thecontinuity requirement on our choice of normal vectors settles the orientation of all normal vectorsnearP0. Of course, once we have decided on the choice of normal vectors at points nearP0, wecan repeat the same procedure to settle the orientation of the normal vectors at points that may notbe nearP0 but are near points that are nearP0; after that, we can settle the choices at points thatmay not be nearP0 or near points that are nearP0, but are near points that are near points that arenearP0; etc. Ultimately, after we choose between the two unit normal vectorsn0 and−n0 at P0,we set a domino effect that yields a consistent choice between the unit normal vectors at all othernonboundary points ofΣ.
Example 10.7.Let Σ be the graph of a functionz= f (x, y), (x, y) ∈ D. In Example 10.4, we usedthe parametrization
r (x, y) = xe1 + ye2 + f (x, y)e3 ((x, y) ∈ D)
to show thatr x × r y = − fxe1 − fye2 + e3
is a normal vector to the surface at the point (x, y, z) of the surface. Note that this vector alwayshas a positive third component. Thus, the normalized cross product
n =r x × r y
‖r x × r y‖=− fxe1 − fye2 + e3(
1+ f 2x + f 2
y
)1/2 ,
will pick consistently one of the two unit normal vectors to the surface: namely, it will always pickthe unit normal that points upward. Consequently,n represents an orientation of the surface; thisorientation is often referred to as theupward orientationof the surface. �
Example 10.8.Let Σ be the spherex2 + y2 + z2 = 4. Its spherical equation isρ = 2, so we candescribeΣ as the set of points (x, y, z) in space whose Cartesian coordinates are related to theirspherical coordinates via the equations
x = 2 cosθ sinφ, y = 2 sinθ sinφ, z= 2 cosφ,
where 0≤ θ ≤ 2π, 0 ≤ φ ≤ π. This provides a parametrization ofΣ:
r (θ,φ) = 2 cosθ sinφe1 + 2 sinθ sinφe2 + 2 cosφe3,66
where 0≤ θ ≤ 2π, 0 ≤ φ ≤ π. Then
r θ = −2 sinθ sinφe1 + 2 cosθ sinφe2, rφ = 2 cosθ cosφe1 + 2 sinθ cosφe2 − 2 sinφe3,
and
r θ × rφ = (−2 sinθ sinφe1 + 2 cosθ sinφe2) × (2 cosθ cosφe1 + 2 sinθ cosφe2 − 2 sinφe3)
= −4 sin2 θ sinφ cosφ(e1 × e2) + 4 sinθ sin2φ(e1 × e3)
+ 4 cos2 θ sinφ cosφ(e2 × e1) − 4 cosθ sin2φ(e2 × e3)
= −4 cosθ sin2φe1 − 4 sinθ sin2φe2 − 4(
sin2 θ + cos2 θ)
sinφ cosφe3
= −4 sinφ(
cosθ sinφe1 + sinθ sinφe2 + cosφe3
)
= −2 sinφ r .
Hence, the normal vectors at a point of the sphere are parallel to the position vectorr of the point.Since the length ofr is the distance from the origin to a point onΣ, we have‖r‖ = 2. Thus,n = 1
2ris a unit normal vector at the pointP = r (u, v) of Σ. Note that sincen has the same direction as
the−−→OP, it will consistently be the unit normal vector to the spherethat is pointed away from the
origin. Therefore,n yields an orientation ofΣ that is often called theoutward orientationof thesphere. �
Exercises
1. Find the equations of the tangent plane and of the normal line to the given surface at the given point:(a)z= xyex+2y, P(2,−1,−2)(c) x2 − 2y2 + 2z2 = 10, P(2,−1,−2)
(b) z= log2
(
xy2 + x2y)
, P(1,−2,1)(d) ln(xyz) = x+ yz−1, P(1,e,e)
2. (a) Find the equation of the normal line to the spherex2 + y2 + z2 = r2 at a point (x0, y0, z0) of the sphere.(b) Show that the normal line from part (a) passes through theorigin.
3. Find the points on the hyperboloidx2 − 2y2 − z2 = −2 at which the tangent plane is parallel to the plane2x− 3y+ 2z+ 7 = 0.
4. Find the points on the hyperbolic paraboloidx = y2 − 4z2 at which the normal line is perpendicular to theplanex+ y+ z= 1.
5. Show that the hyperboloidx2−2y2−z2 = −2 and the ellipsoid(
x+ 43
)2+(
y+ 13
)2+ 1
4
(
z+ 13
)2= 1 are tangent
to each other at the point (−1,−1,1), that is, show that they have the same tangent plane at thatpoint.
6. Find an equation of the tangent plane to the given parametric surface at the given point:(a) r (u, v) = 〈uv,usinv,ucosv〉, (u, v) = (1, π/2)(c) r (u, v) =
⟨
u2 − v2,u+ v,u2 + 2v⟩
, (0,2,3)(b) r (y, z) =
⟨√
4+ y2 + z2, y, z⟩
, (y, z) = (1,1)(d) r (u, v) = 〈ucos 2v,usin 2v, v〉 , (1,
√3, π/6)
7. Determine whether the given surface is smooth, piecewisesmooth, or neither:(a) x2 + 3y2 − 2xyz= 6(c) r (u, v) = 〈u,uv, sinuv〉 , (u, v) ∈ R2
(b) r (u, v) =⟨
3u,u2 − 2v,u3 + v2⟩
, (u, v) ∈ R2
(d) z2 + sin(xy) = 1
67
11. Extremal values of multivariable functions
In this lecture, we shall discuss the local and absolute (or global) maximum and minimum valuesof a function of two variablesf (x, y).
11.1. Review of extremal values of single-variable functions.We start with a review of theextrema of single-variable functions. For example, let us classify the local and global extrema ofthe function
f (x) = x4 − 8x3 + 18x2 − 10
in the interval [−1,2]. First, we find the critical points off : these are the zeros of
f ′(x) = 4x3 − 24x2 + 36x = 4x(x2 − 6x+ 9) = 4x(x− 3)2.
We have
4x(x− 3)2 = 0 ⇐⇒ x = 0 or x = 3.
We determine whether the critical points yield local extrema by the second-derivative test:
f ′′(x) = 12x2 − 48x+ 36 =⇒ f ′′(0) = 36> 0, f ′′(3) = 0,
that is, f has a local minimum atx = 0, (0,−10). (We can also show thatf has no extremum atx = 3, but that does not follow from the second derivative test.)Finally, to find the absolute extremaof f in [−1,2], we compare the local extrema and the boundary values:f (−1) = 17, f (0) = −10,and f (2) = 14. We find that the absolute maximum off is 17 and its absolute minimum is−10.
11.2. Local extrema of a function of two variables.
Definition. Let f be a function of two variables. If (a,b) is a point in the domain off , then f hasa local maximum at(a,b) if f (x, y) ≤ f (a,b) for all points (x, y) near (a,b); f has alocal minimumat (a,b) if f (x, y) ≥ f (a,b) for all points (x, y) near (a,b). A local maximum or minimum is calledalso alocal extremum.
Theorem 11.1. If f has a local extremum at(a,b) and its first-order partials fx(a,b) and fy(a,b)exist, then
fx(a,b) = fy(a,b) = 0.
Proof. If f (x, y) has a local maximum/minimum at (a,b), then the single-variable functiong(x) =f (x,b) also has a local maximum/minimum atx = a. Hence,
g′(a) = 0 =⇒ fx(a,b) = 0.
Similarly, fy(a,b) = 0 by consideringh(y) = f (a, y). �
Definition. Let f be a function of two variables. Acritical point of f is a point (a,b) such thateither fx(a,b) = fy(a,b) = 0, or one of these partial derivatives does not exist. Those critical points(a,b) of f for which fx(a,b) = fy(a,b) = 0 are sometimes calledstationary points.
Example 11.1.Find all critical points of the function
f (x, y) = x3 + y2 − 6xy+ 6x+ 3y− 2.68
Solution.This function has first-order (and also second-, third-, andhigher-order) partial deriva-tives at any point (x, y) in theR2, so the only critical points will be the stationary points off .Since
fx(x, y) = 3x2 − 6y+ 6 and fy(x, y) = 2y− 6x+ 3,the stationary points off are the solutions of the system
{
3x2 − 6y+ 6 = 0
2y− 6x+ 3 = 0⇐⇒
{
x2 − 2y+ 2 = 0
2y = 6x− 3.
Substituting 2y from the second equation into the first, we find thatx must satisfy the equation
x2 − (6x− 3)+ 2 = 0 ⇐⇒ x2 − 6x+ 5 = 0 ⇐⇒ x = 1 or x = 5.
When we substitutex = 1 in the second equation of the system, we obtain 2y = 3 soy = 1.5; whenwe substitutex = 5, we obtain 2y = 27 soy = 13.5. Thus, the critical points off are (1,1.5) and(5,13.5). �
11.3. The second-derivative test.Note that the theorem in the previous section says that (justasin the case of functions of a single variable) a function of two variables can have a local extremumonly at a critical point. However, we know that for a functionof one variable a critical point is notnecessarily an extremal point. Is this also the case for functions of two variables? And if so, thenhow can we determine whether a critical point is an extremal point or not? The following theoremis often sufficient to answer these questions. It is the two-variable version of the second-derivativetest.
Theorem 11.2(Second-derivative test). Suppose that f(x, y) has continuous second-order partialderivatives near the stationary point(a,b) and define
D = D(a,b) = fxx(a,b) fyy(a,b) − fxy(a,b)2 =
∣
∣
∣
∣
fxx(a,b) fxy(a,b)fyx(a,b) fyy(a,b)
∣
∣
∣
∣
.
Then:i) if D > 0 and fxx(a,b) > 0, f has a local minimum at(a,b);ii) if D > 0 and fxx(a,b) < 0, f has a local maximum at(a,b);iii) if D < 0, f does not have a local extremum at(a,b).
In case iii), the point (a,b) is called asaddle point, since under those conditions the graph offnear (a,b) resembles a saddle.
Example 11.2.Find the local extrema of the function
f (x, y) = x3 + y2 − 6xy+ 6x+ 3y− 2.
Solution.We know from Example 11.1 that the critical points off are (1,1.5) and (5,13.5). Thesecond-order partials off are:
fxx(x, y) = 6x, fyy(x, y) = 2, fxy(x, y) = −6,
so fxx(5,13.5) = 30 and
D(1,1.5) =
∣
∣
∣
∣
6 −6−6 2
∣
∣
∣
∣
= −24, D(5,13.5) =
∣
∣
∣
∣
30 −6−6 2
∣
∣
∣
∣
= 24.
Hence, by the second-derivative test, (1,1.5) is a saddle point off and f (5,13.5) = −29 is a localminimum. �
69
Example 11.3.Find the point on the plane 2x− y+ 2z− 3 = 0 that is closest to the origin. Whatis the distance between the plane and the origin?
Solution.We want to minimize the function
d(x, y, z) = x2 + y2 + z2
for (x, y, z) subject to the constraint
2x− y+ 2z− 3 = 0.
This is the same as minimizing the function
f (x, z) = x2 + (2x+ 2z− 3)2 + z2 = 5x2 + 8xz+ 5z2 − 12x− 12z+ 9
in thexz-plane. The first-order partials off are
fx(x, z) = 10x+ 8z− 12, fz(x, z) = 8x+ 10z− 12,
so the critical points are the solutions of the system{
10x+ 8z− 12= 0
8x+ 10z− 12= 0⇐⇒
{
5x+ 4z= 6
4x+ 5z= 6.
Solving the last system, we find that the only critical point of f is (23,
23). The second-order partials
of f arefxx(x, z) = fzz(x, z) = 10, fxz(x, z) = fzx(x, z) = 8,
so
D = D(23,
23) =
∣
∣
∣
∣
10 88 10
∣
∣
∣
∣
= 36.
SinceD > 0 and fxx > 0, we conclude thatf (23,
23) = 4
3 is a local minimum off . It is intuitivelyclear that this local minimum must actually be an absolute minimum of f , so the minimum distanceoccurs when the point (x, y, z) on the plane is (23,−
13,
23). The distance from the origin to that point
(which is the distance from the origin to the plane) is√
(
23
)2+(
− 13
)2+(
23
)2= 1. �
11.4. Absolute extrema of a function of two variables. In real life, we are more likely to beinterested in absolute (or global) extrema of functions than in local extrema. In the last examplewe used practical considerations (i.e., intuition) to makethe jump from the local to the globalminimum. Is there another (hopefully, more rigorous) way toapproach global extrema? Theanswer is “yes”, if the function is defined on a “closed, bounded set”.
First, we must say what does it mean for a set in the plane to be “closed and bounded”. Instead offormal definitions we will be content with explanations by example. A setR is closedif it containsall its “boundary points”. For example, among the sets on Figure 11.1 (where solid boundary curvebelongs to the set and a dashed curve does not), the sets in (a)and (d) are closed and the sets in (b),(c), and (e) are not. A setR is boundedif it is contained inside some disk centered at the origin.For example, of the sets on Figure 11.1, only (d) is not bounded.
We can now state the algorithm for finding the absolute extrema of a function of two variableson a closed, bounded set. It is a generalization of the closedinterval method for finding absoluteextrema of functions of a single variable on a closed interval. To find the absolute extrema off (x, y) on a closed, bounded setR:
1. Find the critical points off insideR and the respective values off .70
(a) (b) (c)
(d) (e)
Figure 11.1. Examples of closed and bounded sets
2. Find the absolute extrema off on the boundary ofR.3. The absolute maximum off in R is the largest among the numbers found in Steps 1 and 2;
the absolute minimum off in R is the smallest among the numbers found in Steps 1 and 2.
Example 11.4.Find the absolute maximum and minimum values of the function
f (x, y) = x3 − y3 − 3x+ 12y
in the closed and bounded region
R= {(x, y) : 0 ≤ x ≤ 2, −3x ≤ y ≤ 0}.
Solution.1. The first-order partials off are
fx(x, y) = 3x2 − 3, fy(x, y) = −3y2 + 12,
so the critical points off are the solutions of the system{
3x2 − 3 = 0
−3y2 + 12= 0⇐⇒
{
x2 = 1
y2 = 4⇐⇒
{
x = ±1
y = ±2.
That is, f has four critical points: (1,2), (1,−2), (−1,2), and (−1,−2). However, of these onlyP(1,−2) lies insideR. The respective value off is f (1,−2) = −18.
2. The boundary ofR consists of three line segments (see Figure 11.2):
ℓ1 : {(x, y) : 0 ≤ x ≤ 2, y = 0},ℓ2 : {(x, y) : 0 ≤ x ≤ 2, y = −3x},ℓ3 : {(x, y) : x = 2,−6 ≤ y ≤ 0}.
71
xy
bP
ℓ1
ℓ3 ℓ2
Figure 11.2. The boundary ofR
2.1. When (x, y) lies onℓ1, f (x, y) is really only a function ofx:
g1(x) = f (x,0) = x3 − 3x (0 ≤ x ≤ 2).
The critical values ofg1 are the solutions of 3x2 − 3 = 0. These arex = ±1, but we are interestedonly in x = 1 (since−1 lies outside [0,2]). Thus, we need compare
g1(0) = 0, g1(1) = −2, and g1(2) = 2.
We find that the absolute minimum and maximum off on ℓ1 are f (1,0) = −2 and f (2,0) = 2.2.2. When (x, y) lies onℓ2, f (x, y) is again a function ofx only:
g2(x) = f (x,−3x) = 28x3 − 39x (0 ≤ x ≤ 2).
The critical values ofg2 are the solutions of 84x2−39= 0: x = ±√
13/28. Again, we are interestedonly in one of these critical points:x =
√13/28. We compare
g2(0) = 0, g2
(√
13/28)
= −26√
13/28= −17.71. . . , and g2(2) = 146.
We find that the absolute minimum and maximum off on ℓ2 are
f (√
13/28,−3√
13/28)= −17.71. . . and f (2,−6) = 146.
2.3. When (x, y) lies onℓ3, f (x, y) is only a function ofy:
g3(y) = f (2, y) = −y3 + 12y+ 2 (−6 ≤ y ≤ 0).
The critical values ofg3 are the solutions of−3y2 + 12 = 0, y = ±2. After droppingy = 2, wecompare
g3(0) = 2, g3(−2) = −14, and g3(−6) = 146.
We find that the absolute minimum and maximum off on ℓ3 are f (2,−2) = −14 and f (2,−6) =146.
3. Comparing all the values found in Steps 1 and 2, we conclude that
min(x,y)∈R
f (x, y) = min{−18,−2,2,−17.71,146,−14} = −18= f (1,−2),
max(x,y)∈R
f (x, y) = max{−18,−2,2,−17.71,146,−14} = 146= f (2,−6). �
72
Example 11.5.Find the absolute maximum and minimum values of the function
f (x, y) = x2 − 2xy− 2x+ 4y
in the closed region bounded by the ellipseγ with parametric equations
x = 3 cost, y = 2 sint (0 ≤ t ≤ 2π).
Solution.1. The first-order partials off are
fx(x, y) = 2x− 2y− 2, fy(x, y) = −2x+ 4,
so the critical points off are the solutions of the system{
2x− 2y− 2 = 0
−2x+ 4 = 0⇐⇒
{
x = y+ 1
x− 2 = 0⇐⇒
{
y = x− 1 = 1
x = 2
Hence,P(2,1) is the only critical point off (x, y). It lies insideγ (see Figure 11.3) andf (2,1) = 4.
x
y
bPγ
Figure 11.3. The ellipseγ and the critical pointP
2. Onγ, we have
f (x, y) = f (3 cost,2 sint) = 9 cos2 t − 12 cost sint − 6 cost + 8 sint,
so we have to find the maximum and the minimum of
g(t) = 9 cos2 t − 12 cost sint − 6 cost + 8 sint (0 ≤ t ≤ 2π).
We have
g′(t) = −18 cost sint + 12 sin2 t − 12 cos2 t + 6 sint + 8 cost
= 24 sin2 t + 6 sint − 12+ 2 cost(4− 9 sint).
Substitutingu = sint and cost = ±√
1− u2, we can rewrite the equationg′(t) = 0 as
24u2 + 6u− 12± 2√
1− u2(4− 9u) = 0
12u2 + 3u− 6 = ±√
1− u2(9u− 4)(
12u2 + 3u− 6)2= (1− u2)(9u− 4)2
225u4 − 200u2 + 36u+ 20= 0.
The last equation has four real roots, whose approximate values are:
α1 ≈ −0.980, α2 ≈ −0.245, α3 ≈ 0.547, α4 ≈ 0.678.
These are the values of sint at the critical numbers ofg(t). They yield a total of eight possiblecritical numbers: four pairs that have the same sines and opposite cosines. In fact, only one of
73
the numbers in each pair is a critical number ofg, but we will ignore this and will treat all eightnumbers as critical. The values of all these critical (and pseudo-critical) numbers, their sines andcosines, and the respective values ofg(t) are listed in Table 11.1.
αk t cost sint g(t)
−0.980 4.510 −0.201 −0.980 −8.630−0.980 4.915 0.201 −0.980 −6.316−0.245 3.389 −0.970 −0.245 9.470−0.245 6.036 0.970 −0.245 3.533
0.547 0.579 0.837 0.547 0.1660.547 2.563 −0.837 0.547 21.2010.678 0.745 0.735 0.678 −0.1040.678 2.397 −0.735 0.678 20.681
Table 11.1. Values ofg(t) at its critical numbers
We also haveg(0) = g(2π) = 3, so
max0≤t≤2π
g(t) = g(π − arcsinα3) ≈ 21.201, min0≤t≤2π
g(t) = g(π − arcsinα1) ≈ −8.630.
3. Comparing all the values found in Steps 1 and 2, we conclude that the absolute minimum off insideγ is
min{4,−8.630,21.201} = −8.630,and that the absolute maximum off is
max{4,−8.630,21.201} = 21.201. �
Exercises
1. Find and classify the critical points (as local maxima, local minima, or saddle points) of the given function:(a) f (x, y) = 9+x+2y−x2+ 1
2y2
(d) f (x, y) = xy(x+ y+ 1)(b) f (x, y) = 2x− 4y− x2 − 2y2
(e) f (x, y) = e2y sin(xy)(c) f (x, y) = x4 + y4 − 2xy(f) f (x, y) = xye−x2−y2
2. Find the absolute maxima and minima of the given function on the given set:(a) f (x, y) = 1+ 2x+ 3y on the triangle with vertices (0,3), (2,1), and (5,3)(b) f (x, y) = x2 + 4y2 − xyon the square−1 ≤ x ≤ 1,−1 ≤ y ≤ 1(c) f (x, y) = x3 + 2xy2 − 2x on the diskx2 + y2 ≤ 4
3. Find the shortest distance from the point (3,2,−1) to the planex− y+ 2z= 1.
4. Find the point(s) on the cone (z− 2)2 = x2 + y2 that is closest to the point (1,1,2).
5. Among all the rectangular boxes of volume 125 ft3, find the dimensions of that which has the least surfacearea.
6. Find the volume of the largest rectangular box with faces parallel to the coordinate planes that can be inscribedin the ellipsoidx2 + 4y2 + 4z2 = 16.
74
12. Double integrals over rectangles
In this lecture, we generalize the idea of the definite integral (see Appendix B) to functions oftwo variables.
12.1. Definition. Let f (x, y) be a bounded function on the rectangleR,
R= [a,b] × [c,d] ={
(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}
.
For eachn ≥ 1, let
x0 = a, x1 = a+ ∆x, x2 = a+ 2∆x, . . . , xn = a+ n∆x = b, ∆x = (b− a)/n,
be the points that divide the interval [a,b] into n subintervals of equal lengths; and let
y0 = c, y1 = c+ ∆y, y2 = c+ 2∆y, . . . , yn = c+ n∆y = d, ∆y = (d − c)/n,
be the points that divide the interval [c,d] into n subintervals of equal lengths. Also, pickn2 samplepoints (x∗i , y
∗j ) so that
xi−1 ≤ x∗i ≤ xi , yj−1 ≤ y∗j ≤ yj (1 ≤ i, j ≤ n).
Then thedouble integral of f over Ris defined by∫∫
Rf (x, y) dA= lim
n→∞
n∑
i=1
n∑
j=1
f (x∗i , y∗j )∆A, ∆A = ∆x∆y, (12.1)
provided that the limit exists. The sums on the right side of (12.1) are calledRiemann sums of fon R.
Remarks. 1. Note that the vertical linesx = xi andy = yj split the rectangleR into n2 congruentsubrectangles, which we may label according to their upper right corner points: if 1≤ i, j ≤ n,thenRi j is the subrectangle whose upper right corner is the point (xi , yj). With this labelling of thesubrectangles, we can report that each sample point (x∗i , y
∗j ) lies in the subrectangleRi j with the
same indeces and there is exactly one sample point per subrectangle. Furthermore, the quantity∆Ain (12.1) is equal to the common area of the rectanglesRi j : onen2th part of the area ofR. Thus,each term in the Riemann sum off is the product of a sample value off in one of the rectanglesRi j and the area of that rectangle.
2. The limit in (12.1) is supposed to exist and be independentof the choice of the sample points.This is not necessarily true for any old function, but it doeshold for functions that are piecewisecontinuous inR (see§6.7 for the definition of piecewise continuity).
3. If f (x, y) ≥ 0 on R, then∫∫
R f (x, y) dA represents the volume of the solid whose upperboundary is the surfacez= f (x, y) and whose lower boundary isR. That is,
∫∫
Rf (x, y) dA= vol.
{
(x, y, z) : (x, y) ∈ R, 0 ≤ z≤ f (x, y)}
= vol.{
(x, y, z) : a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z≤ f (x, y)}
.
4. From now on, we shall use the notation|D| to denote the area of a plane regionD. Let usdivide both sides of (12.1) by|R| = (b− a)(c− d). Since∆A = |R|/n2, we have∆A/|R| = 1/n2, so
75
we get
1|R|
∫∫
Rf (x, y) dA= lim
n→∞
1n2
n∑
i=1
n∑
j=1
f (x∗i , y∗j ).
Since 1n2
∑
i
∑
j f (x∗i , y∗j ) is the average of the sample valuesf (x∗i , y
∗j ), we interpret the last equation
in terms of the average value off :
average(x,y)∈R
f (x, y) =1|R|
∫∫
Rf (x, y) dA. (12.2)
Example 12.1.Evaluate the double integral of the function
f (x, y) =
{
5− 2x− 3y if 5 − 2x− 3y ≥ 0,
0 if 5 − 2x− 3y ≤ 0,
over the rectangleR= [0,3] × [0,2].
Solution.We shall use the interpretation of the double integral as volume. By Remark 3 above, thevalue of the integral
∫∫
Rf (x, y) dA
is the volume of the solidE that lies betweenRand the part of the surfacez= f (x, y) with (x, y) ∈ R.The graph off (x, y) (before the restriction toR) consists of the part of the planez = 5 − 2x − 3ywith 5−2x−3y ≥ 0 and the part of thexy-planez= 0 with 5−2x−3y ≤ 0. The lineℓ with equation5− 2x− 3y = 0 splits the rectangleR in two parts: a triangle with vertices (0,0), (5
2,0), and (0, 53)
where 5− 2x − 3y ≥ 0; and a pentagon with vertices (52,0), (3,0), (3,2), (0,2), and (0, 5
3) where5− 2x− 3y ≤ 0 (see Figure 12.1). For points (x, y) lying in the pentagon, the graphz= f (x, y) and
(0,0)
(0,2)(0, 5
3)
(52,0)
(3,0)
(3,2)
ℓ
Figure 12.1. The base of the pyramid
the xy-plane coincide, and for (x, y) in the triangle, the graphz = f (x, y) is a triangle in the planez= 5− 2x− 3y. Thus, the solidE is the pyramid with verticesO(0,0,0), A(5
2,0,0), B(0, 53,0), and
C(0,0,5). Its volume is
13 |OC| · area(△OAB) = 1
3 |OC| · 12 |OA| |OB| = 1
3 5 · 12
52
53 =
12536 .
Hence,∫∫
Rf (x, y) dA=
12536. �
76
12.2. Iterated integrals. In practice, we want to reverse the order of things in the above example,so that we are able to compute volumes, averages, etc. by interpreting them as double integrals andthen evaluating the latter using some “standard” machineryindependent of our interpretation. Onesuch method is the method of iterated integrals. Aniterated integralis an expression of one of theforms
∫ d
c
[∫ b
af (x, y) dx
]
dy or∫ b
a
[∫ d
cf (x, y) dy
]
dx.
Sometimes, we write these integrals as∫ d
c
∫ b
af (x, y) dxdy and
∫ b
a
∫ d
cf (x, y) dydx,
respectively. Note that these are definite integrals of functions defined by means of definite inte-grals andNOT double integrals. Thus, we can evaluate such integrals using the familiar methodsfrom single-variable calculus.
Example 12.2.Evaluate∫ 1
0
∫ 3
0x3y dydx and
∫ 3
0
∫ 1
0x3y dxdy.
Solution.We have∫ 1
0
∫ 3
0x3y dydx=
∫ 1
0
[
x3y2
2
]3
0
dx=92
∫ 1
0x3dx=
[
9x4
8
]1
0
=98,
and∫ 3
0
∫ 1
0x3y dxdy=
∫ 3
0
[
x4y4
]1
0
dy=14
∫ 3
0y dy=
[
y2
8
]3
0
=98. �
12.3. Fubini’s theorem. Notice that in the last example, we evaluated the two possible iteratedintegrals ofx3y over [0,1]× [0,3] and obtained the same value for both despite following differentroutes. Is this always the case? The answer to this question is in the affirmative. Moreover,the common value of the two iterated integrals is the value ofthe double integral off over therectangle. This fact is known asFubini’s theorem.
Theorem 12.1(Fubini). Let f(x, y) be piecewise continuous in the rectangle R= [a,b] × [c,d].Then
∫∫
Rf (x, y) dA=
∫ b
a
∫ d
cf (x, y) dydx=
∫ d
c
∫ b
af (x, y) dxdy.
Example 12.3.Find the volume of the solid bounded above by the graph off (x, y) = x3y andbelow by the rectangleR= [0,1] × [0,3].
Solution.We can express this volume by the double integral∫∫
Rx3y dA.
By Fubini’s theorem and by Example 12.2,∫∫
Rx3y dA=
∫ 1
0
∫ 3
0x3y dydx=
∫ 3
0
∫ 1
0x3y dxdy=
98,
so the volume of the solid is 9/8. �
77
Example 12.4.Evaluate∫∫
Rey√
x+ ey dA, where R= [0,4] × [0,1].
Solution.By Fubini’s theorem,∫∫
Rey√
x+ ey dA=∫ 4
0
∫ 1
0ey√
x+ ey dydx.
By the substitutionu = x+ ey, du= ey dy, the inner integral is∫ 1
0ey√
x+ ey dy=∫ e+x
1+xu1/2du=
[
u3/2
3/2
]e+x
1+x
=23
(
(x+ e)3/2 − (x+ 1)3/2)
.
Hence,∫∫
Rey√
x+ ey dA=23
∫ 4
0
(
(x+ e)3/2 − (x+ 1)3/2)
dx
=23
25
[
(x+ e)5/2 − (x+ 1)5/2]4
0
=415
(
(e+ 4)5/2 − e5/2 − 55/2 + 1)
≈ 13.308. �
Second solution.Let us try to see how things change, if we use the other iterated integral. We have∫ 4
0
√x+ ey dx=
∫ 4+ey
ey
√u du=
[
u3/2
3/2
]4+ey
ey
=23
(
(4+ ey)3/2 − e3y/2)
.
Hence, Fubini’s theorem yields∫∫
Rey√
x+ ey dA=∫ 1
0
∫ 4
0ey√
x+ ey dxdy=∫ 1
0ey
[∫ 4
0
√x+ ey dx
]
dy
=23
∫ 1
0ey(
(4+ ey)3/2 − e3y/2)
dy=23
∫ e
1
(
(4+ u)3/2 − u3/2)
du
=23
25
[
(4+ u)5/2 − u5/2]e
1=
415
(
(4+ e)5/2 − e5/2 − 55/2 + 1)
. �
Exercises
1. Use a Riemann sum withn = 3 to estimate the value of∫∫
R
(
xy+ y2)
dA, whereR= [1,2]× [−1,1]. Take thesample points to be:
(a) the lower right corners of the subrectangles (b) the centers of the subrectangles.
2. Evaluate the given double integral by identifying it as the volume of a solid and calculating that volume usinggeometric means:
(a)∫∫
R2dA, R= [−1,2] × [0,3] (b)
∫∫
R(3− x) dA, R= [0,1] × [0,1]
3. Evaluate the given iterated integral:
(a)∫ 2
1
∫ 3
1
(
1+ 2xy2)
dxdy (b)∫ 4
1
∫ 3
1
(
xy+ y ln x
)
dydx (c)∫ 1
0
∫ 1
03x+2y dydx
78
4. Evaluate the given double integral:
(a)∫∫
R
(
xy+ y2)
dA, R= [1,2] × [−1,1]
(c)∫∫
Rsin(3x+ 2y) dA, R= [−π,0] × [− 1
2π,14π]
(e)∫∫
R(x+ y)e2xy dA, R= [1,2] × [0,1]
(b)∫∫
R
xyx2 + y2 + 1
dA, R= [0,1] × [0,1]
(d)∫∫
R
2+ y1+ x2
dA, R= [0,1] × [−2,3]
(f)∫∫
Rxycos
(
x2+y2)
dA, R=[
0,√π]
×[
0,√π]
79
13. Double integrals over general regions
To define the double integral of a bounded functionf (x, y) on a bounded regionD in the plane,we use that the definition given in§12.1 applies to any piecewise continuous functionF(x, y)defined on a rectangle (recall Remark 2 in§12.1).
13.1. Definition. Let D be the closed region inR2 bounded by a piecewise smooth curveγ and letf (x, y) be piecewise continuous insideD. Define the function
F(x, y) =
{
f (x, y) if ( x, y) ∈ D,
0 otherwise.
Then thedouble integral of f over Dis defined by∫∫
Df (x, y) dA=
∫∫
RF(x, y) dA,
whereR is any rectangle containingD.
Remark. Note that with the assumption thatf (x, y) is piecewise continuous inD, F(x, y) is con-tinuous everywhere except possibly at the points ofγ and the (possible) points insideD wheref (x, y) is discontinuous. Thus, iff (x, y) is actually continuous inD, the only discontinuities ofF(x, y) lie onγ. If f (x, y) is a genuine piecewise continuous function, with discontinuities on somesmooth curvesγ1, . . . , γn insideD, then the discontinuities ofF(x, y) lie on the curvesγ, γ1, . . . , γn.In either case,F(x, y) is piecewise continuous and
∫∫
R F(x, y) dA is defined by (12.1).
13.2. Properties of the double integral. We now state several properties of the double integralwhich are two-dimensional versions of the properties of thedefinite integral listed in Appendix B.
Theorem 13.1.Let f(x, y),g(x, y) be piecewise continuous functions and a,b denote constants.Then
1.∫∫
D[a f(x, y) + bg(x, y)] dA= a
∫∫
Df (x, y) dA+ b
∫∫
Dg(x, y) dA.
2. If f (x, y) ≤ g(x, y), then∫∫
Df (x, y) dA≤
∫∫
Dg(x, y) dA.
3. If D = D1 ∪ D2, D1 ∩ D2 = ∅, then∫∫
Df (x, y) dA=
∫∫
D1
f (x, y) dA+∫∫
D2
f (x, y) dA.
4.∫∫
D1dA= |D|, |D| being the area of D.
Furthermore, similar to the double integral over a rectangle, the double integral over a generalregionD can be interpreted as the volume of a solid or the average value of the function. Whenf (x, y) ≥ 0,
∫∫
Df (x, y) dA= vol.
{
(x, y, z) : (x, y) ∈ D, 0 ≤ z≤ f (x, y)}
.
80
Also, for any functionf (x, y) that is piecewise continuous onD, we have
average(x,y)∈D
f (x, y) =1|D|
∫∫
Df (x, y) dA. (13.1)
13.3. Evaluation of double integrals. We now state two formulas which follow from Fubini’stheorem and which we shall use to evaluate double integrals.First, consider a regionD of the form
D ={
(x, y) : a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}
,
whereg1 and g2 are continuous functions ofx. We call such regionstype I regions. If f is acontinuous function onD, we have
∫∫
Df (x, y) dA=
∫ b
a
[∫ g2(x)
g1(x)f (x, y) dy
]
dx. (13.2)
We also definetype II regions, which a regions of the form
D ={
(x, y) : c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)}
,
whereh1 andh2 are continuous functions ofy. If f is a continuous function on such a region, wehave
∫∫
Df (x, y) dA=
∫ d
c
[∫ h2(y)
h1(y)f (x, y) dx
]
dy. (13.3)
Example 13.1.Evaluate∫∫
Dyex dA,
whereD ={
(x, y) : 0 ≤ x ≤ 1,√
x ≤ y ≤ 1}
.
Solution.SinceD is a type I region, (13.2) yields∫∫
Dyex dA=
∫ 1
0
∫ 1
√xyex dydx=
∫ 1
0
[
y2ex
2
]1
√x
dx=12
∫ 1
0(1− x)ex dx
=12
(
[
(1− x)ex]1
0+
∫ 1
0ex dx
)
=12
(
− 1+[
ex]1
0
)
=e− 2
2. �
Second solution.Alternatively, we can try to describeD as a type II region and then use (13.3).We start by sketchingD: see Figure 13.1. In our type I description ofD, we first letx vary in therange 0≤ x ≤ 1, and then for each fixedx, we restricty to the range
√x ≤ y ≤ 1. To obtain a
type II description, we interchange the roles ofx andy: first, we lety vary in 0≤ y ≤ 1, and thenfor each fixedy, we restrictx to the range 0≤ x ≤ y2. Here, the limity2 comes from the descriptionof the curvey =
√x in the formx = x(y):
y =√
x ⇐⇒ x = y2.
Thus,
D ={
(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y2}
,81
(0,0)
(0,1) (1,1)
y =√
xD
Figure 13.1. The regionD
and (13.3) yields∫∫
Df (x, y) dA=
∫ 1
0
∫ y2
0yex dxdy=
∫ 1
0y[
ex]y2
0dy=
∫ 1
0y(
ey2 − 1)
dy
=
∫ 1
0
(
eu − 1) (
12du
)
=
[
eu − u2
]1
0
=e− 2
2. �
Example 13.2.Evaluate the double integral∫∫
Dey2
dA,
whereD ={
(x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1}
.
Solution.Our initial impulse is to write∫∫
Dey2
dA=∫ 1
0
∫ 1
xey2
dydx,
but this approach is doomed, because the indefinite integral∫
ey2dy is not solvable in elementary
functions. On the other hand, if we observe that
D ={
(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y}
,
we have∫∫
Dey2
dA=∫ 1
0
∫ y
0ey2
dxdy=∫ 1
0ey2[
x]y
0dy=
∫ 1
0ey2
y dy
=
∫ 1
0eu(
12du
)
=
[
eu
2
]1
0
=e− 1
2. �
Example 13.3.Evaluate the double integral∫∫
D2xy dA,
whereD is the triangle with vertices (0,0), (0,3), and (1,2).
Solution.The sides of the triangleD are the lines in the plane determined by the pairs of points(0,0) and (0,3), (0,0) and (1,2), and (0,3) and (1,2). These are the lines with equations
x = 0, y = 2x, and y = 3− x,
respectively (see Figure 13.2). Hence, we can expressD as a type I region:
D ={
(x, y) : 0 ≤ x ≤ 1, 2x ≤ y ≤ 3− x}
.82
(0,0)
(0,3)
(1,2)
y = 2x
y = 3− x
D
Figure 13.2. The triangleD
Thus, the given double integral is∫∫
D2xy dA=
∫ 1
0
∫ 3−x
2x2xy dydx=
∫ 1
0x[
(3− x)2 − 4x2]
dx
=
∫ 1
0
(
9x− 6x2 − 3x3)
dx=
[
9x2
2− 2x3 − 3x4
4
]1
0
=74. �
Example 13.4.Find the volume of the solidE in the first octant that is bounded by the cylindery2 + z2 = 4 and by the planex = 2y.
Solution.We start by sketching the solidE: Figure 13.3 shows a joint display of the parts of thecylindery2 + z2 = 4 and the planex = 2y that lie in the first octant and how they intersect and awireframe of the solid. From these displays, we see thatE is bounded from below by a triangle
Figure 13.3. The solidE
D in the xy-plane and from above by the part of the cylinder that lies above D. The triangle is theregion in thexy-plane bounded by they-axis, by the linex = 2y, and by the liney = 2, along whichthe cylinder intersects thaxy-plane. Thus, we see from Figure 13.4 that
D ={
(x, y) : 0 ≤ x ≤ 4, 12x ≤ y ≤ 2
}
={
(x, y) : 0 ≤ y ≤ 2, 0 ≤ x ≤ 2y}
.
(0,0)
(0,2) (4,2)
x = 2yD
Figure 13.4. The triangleD83
Since the surface of the cylinder is the graph ofz=√
4− y2, it follows that
E ={
(x, y, z) : (x, y) ∈ D, 0 ≤ z≤√
4− y2}
.
Hence, by Remark 3 and Fubini’s theorem,
vol.(E) =∫∫
D
√
4− y2 dA=∫ 4
0
∫ 2
x/2
√
4− y2 dydx=∫ 2
0
∫ 2y
0
√
4− y2 dxdy.
We shall use the latter iterated integral, because its evaluation is less technical:
vol.(E) =∫ 2
0
∫ 2y
0
√
4− y2 dxdy=∫ 2
02y
√
4− y2 dy=∫ 0
4
√u (−du) =
[
u3/2
3/2
]4
0
=163. �
Example 13.5.Reverse the order of integration in the iterated integral∫ 1
0
∫ arctanx
0f (x, y) dydx.
First solution. We have∫ 1
0
∫ arctanx
0f (x, y) dydx=
∫∫
Df (x, y) dA,
whereD ={
(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ arctanx}
is the region displayed on Figure 13.5.
(0,0) (1,0)
(1, π/4)y = arctanx
D
Figure 13.5. The regionD
Sincey = arctanx ⇐⇒ x = tany,
a reversal of the roles ofx andy yields a representation ofD as a type II region:
D ={
(x, y) : 0 ≤ y ≤ π/4, tany ≤ x ≤ 1}
.
Thus,∫∫
Df (x, y) dA=
∫ π/4
0
∫ 1
tanyf (x, y) dxdy. �
Second solution.We now demonstrate an alternative way for changingD into a type II set. Initially,D is described by the inequalities
0 ≤ x ≤ 1, 0 ≤ y ≤ arctanx.
Since arctanx ≤ π/4 for all x, 0 ≤ x ≤ 1, we can add to the above inequalities the redundantinequalityy ≤ π/4. Thus, we now haveD described by
0 ≤ x ≤ 1, 0 ≤ y ≤ arctanx, y ≤ π/4,or equivalently, by
0 ≤ x ≤ 1, 0 ≤ y ≤ π/4, y ≤ arctanx.84
We observe that
y ≤ arctanx ⇐⇒ tany ≤ x ⇐⇒ x ≥ tanx,
so we can now replace the definition ofD by
0 ≤ x ≤ 1, 0 ≤ y ≤ π/4, x ≥ tany.
Since tany ≥ 0 for all y in the range 0≤ y < π/2, we find that the conditionx ≥ 0 is superfluousand can be dropped. This yields
D ={
(x, y) : 0 ≤ y ≤ π/4, tany ≤ x ≤ 1}
,
whence∫ 1
0
∫ arctanx
0f (x, y) dydx=
∫ π/4
0
∫ 1
tanyf (x, y) dxdy. �
Example 13.6.Reverse the order of integration in the iterated integral∫ 2
0
∫ 3−x
x/2f (x, y) dydx.
First solution. We have∫ 2
0
∫ 3−x
x/2f (x, y) dydx=
∫∫
Df (x, y) dA,
whereD ={
(x, y) : 0 ≤ x ≤ 2, x/2 ≤ y ≤ 3 − x}
is the region displayed on Figure 13.6. We
(0,0)
(0,3)
(2,1)
y = 3− x↔ x = 3− y
y = x/2↔ x = 2y
D
Figure 13.6. The regionD
can easily convert the equations of the lines to the formx = x(y) (see the figure). However, thistime we have to be careful, because the functionx = x(y) that boundsD from the right is neitherx = 3− y nor x = 2y. The precise representation ofD as a type II region is
D ={
(x, y) : 0 ≤ y ≤ 3, 0 ≤ x ≤ g(y)}
,
where
g(y) =
{
2y if 0 ≤ y ≤ 1,
3− y if 1 ≤ y ≤ 3.
However, this function is not particularly convenient to work with. To obtain a “friendlier” answer,we split D in two using the horizontal liney = 1. Then each of the two subregions we get isbounded from the right by a single line, say:
D1 ={
(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y}
, D2 ={
(x, y) : 1 ≤ y ≤ 3, 0 ≤ x ≤ 3− y}
.85
Thus,∫ 2
0
∫ 3−x
x/2f (x, y) dydx=
∫ 1
0
∫ 2y
0f (x, y) dxdy+
∫ 3
1
∫ 3−y
0f (x, y) dxdy. �
Second solution.Initially, D is described by the inequalities
0 ≤ x ≤ 2, x/2 ≤ y ≤ 3− x.
On noting that 3− x ≤ 3 andx/2 ≥ 0, we can add to the above inequalities the redundant condition0 ≤ y ≤ 3, that is, we haveD described by
0 ≤ x ≤ 2, 0 ≤ y ≤ 3, x/2 ≤ y ≤ 3− x.
Next, we observe that
x/2 ≤ y ⇐⇒ x ≤ 2y, y ≤ 3− x ⇐⇒ x ≤ 3− y,
so we can further transform the definition ofD to
0 ≤ y ≤ 3, 0 ≤ x ≤ 2, x ≤ 2y, x ≤ 3− y,
or equivalently to0 ≤ y ≤ 3, 0 ≤ x ≤ min(2,2y,3− y).
This is the equivalent of the description ofD usingg(y) in our first solution. By comparing 2, 2y,and 3− y, we find that the minimum equals 2y or 3− y according asy ≤ 1 or y ≥ 1. Therefore,Dis the union of the regionD1 described by
0 ≤ y ≤ 1, 0 ≤ x ≤ 2y
and the regionD2 described by
1 ≤ y ≤ 3, 0 ≤ x ≤ 3− y.
Thus,∫ 2
0
∫ 3−x
x/2f (x, y) dydx=
∫ 1
0
∫ 2y
0f (x, y) dxdy+
∫ 3
1
∫ 3−y
0f (x, y) dxdy. �
Exercises
1. Evaluate the given iterated integral:
(a)∫ 2
0
∫ 4−y
y
(
y2 − 3x)
dxdy (b)∫ 2
1
∫ 2x
0
3√4− x2 dydx (c)
∫ 1
0
∫ y
0ycos(x− y) dxdy
2. Evaluate the given double integral:
(a)∫∫
Dyx2 dA, D =
{
(x, y) : 1 ≤ y ≤ 3, y ≤ x ≤ y2}
(b)∫∫
D3xsin 2y dA, D =
{
(x, y) : 0 ≤ x ≤ π, x/2 ≤ y ≤ x}
(c)∫∫
D
2yx2 + 1
dA, D ={
(x, y) : 0 ≤ x ≤ 1,0 ≤ y ≤√
x}
(d)∫∫
Dex/y dA, D =
{
(x, y) : 0 ≤ y ≤ 1, y2 ≤ x ≤ y}
(e)∫∫
Dxsiny dA, D =
{
(x, y) : 0 ≤ x ≤ 1,0 ≤ y ≤ x2}
86
(f)∫∫
D
(
x2 + y3)
dA, D is the region bounded byy = 2x andy = x2
(g)∫∫
D(3x− 4y) dA, D is the triangle with vertices (1,−1), (3,1), and (1,2)
(h)∫∫
Dxy2 dA, D is the right half of the diskx2 + y2 ≤ 4
3. Sketch the region of integration and reverse the order of integration:
(a)∫ 1
0
∫ 1
√x
f (x, y) dydx
(d)∫ 2
0
∫ 4−y
2f (x, y) dxdy
(b)∫ 2
0
∫
√4−x2
−√
4−x2f (x, y) dydx
(e)∫ 2
1
∫ ln y
0f (x, y) dxdy
(c)∫ 2
0
∫ 4−x
x2/2f (x, y) dydx
(f)∫ π/2
0
∫ 1
sinxf (x, y) dydx
4. Evaluate the given integral by reversing the order of integration:
(a)∫ 1
0
∫ 1
y2ye−x2
dxdy (b)∫ 8
0
∫ 2
3√y
dxdyx4 + 1
(c)∫ 4
0
∫ 2
√xey3
dydx
5. Find the volume of the solid lying below the hyperbolic paraboloidz = 4+ x2 − y2 and above the rectangleR= [0,1] × [0,2].
6. Find the volume of the solid in the first octant bounded by the cylinderx2 + z2 = 1, the planey = 3, and thecoordinate planes.
7. Find the volume of the solid lying below the plane 4x − y + 2z = 6 and above the region bounded by theparabolax = y2 − 1 and the right half of the circlex2 + y2 = 1.
8. Find the volume of the solid lying below the parabolic cylinderz = y2 and above the region bounded by theparabolay = x2 andy = x+ 2.
9. Find the volume of the solid bounded by the cylindersx2 + y2 = 1 andx2 + z2 = 1.
10. Evaluate∫∫
D
√
1− x2 − y2 dA, whereD is the unit diskx2+ y2 ≤ 1, by identifying it as the volume of a solidand calculating that volume using geometric means.
11. Evaluate∫∫
D
(
x8 sinx+ y9 + 1)
dA, whereD is the diskx2 + y2 ≤ 4.
87
14. Double integrals in polar coordinates
14.1. The formula.
Theorem 14.1.Let f(x, y) be piecewise continuous in a region D in the (Cartesian) xy-plane andlet R be the same region in polar coordinates, that is,
R={
(r, θ) : (r cosθ, r sinθ) ∈ D}
.
Then∫∫
Df (x, y) dA=
∫∫
Rf (r cosθ, r sinθ)r dA. (14.1)
The formal proof of (14.1) lies beyond the scope of these notes, but we should give at leastsome motivation for this formula. Leta,b,α, β be non-negative numbers such thata < b and0 ≤ α < β ≤ 2π. Consider the case whenD is the region bounded by the circlesx2 + y2 = a2
andx2 + y2 = b2 and by the half-lines through the origin forming anglesα andβ with the positivex-axis (see Figure ??). Note thatD contains the points (x, y) in the plane whose polar coordinates(r, θ) satisfy the inequalitiesa ≤ r ≤ b andα ≤ θ ≤ β. We recall that the area ofD is given by theformula
area(D) = 12(β − α)
(
b2 − a2)
= (β − α)(b− a)
(
b+ a2
)
. (14.2)
For each integern ≥ 1, let
r0 = a, r1 = a+ h, r2 = a+ 2h, . . . , rn = a+ nh= b, h = (b− a)/n,
be the points that divide the interval [a,b] into n subintervals of equal lengths; and let
θ0 = α, θ1 = α + δ, θ2 = α + 2δ, . . . , θn = α + nδ = β, δ = (β − α)/n,be the points that divide the interval [α, β] into n subintervals of equal lengths. We now subdividethe regionD into n2 subregions of similar shapes bounded by the circles of radiia, r1, r2, . . . , rn−1,bcentered at the origin and the half-lines through the originforming anglesα, θ1, θ2, . . . , θn−1, βwiththe positivex-axis (see Figure ??). LetDi j denote the subregion bounded by the circles of radiir i−1
andr i and by the half-lines at anglesθ j−1 andθ j with the positivex-axis: in polar coordinates,Di j
is given by the inequalitiesr i−1 ≤ r ≤ r i , θ j−1 ≤ θ ≤ θ j .
We now pick a sample point (x∗i j , y∗i j ) in each subregionDi j : we choose (x∗i j , y
∗i j ) to be the point in
θ = α
θ = β
θ = θ j
r = a
r = r i
r = b
(x∗i j , y∗i j )
α
θ j
β
a r i br
θ
(r∗i , θ∗j )
88
the plane whose polar coordinates arer∗i = (r i−1+r i)/2 andθ∗j = (θ j−1+θ j)/2. Using the subregionsDi j and the sample points (x∗i j , y
∗i j ), we form the sum
n∑
i=1
n∑
j=1
f (x∗i j , y∗i j )|Di j |, |Di j | = area(Di j ).
Note that this sum bears a strong resemblance to a Riemann sum for f (x, y). Indeed, it can beproved that iff (x, y) is continuous, then
∫∫
Df (x, y) dA= lim
n→∞
n∑
i=1
n∑
j=1
f (x∗i j , y∗i j )|Di j |. (14.3)
We now express the double sum in (14.3) in terms of ther i ’s and theθ j ’s. By (14.2) witha = r i−1,b = r i, α = θ j−1 andβ = θ j, we have
|Di j | = (θ j − θ j−1)(r i − r i−1)( r i + r i−1
2
)
= δhr∗i .
Furthermore, since the polar coordinates of (x∗i j , y∗i j ) are (r∗i , θ
∗j ), we have
x∗i j = r∗i cosθ∗j , y∗i j = r∗i sinθ∗j ,
whencef (x∗i j , y∗i j ) = f (r∗i cosθ∗j , r
∗i sinθ∗j ). Therefore,
n∑
i=1
n∑
j=1
f (x∗i j , y∗i j )|Di j | =
n∑
i=1
n∑
j=1
f (r∗i cosθ∗j , r∗i sinθ∗j )r
∗i δh.
Substituting this expression into the right side of (14.3),we obtain∫∫
Df (x, y) dA= lim
n→∞
n∑
i=1
n∑
j=1
f (r∗i cosθ∗j , r∗i sinθ∗j )r
∗i δh. (14.4)
The double sum on the right side of (14.4) is a Riemann sum for the function f (r cosθ, r sinθ)r onthe rectangleR= [a,b] × [α, β] when the sample points (r∗i , θ
∗j ) are always chosen at the centers of
the subrectangles (see Figure ??). Hence,
limn→∞
n∑
i=1
n∑
j=1
f (r∗i cosθ∗j , r∗i sinθ∗j )r
∗i δh =
∫∫
Rf (r cosθ, r sinθ)r dA.
The latter equality and (14.4) justify the (14.1) in the special case whenD is a sector of the abovekind.
14.2. Examples.
Example 14.1.Evaluate the integral∫∫
D(x+ y) dA,
whereD is the region lying between the circlesx2 + y2 = 1 andx2 + y2 = 4 and to the left of they-axis.
89
Solution.The polar form ofD is
R= {(r, θ) : 1 ≤ r ≤ 2, π/2 ≤ θ ≤ 3π/2}.Hence,
∫∫
D(x+ y) dA=
∫∫
R(r cosθ + r sinθ)r dA =
∫ 2
1
[∫ 3π/2
π/2
(
r2 cosθ + r2 sinθ) dθ
]
dr
=
∫ 2
1r2[
sinθ − cosθ]3π/2
π/2dr = −2
∫ 2
1r2dr = −14
3. �
Example 14.2.Find the area enclosed by the curver = 4+ 3 cosθ.
Solution.Since 4+ 3 cosθ ≥ 1 for all θ, the given curve “loops” around the origin, enclosing theregion (see Figure 14.1)
R={
(r, θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4+ 3 cosθ}
.
b
OR
Figure 14.1. The region enclosed byr = 4+ 3 cosθ
If D is the regionR in Cartesian coordinates, then we know from Property 4 in Theorem 13.1 thatits area is
|D| =∫∫
D1dA.
Hence, by (14.1),
|D| =∫∫
D1dA=
∫∫
Rr dA =
∫ 2π
0
[∫ 4+3 cosθ
0r dr
]
dθ
=12
∫ 2π
0(4+ 3 cosθ)2 dθ =
12
∫ 2π
0
(
16+ 24 cosθ + 9 cos2 θ)
dθ
=
∫ 2π
0
[
8+ 12 cosθ +94
(1+ cos 2θ)
]
dθ =
[
41θ4+ 12 sinθ +
98
sin 2θ
]2π
0
=41π2. �
Example 14.3.Find the volume of the solid that lies inside the spherex2+y2+z2 = 16 and outsidethe cylinderx2 + y2 = 4.
Solution.The given solid is symmetric with respect to thexy-plane, so it is twice as large as thesolid E that lies inside the sphere, outside the cylinder, and abovethe xy-plane (see Figure 14.2).This solid is the part of the space between the upper hemisphere and the annular regionD in thexy-plane bound by the sphere and the cylinder:
D ={
(x, y) : 4 ≤ x2 + y2 ≤ 16}
.90
Since the upper hemisphere ofx2 + y2 + z2 = 16 can be described as the graph of the functionz=
√
16− x2 − y2, we can expressE as
E ={
(x, y, z) : (x, y) ∈ D, 0 ≤ z≤√
16− x2 − y2}
.
Hence,
vol.(E) =∫∫
D
√
16− x2 − y2 dA.
The polar form ofD isR=
{
(r, θ) : 2 ≤ r ≤ 4, 0 ≤ θ ≤ 2π}
,
and√
16− (r cosθ)2 − (r sinθ)2 =√
16− r2(cos2 θ + sin2 θ) =√
16− r2.
Hence, (14.1) yields
vol.(E) =∫∫
D
√
16− x2 − y2 dA=∫∫
Rr√
16− r2 dA
=
∫ 4
2
[∫ 2π
0r√
16− r2 dθ
]
dr = 2π∫ 4
2r√
16− r2 dr.
We evaluate the last integral using the substitutionu = 16− r2. Then
du= −2r dr, u(2) = 12, u(4) = 0,
so
vol.(E) = 2π∫ 0
12u1/2
(
− 12du
)
= π
∫ 12
0u1/2 du=
[
2π3
u3/2
]12
0
= 16√
3π.
Thus, the volume of the original solid is 32√
3π. �
Exercises
1. Evaluate the given integral by changing to polar coordinates:
(a)∫∫
D(x+ 3y) dA, D is the diskx2 + y2 ≤ 4
(b)∫∫
Dex2+y2
dA, D is the right half of the diskx2 + y2 ≤ 1
(c)∫∫
D
√
16− x2 − y2 dA, D is the part of the annulus 1≤ x2 + y2 ≤ 9 lying in the first quadrant
Figure 14.2. The solidE91
(d)∫∫
D
xy4+ x2 + y2
dA, D is the part of the diskx2 + y2 ≤ 9 lying below the liney = x and above the
x-axis
2. Find the volume of the solid enclosed by the paraboloidz= 4− x2 − y2 and thexy-plane.
3. Find the volume of the solid enclosed by the conez2 = x2 + y2 and the planez= 3.
4. Find the volume of the solid lying inside both the cylindery2 + z2 = 4 and the spherex2 + y2 + z2 = 9.
5. Find the volume lying above the paraboloidz= 12
(
x2 + y2)
and inside the ellipsoidx2 + y2 + 4z2 = 2.
6. Find the volume lying below the hemispherez=√
9− x2 − y2 and above the regionD in thexy-plane wherex2 + y2 ≤ 2x.
7. Find the volume of the solid lying inside all three cylinders x2 + y2 = 1, x2 + z2 = 1, andy2 + z2 = 1.
92
15. Triple integrals
15.1. Definition. Let f (x, y, z) be a bounded function on the rectangular boxB,
[a1,b1] × [a2,b2] × [a3,b3] ={
(x, y, z) : a1 ≤ x ≤ b1, a2 ≤ y ≤ b2, a3 ≤ z≤ b3
}
.
For eachn ≥ 1, let
x0 = a1, x1 = a1 + ∆x, x2 = a1 + 2∆x, . . . , xn = a1 + n∆x = b1, ∆x = (b1 − a1)/n,
be the points that divide the interval [a1,b1] into n subintervals of equal lengths; let
y0 = a2, y1 = a2 + ∆y, y2 = a2 + 2∆y, . . . , yn = a2 + n∆y = b2, ∆y = (b2 − a2)/n,
be the points that divide the interval [a2,b2] into n subintervals of equal lengths; and let
z0 = a3, z1 = a3 + ∆z, z2 = a3 + 2∆z, . . . , zn = a3 + n∆z= b3, ∆z= (b3 − a3)/n,
be the points that divide the interval [a3,b3] into n subintervals of equal lengths. Also, pickn3
sample points (x∗i , y∗j , z∗k) so that
xi−1 ≤ x∗i ≤ xi , yj−1 ≤ y∗j ≤ yj , zk−1 ≤ z∗k ≤ zk (1 ≤ i, j, k ≤ n).
Then thetriple integral of f over Bis defined by∫∫∫
Bf (x, y, z) dV = lim
n→∞
n∑
i=1
n∑
j=1
n∑
k=1
f (x∗i , y∗j , z∗k)∆V, ∆V = ∆x∆y∆z, (15.1)
provided that the limit exists. The sums on the right side of (15.1) are calledRiemann sums of fon B.
More generally, ifE is a region inR3 bounded by a piecewise smooth surfaceΣ and f is apiecewise continuous function onE, we define thetriple integral of f over Eas
∫∫∫
Ef (x, y, z) dV =
∫∫∫
BF(x, y, z) dV,
whereB is any rectangular box containingE andF is the function
F(x, y, z) =
{
f (x, y, z) if ( x, y, z) ∈ E,
0 otherwise.
Remark. As in the cases of definite and double integrals, the above definitions define the tripleintegral for all piecewise continuous functions over solids bounded by piecewise smooth surfaces.Furthermore, similarly to the definite and double integrals, the triple integral of a function is relatedto its average in the region of integration:
average(x,y,z)∈E
f (x, y, z) =1
vol.(E)
∫∫∫
Ef (x, y, z) dV. (15.2)
15.2. Properties of the triple integral. The following theorem lists the three-dimensional ver-sions of the properties of double integrals from Theorem 13.1.
Theorem 15.1.Let f(x, y, z),g(x, y, z) be piecewise continuous functions and a,b denote constants.Then
1.∫∫∫
E[a f(x, y, z) + bg(x, y, z)] dV = a
∫∫∫
Ef (x, y, z) dV+ b
∫∫∫
Eg(x, y, z) dV.
93
2. If f (x, y, z) ≤ g(x, y, z), then∫∫∫
Ef (x, y, z) dV ≤
∫∫∫
Eg(x, y, z) dV.
3. If E = E1 ∪ E2, E1 ∩ E2 = ∅, then∫∫∫
Ef (x, y, z) dV =
∫∫∫
E1
f (x, y, z) dV+∫∫∫
E2
f (x, y, z) dV.
4.∫∫∫
E1dV = vol.(E).
There is also a three-dimensional version of Fubini’s theorem, which we shall use to evaluatetriple integrals. Before stating it, we need to introduce some terminology. We say that a regionEin R3 is of type I, if it can be written in the form
E ={
(x, y, z) : (x, y) ∈ D, u1(x, y) ≤ z≤ u2(x, y)}
for some regionD in thexy-plane and some continuous functionsu1 andu2 on D. We say thatE isa type II region, if it can be written in the form
E ={
(x, y, z) : (y, z) ∈ D, u1(y, z) ≤ x ≤ u2(y, z)}
for some regionD in theyz-plane. We say thatE is atype III region, if it can be written in the form
E ={
(x, y, z) : (x, z) ∈ D, u1(x, z) ≤ y ≤ u2(x, z)}
for some regionD in thexz-plane.
Theorem 15.2.Let f(x, y, z) be piecewise continuous functions in a region E inR3. Then
i) if E is of type I, then∫∫∫
Ef (x, y, z) dV =
∫∫
D
[∫ u2(x,y)
u1(x,y)f (x, y, z) dz
]
dA; (15.3)
ii) if E is of type II, then∫∫∫
Ef (x, y, z) dV =
∫∫
D
[∫ u2(y,z)
u1(y,z)f (x, y, z) dx
]
dA; (15.4)
iii) if E is of type III, then∫∫∫
Ef (x, y, z) dV =
∫∫
D
[∫ u2(x,z)
u1(x,z)f (x, y, z) dy
]
dA. (15.5)
15.3. Evaluation of triple integrals.
Example 15.1.Evaluate the triple integral∫∫∫
Eyzcos(x5) dV,
whereE ={
(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x, x ≤ z≤ 2x}
.94
Solution.Let D ={
(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x}
. By (15.3),∫∫∫
Eyzcos(x5) dV =
∫∫
D
[∫ 2x
xyzcos(x5) dz
]
dA
=12
∫∫
Dycos(x5)
[
(2x)2 − x2]
dA=32
∫∫
Dyx2 cos(x5) dA
=32
∫ 1
0
[∫ x
0yx2 cos(x5) dy
]
dx=34
∫ 1
0x4 cos(x5) dx
=320
∫ 1
0cosu du=
320
[
sinu]1
0=
320
sin 1≈ 0.126. �
Example 15.2.Evaluate the triple integral∫∫∫
E
√
x2 + z2 dV,
whereE is the solid bounded by the paraboloidy = x2 + z2 and the planey = 4.
Solution.We can writeE as a type III region:
E ={
(x, y, z) : x2 + z2 ≤ 4, x2 + z2 ≤ y ≤ 4}
.
Hence, by (15.5),∫∫∫
E
√
x2 + z2 dV =∫∫
x2+z2≤4
[∫ 4
x2+z2
√
x2 + z2 dy
]
dA=∫∫
x2+z2≤4
(
4− x2 − z2)
√
x2 + z2 dA.
We can easily compute the last integral by changing to polar coordinates in thexz-plane:∫∫
x2+z2≤4
(
4− x2 − z2)
√
x2 + z2 dA=∫∫
0≤r≤2
(
4− r2)
r r dA =∫ 2
0
∫ 2π
0r2(
4− r2)
dθdr
= 2π∫ 2
0
(
4r2 − r4)
dr = 2π
[
4r3
3− r5
5
]2
0
=128π15. �
Example 15.3.Compute the volume of the solidT described by the inequalities
x ≥ 0, z≥ 0, y ≥ 2x, x+ 2y+ 2z≤ 2.
Solution. In fact,T is the tetrahedron with vertices (0,0,0), (0,1,0), (0,0,1), and (25,45,0), but we
shall ignore this fact and make a point of using triple integrals and inequalities. By solving theinequlity x+ 2y+ 2z≤ 2 for z, we can rewrite the definition ofT as
x ≥ 0, y ≥ 2x, 0 ≤ z≤ 1− x/2− y. (15.6)
Note that the implicit in the inequality forz is the condition
1− x/2− y ≥ 0 ⇐⇒ y ≤ 1− x/2.
Adding the latter inequality to those in (15.6), we get
x ≥ 0, 2x ≤ y ≤ 1− x/2, 0 ≤ z≤ 1− x/2− y95
We now note that the conditionx ≤ 25 is implicit in the inequalities fory. Making this condition
explicit, we finally haveT described as a type I set:
T ={
(x, y, z) : 0 ≤ x ≤ 25, 2x ≤ y ≤ 1− x/2, 0 ≤ z≤ 1− x/2− y
}
.
We can now compute the volume ofT:
vol.(E) =∫∫∫
T1dV =
∫ 2/5
0
∫ 1−x/2
2x
∫ 1−x/2−y
01dzdydx
=
∫ 2/5
0
∫ 1−x/2
2x(1− x/2− y) dydx=
∫ 2/5
0
[
(1− x/2)y− 12y2
]1−x/2
2xdx
=
∫ 2/5
0
(
(1− x/2)2 − 12(1− x/2)2 − 2x(1− x/2)+ 2x2
)
dx
=
∫ 2/5
0
(
12 −
52x+ 25
8 x2)
dx=∫ 1
0
(
12 − u+ 1
2u2) (
25du
)
=115. �
Exercises
1. Evaluate the given iterated integral:
(a)∫ 1
0
∫ 1
√x
∫ xy+1
x+y6xyz dzdydx (b)
∫ 2
0
∫ z
0
∫ yz
04e−z4
dxdydz
2. Evaluate the given triple integral:
(a)∫∫∫
E
(
2x2 + 3y+ 2z)
dV, E ={
(x, y, z) : 0 ≤ x ≤ 1,0 ≤ y ≤ x, y ≤ z≤ 2y}
(b)∫∫∫
Ee−y dV, E =
{
(x, y, z) : 1 ≤ z≤ 2,0 ≤ x ≤ 1/z,0 ≤ y ≤ xz}
(c)∫∫∫
Exy dV, E lies below the planez= x+ y+ 2 and above the right half of the diskx2 + y2 ≤ 4
(d)∫∫∫
E4xz dV, E is the tetrahedron bounded by the plane 3x+ 2y+ z= 6 and the coordinate planes
(e)∫∫∫
Excosz dV, E is the solid bounded by the cylindery = x2 and the planesz = 0, y = 0, z = y,
x = 1
(f)∫∫∫
Ey(
x2 + z2)
dV, E is the solid bounded by the cylinderx2 + z2 = 2 and the planesy = 1 andy = 3
3. Use a triple integral to find the volume of the solid boundedby the cylinder 2y = x2 and the planesz = −1,z= 1
2 x, andy = 2.
96
16. Triple integrals in cylindrical and spherical coordinates
16.1. Triple integrals in cylindrical coordinates.
Theorem 16.1.Let f(x, y, z) be piecewise continuous in a region E in (Cartesian) space and let Rbe the same region in cylindrical coordinates, that is,
R={
(r, θ, z) : (r cosθ, r sinθ, z) ∈ E}
.
Then∫∫∫
Ef (x, y, z) dV =
∫∫∫
Rf (r cosθ, r sinθ, z)r dV. (16.1)
The target applications of this theorem are triple integrals over type I regions
E ={
(x, y, z) : (x, y) ∈ D, u1(x, y) ≤ z≤ u2(x, y)}
,
whereD is a two-dimensional region with a simple polar form. In suchsituations, we can appealto (15.3) to get
∫∫∫
Ef (x, y, z) dV =
∫∫
D
[∫ u2(x,y)
u1(x,y)f (x, y, z) dz
]
dA.
Then, after evaluating the integral overz, we can try to evaluate the resulting double integral usingpolar coordinates in thexy-plane. What Theorem 16.1 does is change coordinates in thexy-planefrom Cartesian to polar before the integration overz has been performed.
Example 16.1.Find the volume of the solid that lies within the cylinderx2+y2 = 4 and the spherex2 + y2 + z2 = 9.
Solution.The given solid is symmetric with respect to thexy-plane, so its volume is twice as largeas that of the solidE that lies within the cylinder, below the sphere, and above the xy-plane (seeFigure 16.1). Since the upper hemisphere is the graph of the functionz =
√
9− x2 − y2, we canwrite E as a type I region:
E ={
(x, y, z) : x2 + y2 ≤ 4, 0 ≤ z≤√
9− x2 − y2}
.
Figure 16.1. The solidE
The cylindrical coordinates description of the solid is
R={
(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ z≤√
9− r2}
.97
Therefore, by (16.1),
vol.(E) =∫∫∫
E1dV =
∫∫∫
Rr dV =
∫∫
D
[∫
√9−r2
0r dz
]
dA=∫∫
Dr√
9− r2 dA,
whereD is the polar rectangle (that is, a disk)D = [0,2] × [0,2π]. Applying Fubini’s theorem, weget
∫∫
Dr√
9− r2 dA=∫ 2
0
∫ 2π
0r√
9− r2 dθdr =∫ 2
02πr√
9− r2 dr.
We now make a substitutionu = 9− r2. We have
du= −2r dr, u(0) = 9, u(2) = 5,
so the volume ofE is∫ 2
02πr√
9− r2 dr =∫ 5
9πu1/2(−du) =
[
πu3/2
3/2
]9
5
=2π(27− 5
√5)
3≈ 33.133.
Thus, the volume of the original solid is≈ 66.265. �
Example 16.2.Evaluate∫∫∫
B
√
x2 + y2 dV,
whereB is the ballx2 + y2 + z2 ≤ 4.
Solution. In cylindrical coordinates, the ball is
R={
(r, θ, z) : r2 + z2 ≤ 4}
={
(r, θ, z) : 0 ≤ r ≤ 2, −√
4− r2 ≤ z≤√
4− r2}
.
Thus, by (16.1) and (15.3),∫∫∫
B
√
x2 + y2 dV =∫∫∫
Rr r dV =
∫∫
D
[∫
√4−r2
−√
4−r2r2 dz
]
dA=∫∫
D2r2√
4− r2 dA
=
∫ 2
0
∫ 2π
02r2√
4− r2 dθdr = 2π∫ 2
02r2√
4− r2 dr.
We now substituter = 2 sint. Since√
4− r2 =
√
4− 4 sin2 t = 2 cost, dr = (2 cost)dt, t(0) = 0, t(2) = π/2,
we get
2π∫ 2
02r2√
4− r2 dr = 4π∫ π/2
0(2 sint)2(2 cost) (2 cost)dt
= 16π∫ π/2
0(2 sint cost)2 dt = 16π
∫ π/2
0(sin 2t)2 dt
= 8π∫ π/2
0(1− cos 4t) dt = 8π
[
t − 14 sin 4t
]π/2
0= 4π2. �
98
16.2. Triple integrals in spherical coordinates.
Theorem 16.2.Let f(x, y, z) be piecewise continuous in a region E in (Cartesian) space and let Rbe the same region in spherical coordinates, that is,
R={
(ρ, θ,φ) : (ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ) ∈ E}
.
Then∫∫∫
Ef (x, y, z) dV =
∫∫∫
Rf (ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ)ρ2 sinφdV. (16.2)
Example 16.3.Evaluate∫∫∫
B
√
x2 + y2 + z2e−x2−y2−z2dV,
whereB is the ballx2 + y2 + z2 ≤ a2.
Solution. In spherical coordinates,x2 + y2 + z2 = ρ2 and
B ={
(ρ, θ,φ) : ρ ≤ a}
= [0,a] × [0,2π] × [0, π],
so (16.2) yields
∫∫∫
B
√
x2 + y2 + z2e−x2−y2−z2dV =
∫ a
0
∫ 2π
0
∫ π
0ρe−ρ
2ρ2 sinφdφdθdρ
=
∫ a
0
∫ 2π
0ρ3e−ρ
2[ − cosφ]π
0dθdρ
=
∫ a
0
∫ 2π
02ρ3e−ρ
2dθdρ = 4π
∫ a
0ρ3e−ρ
2dρ.
We substituteu = ρ2 in the integral overρ and get
∫∫∫
B
√
x2 + y2 + z2e−x2−y2−z2dV = 2π
∫ a2
0ue−u du= 2π
∫ a2
0u d
(
− e−u)
=[
− 2πue−u]a2
0+ 2π
∫ a2
0e−u du= 2π
(
1− e−a2 − a2e−a2)
. �
Example 16.4.Evaluate∫∫∫
Ey2z dV,
whereE is the solid that lies between the spheresρ = 1 andρ = 2 and below the coneφ = π/4.
Solution. In spherical coordinates, the conditions describingE can be expresses as
1 ≤ ρ ≤ 2, π/4 ≤ φ ≤ π,99
so in spherical coordinatesE is the rectangular box [1,2] × [0,2π] × [π/4, π]. Then, by (16.2),∫∫∫
Ey2z dV=
∫ 2
1
∫ 2π
0
∫ π
π/4(ρ sinφ sinθ)2(ρ cosφ)ρ2 sinφdφdθdρ
=
∫ 2π
0
∫ π
π/4
[
sin3φ cosφ sin2 θ
∫ 2
1ρ5 dρ
]
dφdθ
=
∫ 2π
0
∫ π
π/4sin3φ cosφ sin2 θ
[
16ρ
6]2
1dφdθ
=212
∫ 2π
0
[
sin2 θ
∫ π
π/4sin3φ cosφdφ
]
dθ.
We substituteu = sinφ in the integral overφ and get
∫ π
π/4sin3φ cosφdφ =
∫ 0
1/√
2u3 du= −
[
u4
4
]1/√
2
0
= − 116.
Hence,∫∫∫
Ey2z dV= −21
32
∫ 2π
0sin2 θdθ = −21
64
∫ 2π
0(1− cos 2θ) dθ = −21π
32. �
Example 16.5.Find the volume of the solidE that lies inside the spherex2 + y2 + z2 = 4, abovethexy-plane, and below the conez=
√
x2 + y2.
Solution.We first derive the spherical description ofE. In Cartesian coordinates, the solid isdefined by the inequalities
x2 + y2 + z2 ≤ 4, 0 ≤ z≤√
x2 + y2.
In spherical coordinates, these become
ρ2 ≤ 4, 0 ≤ ρ cosφ ≤ ρ sinφ,
or equivalently,
0 ≤ ρ ≤ 2, 0 ≤ cosφ ≤ sinφ.
The conditions onφ restrict it to the rangeπ/4 ≤ φ ≤ π/2, so the spherical description ofE is[0,2] × [0,2π] × [π/4, π/2]. The solid is displayed at Figure 16.2. We now use Theorem 15.1 toexpress the volume ofE as a triple integral and then appeal to (16.2) to convert the triple integral
Figure 16.2. The solidE100
to spherical coordinates. We obtain
vol.(E) =∫∫∫
E1dV =
∫ π/2
π/4
∫ 2π
0
∫ 2
0ρ2 sinφdρdθdφ
=83
∫ π/2
π/4
∫ 2π
0sinφdθdφ =
16π3
∫ π/2
π/4sinφdφ
=16π3
[
− cosφ]π/2
π/4=
8√
2π3. �
Exercises
1. Evaluate the given integral by changing to cylindrical coordinates:
(a)∫∫∫
E
√
x2 + y2 dV, E is bounded by thexy-plane, the cylinderx2 + y2 = 1, and by the paraboloid
z= 4− x2 − y2
(b)∫∫∫
Ey dV, E lies to the right of thexz-plane and is bounded by the cylindersx2+y2 = 1 andx2+y2 = 9
and by the planesz= 1 andz= 4+ x
(c)∫∫∫
E1dV, E is the smaller solid bounded by the planez= 1 and the spherex2 + y2 + z2 = 2
2. Evaluate the given integral by changing to spherical coordinates:
(a)∫∫∫
E
√
x2 + y2 dV, E lies above thexy-plane and within the spherex2 + y2 + z2 = 4
(b)∫∫∫
Exz dV, E lies between the spheresx2 + y2 + z2 = a2 andx2 + y2 + z2 = 2a2 in the first octant
(c)∫∫∫
E1dV, E lies above the conez2 = x2 + y2 and below the spherex2 + y2 + (z− 2)2 = 4
(d)∫∫∫
E
(
xy+ z)
dV, E lies above the conez=√
x2 + y2 and below the spherex2 + y2 + z2 = 4
3. Evaluate the given iterated integral by changing to cylindrical or spherical coordinates:
(a)∫ 1
0
∫
√1−x2
0
∫ 2−x2−y2
x2+y2
(
x2 + y2)3/2
dzdydx
(c)∫ 2
0
∫
√4−x2
0
∫
√9−x2−y2
√4−x2−y2
xyz dzdydx
(b)∫ 2
0
∫
√4−y2
0
∫
√4−x2−y2
0z dzdxdy
4. Use a triple integral to find the volume of the given solid:
(a) the solid bounded by the paraboloidy = −x2 − z2 and the planey = −9
(b) the smaller solid bounded by the planez= 1 and the spherex2 + y2 + z2 = 2
(c) the solid above the coneφ = π/6 and below the sphereρ = 4 cosφ
101
17. Applications of double and triple integrals
We have already seen one application of the double and tripleintegrals: both can be used tocompute volumes of solids. In this lecture, we describe someother applications.
17.1. Surface area. Let Σ be a smooth parametric surface, given by
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D).
We can use a double integral to calculate the surface area ofΣ. However, we first shoulddefinewhat “surface area” is. For that, we draw inspiration from the definition of arc length in§5.6.
For simplicity, we shall assume thatD is a rectangle in theuv-plane, say: [a,b] × [c,d]. We cantreat the general case similarly, but there are more technical details to pay attention to and littleextra insight to gain. For eachn ≥ 1, leta = u0 < u1 < · · · < un = b be the points that divide theinterval [a,b] into n subintervals of equal lengths; and letc = v0 < v1 < · · · < vn = d be the pointsthat divide the interval [c,d] into n subintervals of equal lengths. The points
(un, v0) (un, v1) · · · (un, vj) · · · (un, vn),...
......
...(ui , v0) (ui , v1) · · · (ui , vj) · · · (ui , vn),...
......
...(u1, v0) (u1, v1) · · · (u1, vj) · · · (u1, vn),(u0, v0) (u0, v1) · · · (u0, vj) · · · (u0, vn),
define a partition of the rectangleD into n2 congruent subrectangles. The vector functionr mapseach of those small rectangles to a small “patch” onΣ and each grid point (ui , vj) to a pointPi j onthe surface. The patches form a partition ofΣ into n2 pieces, which we label asΣi j , 1 ≤ i, j ≤ n,so thatΣi j is the image of the rectangle that has (ui , vj) as its upper right corner point. We canapproximate the area ofΣi j by the area of the parallelogramΠi j three of whose vertices are thepointsPi−1, j, Pi, j−1, andPi j . Recall that the area of the parallelogramΠi j is given by
|Πi j | =∥
∥
−−−→Pi j Pi−1, j ×
−−−→Pi j Pi, j−1
∥
∥.
It follows that the total areaAn of all the parallelogramsΠi j is
An =
n∑
i=1
n∑
j=1
|Πi j | =n
∑
i=1
n∑
j=1
∥
∥
−−−→Pi j Pi−1, j ×
−−−→Pi j Pi, j−1
∥
∥.
We define thesurface areaof Σ as the limit of the sequence{An}∞n=1, if it exists. Using the abovedefinition and the properties of differentiable functions and double integrals, we can prove thefollowing formula for the surface area of a piecewise smoothsurface.
Theorem 17.1.LetΣ be a smooth parametric surface, given by
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D).
The surface area ofΣ is given by
area(Σ) =∫∫
D‖ru × r v‖dA. (17.1)
102
If Σ is the graph ofz= f (x, y), (x, y) ∈ D, then using the parametrization
r (x, y) = xe1 + ye2 + f (x, y)e3 ((x, y) ∈ D),
we can express the area ofΣ as
area(Σ) =∫∫
D
√
1+ f 2x + f 2
y dA. (17.2)
Example 17.1.Find the area of the part of the plane 2x+ 5y+ z= 10 that lies inside the cylinderx2 + y2 = 9.
Solution.We apply (17.2) withf (x, y) = 10− 2x− 5y andD the diskx2+ y2 ≤ 9. The desired areais then
A =∫∫
D
√
1+ (−2)2 + (−5)2 dA=∫∫
D
√30dA=
√30 · 9π,
where 9π is the area of the diskx2 + y2 ≤ 9. �
Example 17.2.Find the area of the surfaceΣ with parametrization
r (u, v) = cosu(4+ 2 cosv)e1 + sinu(4+ 2 cosv)e2 + 2 sinve3 (0 ≤ u, v ≤ 2π).
This surface is called atorusand resembles a donut; it is displayed on Figure 17.1
Figure 17.1. A torus
Solution.We have
ru(u, v) = − sinu(4+ 2 cosv)e1 + cosu(4+ 2 cosv)e2,
r v(u, v) = −2 cosusinve1 − 2 sinusinve2 + 2 cosve3,
so
ru × r v = 2 sin2 usinv(4+ 2 cosv)(e1 × e2) − 2 sinucosv(4+ 2 cosv)(e1 × e3)
− 2 cos2 usinv(4+ 2 cosv)(e2 × e1) + 2 cosucosv(4+ 2 cosv)(e2 × e3)
= 2 cosucosv(4+ 2 cosv)e1 + 2 sinucosv(4+ 2 cosv)e2 + 2 sinv(4+ 2 cosv)e3
= (8+ 4 cosv)(
cosucosve1 + sinucosve2 + sinve3
)
,
‖ru × r v‖ = |8+ 4 cosv|√
cos2 ucos2 v+ sin2 ucos2 v+ sin2 v = 8+ 4 cosv.
Thus, the surface area ofΣ is∫ 2π
0
∫ 2π
0(8+ 4 cosv) dudv= 2π
∫ 2π
0(8+ 4 cosv) dv= 32π2. �
103
Example 17.3.Express the area of the ellipsoidΣ with equationx2 + y2 + 4z2 = 1 in terms of thenumber
µ =
∫ 1
0
√1+ 3w2 dw= 1+
ln(2+√
3)
2√
3.
Solution.First, we have to parametrizeΣ. The unit spherex2 + y2 + z2 = 1 is the set of all points(x, y, z) whose spherical coordinateρ is 1:
x = cosusinv, y = sinusinv, z= cosv,
where 0≤ u ≤ 2π and 0 ≤ v ≤ π (u and v play the roles of the spherical coordinatesθ andφ). Since (x, y, z) lies onΣ exactly when (x, y,2z) lies on the unit sphere, we obtain the followingparametrization for the ellipsoid:
r (u, v) = cosusinve1 + sinusinve2 +12 cosve3,
where 0≤ u ≤ 2π and 0≤ v ≤ π. Then
ru(u, v) = − sinusinve1 + cosusinve2 + 0e3,
r v(u, v) = cosucosve1 + sinucosve2 − 12 sinve3;
ru × r v = − sin2 usinvcosv(e1 × e2) + 12 sinusin2 v(e1 × e3)
+ cos2 usinvcosv(e2 × e1) − 12 cosusin2 v(e2 × e3)
= −12 cosusin2 ve1 − 1
2 sinusin2 ve2 − sinvcosve3
= −12 sinv
(
cosusinve1 + sinusinve2 + 2 cosve3
)
;
‖ru × r v‖ = 12 | sinv|
√
cos2 usin2 v+ sin2 usin2 v+ 4 cos2 v
= 12 sinv
√
sin2 v+ 4 cos2 v = 12 sinv
√1+ 3 cos2 v.
Thus, the surface area ofΣ is
A =∫ π
0
∫ 2π
0
12 sinv
√1+ 3 cos2 v dudv= π
∫ π
0sinv√
1+ 3 cos2 v dv
= π
∫ −1
1
√1+ 3w2 (−dw) = 2π
∫ 1
0
√1+ 3w2 dw= 2πµ. �
Example 17.4. Find the volume and the surface area of the solid bounded by the paraboloidsz= 3x2 + 3y2 andz= 4− x2 − y2.
Solution.The given solid has the paraboloidz= 4−x2−y2 as its upper boundary and the paraboloidz= 3x2 + 3y2 as its lower boundary, so it can be described as
E ={
(x, y, z) : (x, y) ∈ D, 3x2 + 3y2 ≤ z≤ 4− x2 − y2}
,
whereD is the projection of the solid on thexy-plane (see Figure 17.2).Since the intersection curve of the two paraboloids is
γ ={
(x, y, z) : z= 3x2 + 3y2 = 4− x2 − y2}
={
(x, y, z) : x2 + y2 = 1, z= 3}
,
104
Figure 17.2. The solidE
we find thatD is the unit diskx2 + y2 ≤ 1. Hence, the volume ofE is
vol.(E) =∫∫∫
E1dV =
∫∫
D
[∫ 4−x2−y2
3x2+3y21dz
]
dA=∫∫
D
(
4− 4x2 − 4y2)
dA.
The surface areaA of the solid is the sum of the surface areas of the two paraboloids, each of whichcan be computed using (17.2):
A =∫∫
D
√
1+ (−2x)2 + (−2y)2 dA+∫∫
D
√
1+ (6x)2 + (6y)2 dA.
Let R = [0,1] × [0,2π] be the polar form of the diskD. Converting all three integrals to polarcoordinates, we obtain
vol.(E) =∫∫
D
(
4− 4(x2 + y2))
dA=∫∫
Rr(
4− 4r2)
dA
=
∫ 1
0
[∫ 2π
0
(
4r − 4r3)
dθ
]
dr = 2π∫ 1
0
(
4r − 4r3)
dr = 2π[
2r2 − r4]1
0= 2π,
A =∫∫
D
(
√
1+ 4(x2 + y2) +√
1+ 36(x2 + y2))
dA
=
∫∫
Rr(√
1+ 4r2 +√
1+ 36r2)
dA=∫ 1
0
[∫ 2π
0r(√
1+ 4r2 +√
1+ 36r2)
dθ
]
dr
= 2π∫ 1
0r(√
1+ 4r2 +√
1+ 36r2)
dr = 2π
[∫ 5
1u1/2
(
18du
)
+
∫ 37
1u1/2
(
172du
)
]
= 2π
([
u3/2
12
]5
1
+
[
u3/2
108
]37
1
)
=π
54
(
37√
37+ 45√
5− 10)
≈ 18.366. �
17.2. Density and mass.Double and triple integrals also represent various physical quantities. Inthis section, we discuss applications related to the computation of mass and the coordinates of thecenter of mass of laminas and solids. Suppose that a lamina occupies a regionD in the xy-planeand hasdensityδ(x, y) at a point (x, y) ∈ D. If δ(x, y) is the constant functionδ(x, y) = c, the totalmass of the lamina isc|D|, where|D| denotes the area ofD. Note that we can think of this numberalso as
∫∫
Dc dA=
∫∫
Dδ(x, y) dA.
105
Using the definition of the double integral as the limit of Riemann sums, we can prove that the latterformula applies to any continuous functionδ(x, y). That is, if the density functionδ is continuousin D, then the massmof the lamina is given by the formula
m=∫∫
Dδ(x, y) dA.
We can also give formulas for the coordinates of thecenter of massof the lamina: if (x, y) is thecenter of mass, then
x =1m
∫∫
Dxδ(x, y) dA, y =
1m
∫∫
Dyδ(x, y) dA.
Similarly, if a solid occupies a regionE in space and hasdensityδ(x, y, z) at a point (x, y, z) ∈ E,then the massmof the solid is given by the integral
m=∫∫∫
Eδ(x, y, z) dV.
We can also give formulas for the coordinates of thecenter of massof the solid: if (x, y, z) is thecenter of mass ofE, then
x =1m
∫∫∫
Exδ(x, y, z) dV, y =
1m
∫∫∫
Eyδ(x, y, z) dV, z=
1m
∫∫∫
Ezδ(x, y, z) dV.
Example 17.5.A flat lamina occupies the diskx2+y2 ≤ 4 and has densityδ(x, y) = x+y+ x2+y2.Find its mass and the coordinates of its center of mass.
Solution.Let R denote the polar form of the diskD, that is,R = [0,2] × [0,2π]. By (14.1), themassmof the lamina is
m=∫∫
D
(
x+ y+ x2 + y2)
dA=∫∫
Rr(
r cosθ + r sinθ + r2)
dA
=
∫ 2π
0
[∫ 2
0
(
r2(cosθ + sinθ) + r3)
dr
]
dθ =∫ 2π
0
(
83(cosθ + sinθ) + 4
)
dθ = 8π.
Similarly, the coordinates of the center of mass are:
x =1m
∫∫
Dx(x+ y+ x2 + y2) dA=
18π
∫∫
Rr2 cosθ
(
r cosθ + r sinθ + r2)
dA
=18π
∫ 2π
0
[∫ 2
0cosθ
(
r3(cosθ + sinθ) + r4)
dr
]
dθ =18π
∫ 2π
0cosθ
(
4(cosθ + sinθ) + 325
)
dθ
=18π
∫ 2π
0
(
2(1+ cos 2θ) + 2 sin 2θ + 325 cosθ
)
dθ =12,
y =1m
∫∫
Dy(x+ y+ x2 + y2) dA=
18π
∫∫
Rr2 sinθ
(
r cosθ + r sinθ + r2)
dA
=18π
∫ 2π
0
[∫ 2
0sinθ
(
r3(cosθ + sinθ) + r4)
dr
]
dθ =18π
∫ 2π
0sinθ
(
4(cosθ + sinθ) + 325
)
dθ
=18π
∫ 2π
0
(
2 sin 2θ + 2(1− cos 2θ) + 325 sinθ
)
dθ =12. �
106
Example 17.6. Find the center of mass of the solidH that occupies the upper half of the ballx2 + y2 + z2 ≤ 4 and has densityδ(x, y, z) = z
√
x2 + y2 + z2.
Solution.The solidH is described by inequalities
x2 + y2 + z2 ≤ 4, z≥ 0.
In spherical coordinates these becomeρ ≤ 2 and 0≤ φ ≤ π/2, respectively. Thus, by (16.2), themass of the solid is
m=∫∫∫
Hz√
x2 + y2 + z2 dV =∫ 2
0
∫ 2π
0
∫ π/2
0(ρ2 cosφ)(ρ2 sinφ) dφdθdρ
=
∫ π/2
0
∫ 2π
0sinφ cosφ
[∫ 2
0ρ4 dρ
]
dθdφ =325
∫ π/2
0
∫ 2π
0sinφ cosφdθdφ
=64π5
∫ π/2
0sinφ cosφdφ =
64π5
∫ 1
0u du=
32π5.
The coordinates of the center of mass then are
x =1m
∫∫∫
Hxz
√
x2 + y2 + z2 dV
=5
32π
∫ 2
0
∫ 2π
0
∫ π/2
0(ρ3 cosθ sinφ cosφ)(ρ2 sinφ) dφdθdρ
=5
32π
∫ 2
0
∫ π/2
0ρ5 sin2φ cosφ
[∫ 2π
0cosθdθ
]
dφdρ = 0,
y =1m
∫∫∫
Hyz√
x2 + y2 + z2 dV
=5
32π
∫ 2
0
∫ 2π
0
∫ π/2
0(ρ3 sinθ sinφ cosφ)(ρ2 sinφ) dφdθdρ
=5
32π
∫ 2
0
∫ π/2
0ρ5 sin2φ cosφ
[∫ 2π
0sinθdθ
]
dφdρ = 0,
z=1m
∫∫∫
Hz2√
x2 + y2 + z2 dV =5
32π
∫ 2
0
∫ 2π
0
∫ π/2
0(ρ3 cos2φ)(ρ2 sinφ) dφdθdρ
=5
32π
∫ π/2
0
∫ 2π
0sinφ cos2φ
[∫ 2
0ρ5 dρ
]
dθdφ =53π
∫ π/2
0
∫ 2π
0sinφ cos2φdθdφ
=103
∫ π/2
0sinφ cos2φdφ =
103
∫ 0
1u2 (−du) =
109. �
17.3. An improper integral*. In this section, we use double integrals and polar coordinates toshow that
∫ ∞
0e−x2
dx=
√π
2.
107
Proof. Let Db denote the diskx2 + y2 ≤ b2 and letSb denote the square with vertices (±b,±b) (seeFigure 17.3). Asb→ ∞, bothDb andSb expand to fill the entirexy-plane, so we figure that
limb→∞
∫∫
Db
e−x2−y2dA= lim
b→∞
∫∫
Sb
e−x2−y2dA=
∫∫
R2e−x2−y2
dA. (17.3)
(This can be proved rigorously, but it is also intuitively clear, so we won’t dwell on it.) We nowconsider each limit separately.
(b,b)
(−b,−b)
y
x
Sb
Db
Figure 17.3. The diskDb and the squareSb
First, we consider the integral over the diskDb. In polar coordinates,Db is [0,b] × [0,2π], so∫∫
Db
e−x2−y2dA=
∫ 2π
0
[∫ b
0re−r2
dr
]
dθ =∫ 2π
0
[
12
∫ b2
0e−u du
]
dθ = π(
1− e−b2)
.
Taking limits asb→ ∞, we deduce that
limb→∞
∫∫
Db
e−x2−y2dA= lim
b→∞π(
1− e−b2)
= π. (17.4)
On the other hand, if we define
I (b) =∫ b
−be−x2
dx,
we have∫∫
Sb
e−x2−y2dA=
∫ b
−b
[∫ b
−be−x2−y2
dx
]
dy=∫ b
−bI (b)e−y2
dy= I (b)I (b) = I (b)2.
Furthermore, since the functione−x2is even, we have
I (b) =∫ b
−be−x2
dx= 2∫ b
0e−x2
dx.
Hence,
limb→∞
∫∫
Sb
e−x2−y2dA= lim
b→∞
(
2∫ b
0e−x2
dx
)2
= 4
(∫ ∞
0e−x2
dx
)2
. (17.5)
Combining (17.3)–(17.5), we obtain that
4
(∫ ∞
0e−x2
dx
)2
= π,
which is equivalent to desired conclusion. �
108
Remark. Note that in the above proof we used that the limit
limb→∞
∫ b
0e−x2
dx
exists, or equivalently, that the improper integral∫ ∞
0 e−x2dx converges. This can be proved using
the comparison test for improper integrals.
Exercises
1. Find the area of the given surfaceΣ:
(a)Σ is the part of the planex+ 2y+ z= 3 that lies in the first octant
(b) Σ is the part of the paraboloidz= x2 − y2 that lies between the cylindersx2 + y2 = 1 andx2 + y2 = 4
(c) Σ is the part of the spherex2 + y2 + z2 = 4 that lies inside the cylinderx2 + y2 = 2
(d) Σ is the part of the surfacez= xy that lies inside the cylinderx2 + y2 = 1
(e)Σ is the parametric surface given byr (u, v) = (uv)e1 + (u+ v)e2 + (u− v)e3, u2 + v2 ≤ 1
(f) Σ is the parametric surface given byr (u, v) = (ucosv)e1 + (usinv)e2 + ve3, 0 ≤ u ≤ 1,0 ≤ v ≤ π
2. Find the mass and the center of mass of the lamina that occupies the regionD and has density given by thefunctionδ(x, y):
(a) D is bounded by the curvey = ex, the linex = 1, and the coordinate axes;δ(x, y) = y
(b) D is the diskx2 + y2 ≤ 2y; δ(x, y) is twice the distance from (x, y) to the origin
3. Find the mass and the center of mass of the given solidE if its density is given by the functionδ(x, y, z):
(a) E is the cube 0≤ x, y, z≤ 2; δ(x, y, z) = x2 + y2 + z2
(b) E is the hemispherex2 + y2 + z2 ≤ 4, z≥ 0; δ(x, y, z) is half the distance from (x, y, z) to thez-axis
(c) E is the hemispherex2 + y2 + z2 ≤ 1, z≥ 0; δ(x, y, z) = 4z
4. Use the value of the integral computed in§17.3 to evaluate the integrals
In =
∫ ∞
0x2ne−x2
dx (n = 1,2,3, . . . ).
109
18. Vector fields
18.1. Definition. If D is a set inR2, avector fieldon D is a functionF that assigns to each point(x, y) in D a two-dimensional vectorF(x, y). If E is a set inR3, avector fieldon E is a functionFthat assigns to each point (x, y, z) in E a three-dimensional vectorF(x, y, z).
Note that we can express each two-dimensional vector fieldF in terms of a pair of functions oftwo variables:
F(x, y) = 〈P(x, y),Q(x, y)〉 = P(x, y)e1 + Q(x, y)e2.
Similarly, we can express each three-dimensional vector field F in terms of a triple of functions ofthree variables:
F(x, y, z) = 〈P(x, y, z),Q(x, y, z),R(x, y, z)〉 = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3.
Example 18.1.The functionF(x, y, z) = x2e1 + y2e2 + z2e3 is a vector field onR3. The functionF(x, y) = cos(xy)e1 + sin(xy)e2 is a vector field onR2. �
The standard way to visualize a vector field is to draw the arrows representingF(x, y) (orF(x, y, z)) applied at the points (x, y) (or (x, y, z)) for a number of choices for (x, y) ∈ D (or(x, y, z) ∈ E).
Example 18.2.Sketch twenty vectors of the vector fieldF(x, y) = 12e1 − xe2.
Answer.See Figure 18.1. �
1
1O
Figure 18.1. The vector fieldF(x, y) = 12e1 − xe2
18.2. Gradient fields and potential functions. We are already familiar with an important classof vector fields. Suppose thatf (x, y) is a differentiable function onD. In Lecture #9, we definedthe gradient∇ f of f :
∇ f (x, y) = fx(x, y)e1 + fy(x, y)e2.
We now see that∇ f is a two-dimensional vector field onD. Similarly, if f (x, y, z) is a differentiablefunction on a setE in R3, then∇ f (x, y, z) is a three-dimensional vector field onE. From now on,we will refer to the gradient of a functionf as thegradient vector fieldof f .
Example 18.3.Find the gradient vector field of the functionf (x, y) = x2 + y2.110
Solution.We have∇ f (x, y) = fx(x, y)e1 + fy(x, y)e2 = 2xe1 + 2ye2. �
In the subsequent lectures, we will see that vector fieldsF which are the gradient fields of somefunction play an important role in vector calculus. Such vector fields are known asconservativevector fields, that is,F is conservative if there is a functionf such thatF = ∇ f . If F = ∇ f , then fis called apotential functionof F, or simply apotential.
Example 18.4.By Example 18.3, the vector fieldF(x, y) = 2xe1 + 2ye2 is conservative:F = ∇ f ,where f (x, y) = x2 + y2. Similarly, F(x, y, z) = 2ye1 + 2xe2 + e3 is conservative, becauseF = ∇g,whereg(x, y, z) = 2xy+ z. �
Not every vector field is conservative, and it is not very difficult to give an example of a vectorfield that is not conservative.
Example 18.5.Show that the vector fieldF(x, y) = (x2 + y)e1 + y3e2 is not conservative.
Solution.Suppose thatF is conservative, that is, there is some functionf (x, y) such that
fx(x, y) = x2 + y, fy(x, y) = y3.
Thenfxy(x, y) = ∂y(x
2 + y) = 1, fyx(x, y) = ∂x(y3) = 0.
Since both mixed partials are constants, they are continuous everywhere. Thus, by Theorem 8.1,we must havefxy = fyx. However, they are not equal, a contradiction. Therefore, our assumptionthatF is conservative must be false. �
We shall use this example as a starting point of our discussion of conservative vector fields.However, before delving into that, we need to fix some terminology.
18.3. Terminology. Recall that in§11.4 we defined a closed region in the plane to be a regioncontaining all of its boundary. We call a regionD in the plane or in spaceopen, if it contains noneof the points on its boundary. Some examples of open and non-open sets in the plane are shownon Figure 18.2. On the figure, solid and dashed curves indicate boundary that does and does notbelong to the set, respectively. The first set is open, because it does not contain any of its boundary;the second set is not open, because it contains all of its boundary (it is closed); and the last set isnot open, because it contains a part of its boundary.
not openopen not open
Figure 18.2. Open and non-open sets
We say that a setD, in the plane or in space, isconnected, if we can connect every two pointsof D with a continuous path that lies entirely inD. Geometrically, this means thatD “has only onepiece”. Several examples of connected and disconnected plane sets are displayed on Figure 18.3.
We say that a setD in the plane issimply-connected, if it is connected and every curveγ thatlies entirely inD can be shrunk to a point inD without leavingD. We say that a setD in space issimply-connected, if it is connected and every surfaceΣ that lies entirely inD can be shrunk to a
111
disconnected connected connected
b
disconnected disconnected connected
Figure 18.3. Connected and disconnected sets
point inD without leavingD. In both cases, it is “closed” curves and surfaces that we really need totest. Both in two and in three dimensions, simple-connectedness means thatD “has only one pieceand no holes”. Also, in both cases, we can check whether a connected set is simply-connected bylooking at the compliment ofD to the whole plane or the whole space. If we denote that set byD′,then we must be able to connect each point ofD′ to any other point ofD′ and to “infinity” by apath that lies entirely inD′. Some examples of simply-connected and not simply-connected planesets are displayed on Figure 18.4.
not simply-connectedsimply-connected not simply-connected
Figure 18.4. Simply-connected and not simply-connected sets
18.4. Conservative vector fields inR2. We can easily generalize the solution of Example 18.5.Let F(x, y) = P(x, y)e1+Q(x, y)e2 be a conservative vector field and letf (x, y) be its potential, thatis,
P(x, y) = fx(x, y), Q(x, y) = fy(x, y).
ThenPy = fxy and Qx = fyx =⇒ Py = Qx.
This is the basic idea underlying the following theorem.
Theorem 18.1. If F(x, y) = P(x, y)e1 + Q(x, y)e2 is a conservative vector field and P(x, y) andQ(x, y) have continuous first order partials in a domain D, then throughout D we have
∂yP = ∂xQ. (18.1)
Note that we can use this theorem to show that a given vector field is not conservative. However,we would rather have a tool for establishing that a vector field is conservative. The next theoremsays that equation (18.1) implies thatF is conservative, provided that the regionD has a specialproperty.
112
Theorem 18.2. If F(x, y) = P(x, y)e1 + Q(x, y)e2 is a vector field that satisfies(18.1)on an open,simply-connected region D, thenF is conservative.
Example 18.6.Determine whether or not the vector field is conservative:
(a) F(x, y) = (2x+ y2)e1 + (2y+ x2)e2;(b) F(x, y) = (x3 + 3xy2)e1 + (y3 + 3x2y)e2;
(c) F(x, y) =−y
x2 + y2e1 +
xx2 + y2
e2.
Solution. (a) LetP(x, y) = 2x+ y2 andQ(x, y) = 2y+ x2. Then
Py = 2y , 2x = Qx,
so the vector field is not conservative.
(b) Let P(x, y) = x3 + 3xy2 andQ(x, y) = y3 + 3x2y. Then
Py = 6xy= Qx for all (x, y) ∈ R2.
SinceR2 is open and simply-connected, it follows that the vector field is conservative.
(c) We have
∂
∂y
(
−yx2 + y2
)
=(−1)(x2 + y2) − (−y)(2y)
(x2 + y2)2=
y2 − x2
(x2 + y2)2,
∂
∂x
(
xx2 + y2
)
=(1)(x2 + y2) − (x)(2x)
(x2 + y2)2=
y2 − x2
(x2 + y2)2,
for all (x, y) , (0,0). However, the set of all (x, y) , (0,0) is not simply-connected. Thus,Theorem 18.2 says nothing aboutF. On the other hand, if we considerF on the setD obtainedby cutting thexy-plane along a half-line containing the origin (e.g., we cancut the plane along thepositivex-axis, or the negativey-axis, or the part of the liney = 2x with x ≤ 1), then the theoremdoes say thatF is conservative onD (which is open and simply-connected). Later we shall showthat this vector field is NOT conservative on the set of all (x, y) , 0. �
18.5. Curl and conservative vector fields inR3.
Definition. If F(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 is a vector field onR3 such thatP,Q,R are differentiable, then thecurl of F is the vector field onR3 defined by
curlF =(
∂R∂y− ∂Q∂z
)
e1 +
(
∂P∂z− ∂R∂x
)
e2 +
(
∂Q∂x− ∂P∂y
)
e3.
Remark. A formula for curlF that is easier to remember is that
curlF = ∇ × F =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
P Q R
∣
∣
∣
∣
∣
∣
.
In reality, this is a terrible abuse of notation, but it helpsremember the formula, just as (2.7) helpsus remember the definition of the cross product.
Example 18.7.Find curlF for F(x, y, z) = xeze1 + yeze2 + (x2z+ y2z)e3.113
Solution.We have
curlF =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
xez yez x2z+ y2z
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∂y ∂z
yez x2z+ y2z
∣
∣
∣
∣
e1 −∣
∣
∣
∣
∂x ∂z
xez x2z+ y2z
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
∂x ∂y
xez yez
∣
∣
∣
∣
e3
=(
∂y(x2z+ y2z) − ∂z(yez)
)
e1 −(
∂x(x2z+ y2z) − ∂z(xez)
)
e2 +(
∂x(yez) − ∂y(xez))
e3
= (2yz− yez)e1 − (2xz− xez)e2 + (0− 0)e3 = y(2z− ez)e1 + x(ez− 2z)e2. �
We can use the curl of a vector field to state three-dimensional versions of the theorems in theprevious section.
Theorem 18.3.If f (x, y, z) has continuous second-order partials, then
curl(∇ f ) = 0.
In particular, if F is a conservative three-dimensional vector field, thencurlF = 0.
Proof. The proof is straightforward:
curl(∇ f ) =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
∂x f ∂y f ∂z f
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∂y ∂z
∂y f ∂z f
∣
∣
∣
∣
e1 −∣
∣
∣
∣
∂x ∂z
∂x f ∂z f
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
∂x ∂y
∂x f ∂y f
∣
∣
∣
∣
e3
=
(
∂2 f∂y∂z
− ∂2 f∂z∂y
)
e1 −(
∂2 f∂x∂z
− ∂2 f∂z∂x
)
e2 +
(
∂2 f∂x∂y
− ∂2 f∂y∂x
)
e3 = 0. �
The converse theorem is not universally true, but as in the case of two-dimensional fields, it canbe proved for special sets.
Theorem 18.4.If F is a three-dimensional vector field that has continuous partials and satisfies
curlF = 0
on an open, simply-connected set inR3, thenF is conservative.
Example 18.8.Determine whether the vector fieldF = ye2xy+z2e1 + xe2xy+z2
e2 + ze2xy+z2e3 is con-
servative.
Solution.We have
curlF =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
ye2xy+z2xe2xy+z2
ze2xy+z2
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∂y ∂z
xe2xy+z2ze2xy+z2
∣
∣
∣
∣
e1 −∣
∣
∣
∣
∂x ∂z
ye2xy+z2ze2xy+z2
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
∂x ∂y
ye2xy+z2xe2xy+z2
∣
∣
∣
∣
e3
= (z(2x) − x(2z))e2xy+z2e1 − (z(2y) − y(2z))e2xy+z2
e2 + (x(2y) − y(2x))e2xy+z2e3 = 0.
Since the partials ofF are continuous everywhere inR3 (a simply-connected set), it follows thatFis conservative. �
114
Remark. Note that whenF(x, y, z) = P(x, y)e1 + Q(x, y)e2 + 0e3 (that is, whenF(x, y, z) is a two-dimensional vector field that “pretends” to be three-dimensional), then
curlF =(
∂Q∂x− ∂P∂y
)
e3,
and the condition curlF = 0 is equivalent to the condition discussed in§18.4.
18.6. Finding potentials.
Example 18.9.Find a potential for the vector fieldF(x, y) = (x3 + 3xy2)e1 + (y3 + 3x2y)e2, if itexists.
Solution.We know that this field is conservative from Example 18.6, so apotential exists. If wedenote it byf (x, y), then f must satisfy the conditions
fx(x, y) = x3 + 3xy2, fy(x, y) = y3 + 3x2y.
In other words, for a fixedy, f (x, y) is an antiderivative ofx3 + 3xy2, and for a fixedx, f (x, y) is anantiderivative ofy3 + 3x2y. Hence,
f (x, y) =∫
x3 + 3xy2 dx=x4
4+
3x2y2
2+ g(y).
Note that our answer infolves an arbtrary function ofy instead of an arbitrary constant. In order tofind out more aboutg(y), we differentiate the above expression forf (x, y) with respect toy:
fy(x, y) = ∂y
(
14x4 + 3
2x2y2 + g(y))
= 3x2y+ g′(y).
Comparing this with the equalityfy = y3 + 3x2y, we find that
3x2y+ g′(y) = y3 + 3x2y ⇐⇒ g′(y) = y3 ⇐⇒ g(y) =y4
4+C.
We conclude thatf (x, y) = 14x4 + 3
2x2y2 + 14y4 +C. �
Note that in the above example we can easily check our answer:direct differentiation gives∇ f (x, y) = F(x, y), so f (x, y) is potential forF(x, y) by the definition of potential.
Example 18.10.Find a potential for the vector fieldF(x, y) =x
x2 + y2e1 +
yx2 + y2
e2, if it exists.
Solution.The potentialf (x, y), if it exists, must satisfy the conditions
fx(x, y) =x
x2 + y2, fy(x, y) =
yx2 + y2
.
Thus,
f (x, y) =∫
xx2 + y2
dx=12
∫
d(x2 + y2)x2 + y2
=ln(x2 + y2)
2+ g(y).
To findg(y), we usefy(x, y):
fy(x, y) =12
2yx2 + y2
+ g′(y) =y
x2 + y2⇐⇒ g′(y) = 0 ⇐⇒ g(y) = C.
We conclude thatf (x, y) = 12 ln(x2 + y2) +C. �
115
Remark. Note that the functionf (x, y) = 12 ln(x2 + y2) +C in the previous example is a potential
for the vector fieldF(x, y) for all (x, y) , (0,0). Therefore,F is conservative in the setR2− {(0,0)}.On the other hand, if we tried to apply Theorem 18.2 to it, we would run into the same trouble asin the solution of Example 18.6, part (c): this set is not simply-connected.
Example 18.11.Find a potential for the vector fieldF(x, y, z) = (2xz+ y2)e1+2xye2+ (x2+3z2)e3,if it exists.
Solution.The potentialf (x, y, z), if it exists, must satisfy the conditions
fx(x, y, z) = 2xz+ y2, fy(x, y, z) = 2xy, fz(x, y, z) = x2 + 3z2.
Thus, arguing similarly to the last two examples, we find that
f (x, y, z) =∫
(2xz+ y2) dx= x2z+ y2x+ g(y, z),
fy(x, y, z) = 2yx+ gy(y, z) = 2xy ⇐⇒ gy(y, z) = 0 ⇐⇒ g(y, z) = h(z),
fz(x, y, z) = x2 + h′(z) = x2 + 3z2 ⇐⇒ h′(z) = 3z2 ⇐⇒ h(z) = z3 +C.
We conclude thatf (x, y, z) = x2z+ y2x+ z3 +C. �
Example 18.12.Find a potential for the vector fieldF(x, y, z) = e2xye1 + e2xye2 + z2e3, if it exists.
Solution.Any potential f (x, y, z) must satisfy
fx(x, y, z) = e2xy, fy(x, y, z) = e2xy, fz(x, y, z) = z2.
From the first of these equations, we find
f (x, y, z) =∫
e2xy dx=e2yx
2y+ g(y, z).
Next, we differentiate the last function with respect toy and get
fy(x, y, z) =2xe2xy
2y− e2xy
2y2+ gy(y, z) = e2xy ⇐⇒ gy(y, z) = e2xy
(
1− xy+
12y2
)
.
We seem to be running in circles, and we indeed are, because the field is non-conservative. Forexample,
∂x
(
e2xy)
= 2ye2xy, 2xe2xy = ∂y
(
e2xy)
,
so the third component of curlF is non-zero. �
18.7. The divergence of a vector field.
Definition. If F(x, y, z) = P(x, y, z)e1 + Q(x, y, z)e2 + R(x, y, z)e3 is a vector field onR3 such thatP,Q,R are differentiable, then thedivergenceof F is the function defined by
div F =∂P∂x+∂Q∂y+∂R∂z.
Example 18.13.Find divF for F = ez sin(xy)e1 + ez cos(xy)e2 + xyze3.
Solution.We have
div F = ∂x(ez sin(xy)) + ∂y(e
z cos(xy)) + ∂z(xyz) = yez cos(xy) − xez sin(xy) + xy. �
116
Theorem 18.5.If F = Pe1 + Qe2 + Re3 has continuous second-order partials, then
div curlF = 0.
Proof. The proof is straightforward:
div curlF =∂
∂x
(
∂R∂y− ∂Q∂z
)
+∂
∂y
(
∂P∂z− ∂R∂x
)
+∂
∂z
(
∂Q∂x− ∂P∂y
)
=∂2R∂x∂y
− ∂2Q∂x∂z
+∂2P∂y∂z
− ∂2R∂y∂x
+∂2Q∂z∂x
− ∂2P∂z∂y
= 0. �
Exercises
1. Sketch the vector fieldF(x, y) = (x− y)e1 + xe2 by drawing arrows in a similar fashion to Example 18.2.
2. Find the gradient vector field of the given function:(a) f (x, y) = e2y sin(xy) (b) f (x, y, z) =
√
x2 + y2 + z2 (c) f (x, y, z) = ln(
x2 + yz)
3. Determine whether or not the given set is open, connected,and simply-connected:(a)
{
(x, y) : x > 0, y > 0}
(c){
(x, y) : 1 < x2 + y2 < 2}
(e){
(x, y) : x2 + y2 < 1 or x2 + y2 > 2}
(b){
(x, y) : x > 0, y ≥ 0}
(d){
(x, y) : x2 + y2 ≤ 1}
(f){
(x, y) : xy≤ 0}
4. Determine whether the vector fieldF(x, y) is conservative. If it is, find a potential function for it:(a)F(x, y) = (x− y)e1 + (x+ y)e2
(c) F(x, y) = eye1 + xeye2
(e)F(x, y) = (4x+ siny)e1 + (siny+ xcosy)e2
(b) F(x, y) = ye1 + (x+ 3y)e2
(d) F(x, y) =y2
1+ x2e1 + (2yarctanx)e2
5. Find the curl and the divergence of the given vector field:(a)F(x, y, z) = x2yze1 − 4xyz2e3
(c) F(x, y, z) = cosxze1 + 2 sinxye2 + ze2xye3
(b) F(x, y, z) = 2 arcsinye1 +2x
√
1− y2e2 − 3z2e3
6. Determine whether the vector fieldF(x, y, z) is conservative. If it is, find a potential function for it:(a)F(x, y, z) = (yz+x2)e1+(xz+4y)e2+(xy−3z2)e3
(c) F(x, y, z) = xeye1 + yexe2 + xyze3
(b) F(x, y, z) = y2 cosze1+2xycosze2− xy2 sinze3
(d) F(x, y, z) = 2 arcsinye1 +2x
√
1− y2e2 − 3z2e3
7. Is there a vector fieldF onR3 such that curlF = xy2e1 + yz2e2 + zx2e3? Explain.
117
19. Line integrals
19.1. Line integrals of scalar functions. We start with a smooth parametric curveγ, given by avector function
r (t) = x(t)e1 + y(t)e2 + z(t)e3 (a ≤ t ≤ b).Recall that the smoothness ofγ means thatr ′(t) = x′(t)e1 + y′(t)e2 + z′(t)e3 is continuous andr ′(t) , 0 whena < t < b (see§5.3). Further, suppose thatf (x, y, z) is a continuous function on asetE in R3 that contains the curveγ. In particular,f (x, y, z) is defined at each point (x, y, z) onγ.
For eachn ≥ 1, set∆n = (b−a)/n and define the numbers that partition [a,b] into n subintervalsof equal lengths:
t0 = a, t1 = a+ ∆n, t2 = a+ 2∆n, . . . , tn = a+ n∆n = b.
We write xi = x(ti), yi = y(ti), zi = z(ti) for the coordinates of the pointPi(xi , yi , zi) that corre-sponds to the valuet = ti of the parameter. The pointsP0,P1,P2, . . . ,Pn partition γ into n arcs:γ1, γ2, . . . , γn, whereγi hasPi−1 andPi as its endpoints. Further, we pickn values of the parametert∗1, t
∗2, . . . , t
∗n so that
t0 ≤ t∗1 ≤ t1, t1 ≤ t∗2 ≤ t2, . . . , tn−1 ≤ t∗n ≤ tn.
These yieldn sample points P∗i (x∗i , y∗i , z∗i ) on the curve with coordinatesx∗i = x(t∗i ), y∗i = y(t∗i ),
z∗i = z(t∗i ). Then theline integral of f alongγ is defined by∫
γ
f (x, y, z) ds= limn→∞
n∑
i=1
f (x∗i , y∗i , z∗i )∆si , ∆si = length(γi), (19.1)
provided that the limit exists.Recall that, by Theorem 5.1,
∆si =
∫ ti
ti−1
√
x′(t)2 + y′(t)2 + z′(t)2 dt.
Using this formula, the above definition, the properties of continuous functions, and the definitionof the definite integral, it can be shown that for continuous functions f the limit (19.1) alwaysexists and the line integral can be expressed as a definite integral.
Theorem 19.1.Letγ be a smooth parametric curve, given by the vector function
r (t) = x(t)e1 + y(t)e2 + z(t)e3 (a ≤ t ≤ b).
If f (x, y, z) is continuous onγ, then∫
γ
f (x, y, z) ds=∫ b
af (x(t), y(t), z(t))
√
x′(t)2 + y′(t)2 + z′(t)2 dt. (19.2)
Remarks. 1. Using more extensively the language of vector functions,we can write (19.2) in theform
∫
γ
f (x, y, z) ds=∫ b
af (r (t))‖r ′(t)‖dt. (19.3)
2. As in the cases of definite and multiple integrals, the lineintegral of a continuous functionalong a curve is related to its average value on the points of the curve:
average(x,y,z)∈γ
f (x, y, z) =1
length(γ)
∫
γ
f (x, y, z) ds.
118
Example 19.1.Evaluate∫
γ
(
x2 + y2 + z2)−1
ds
whereγ is the helixx = cost, y = sint, z= t (0 ≤ t ≤ T).
Solution.We have∫
γ
(
x2 + y2 + z2)−1
ds=∫ T
0
(
(cost)2 + (sint)2 + t2)−1√
(− sint)2 + (cost)2 + 12 dt
=
∫ T
0(1+ t2)−1
√1+ 1dt =
∫ T
0
√2
1+ t2dt =
√2 arctanT. �
We have already seen on several instances that properties ofplane curves or two-dimensionalvectors often appear as special cases of analogous properties of space curves or three-dimensionalvectors. This general principle also applies to line integrals. We just defined the line integral of afunction of three variables along a space curve. Is it possible to define the line integral of a functionof two variables along a plane curve? The answer to this question is in the affirmative. One wayto develop the theory of such line integrals is by repeating the above definitions and reproving theabove theorems with one less variable. Alternatively, we can define two-dimensional line integralsas a special case of three-dimensional line integrals. We take the latter route.
Observe that a two-variable functionf (x, y) is also a three-variable functionf (x, y, z) (whichdoes not really depend onz) and that a plane curve, given by
r (t) = x(t)e1 + y(t)e2 (a ≤ t ≤ b), (19.4)
is also a space curve, given by
r0(t) = x(t)e1 + y(t)e2 + 0e3 (a ≤ t ≤ b). (19.5)
If γ is the plane curve (19.4) andf (x, y) is continuous onγ, we define theline integral of f alongγ by
∫
γ
f (x, y) ds=∫
γ0
g(x, y, z) ds,
whereγ0 is the space curve (19.5) andg(x, y, z) = f (x, y) for all (x, y) in the domain off and allz.In particular, by (19.2), we have
∫
γ
f (x, y) ds=∫ b
ag(x(t), y(t),0)
√
x′(t)2 + y′(t)2 + 02 dt
=
∫ b
af (x(t), y(t))
√
x′(t)2 + y′(t)2 dt.
That is, we have the formula∫
γ
f (x, y) ds=∫ b
af (r (t))‖r ′(t)‖dt =
∫ b
af (x(t), y(t))
√
x′(t)2 + y′(t)2 dt. (19.6)
Note that the vector form of (19.6) and the vector form (19.3)of (19.2) look identical; the onlydifference is that the vector functionr (t) is two-dimensional in (19.6) and three-dimensional in(19.3).
119
Example 19.2.Evaluate∫
γ
xey ds,
whereγ is the arc of the curvex = ey from (1,0) to (e,1).
Solution.The curve can be parametrized by the equations
x = et, y = t (0 ≤ t ≤ 1).
Hence, by (19.6),∫
γ
xey ds=∫ 1
0etet
√
(et)2 + 12 dt =∫ 1
0e2t√
e2t + 1dt.
Using the substitutionu = e2t + 1, we find that
du= 2e2t dt, u(0) = 2, u(1) = e2 + 1,
and∫ 1
0e2t√
e2t + 1dt =∫ e2+1
2u1/2
(
12du
)
=(e2 + 1)3/2 − 23/2
3≈ 7.157. �
We can easily extend the defintion of line integrals to piecewise smooth curves. Ifγ is apiecewise smooth parametric curve that is the union of the smooth curvesγ1, γ2, . . . , γn, γ =γ1 ∪ γ2 ∪ · · · ∪ γn, we define
∫
γ
f (x, y, z) ds=∫
γ1
f (x, y, z) ds+∫
γ2
f (x, y, z) ds+ · · · +∫
γn
f (x, y, z) ds.
Note that all the line integrals on the right side of the formula are along smooth curves and can beevaluated via the limit in (19.1) or via the integral formulas (19.2), (19.3), or (19.6).
Example 19.3.Evaluate the line integral∫
γ
xy2 ds
whereγ is the triangle with vertices (0,1,2), (1,0,3), and (0,−1,0).
Solution.We splitγ into three line segments:
• γ1 is the segment connecting (0,1,2) and (1,0,3):
r (t) = t 〈0,1,2〉 + (1− t) 〈1,0,3〉 = 〈1− t, t,3− t〉 (0 ≤ t ≤ 1);
• γ2 is the segment connecting (0,1,2) and (0,−1,0):
r (t) = t 〈0,1,2〉 + (1− t) 〈0,−1,0〉 = 〈0,2t − 1,2t〉 (0 ≤ t ≤ 1);
• γ3 is the segment connecting (1,0,3) and (0,−1,0):
r (t) = t 〈1,0,3〉 + (1− t) 〈0,−1,0〉 = 〈t, t − 1,3t〉 (0 ≤ t ≤ 1).120
By (19.2),∫
γ1
xy2 ds=∫ 1
0(1− t)t2
√
(−1)2 + 12 + (−1)2 dt =√
3∫ 1
0
(
t2 − t3)
dt =
√3
12;
∫
γ2
xy2 ds=∫ 1
00(2t − 1)2
√02 + 22 + 22 dt = 0;
∫
γ3
xy2 ds=∫ 1
0t(t − 1)2
√12 + 12 + 32 dt =
√11
∫ 1
0
(
t3 − 2t2 + t)
dt =
√11
12.
Hence,∫
γ
xy2 ds=∫
γ1
xy2 ds+∫
γ2
xy2 ds+∫
γ3
xy2 ds=
√3
12+
√11
12≈ 0.421. �
19.2. Line integrals of vector fields. Let F be a continuous vector field defined on a piecewisesmooth curveγ given by a vector functionr (t), a ≤ t ≤ b. We define theline integral ofF alongγby
∫
γ
F · dr =∫
γ
(F · T) ds,
whereT is the unit tangent vector to the curveγ. Recall from§5.4 that the unit tangent vectorT(t)at the point ofγ corresponding to the valuet of the parameter can be expressed in terms ofr (t) bythe formula
T(t) =r ′(t)‖r ′(t)‖
.
Therefore, by (19.3) or (19.6) (depending on the dimension of F andγ),∫
γ
(F · T) ds=∫ b
a
(
F(r (t)) · r ′(t)‖r ′(t)‖
)
‖r ′(t)‖dt =∫ b
aF(r (t)) · r ′(t) dt.
That is, we have∫
γ
F · dr =∫
γ
(F · T) ds=∫ b
aF(r (t)) · r ′(t) dt. (19.7)
What is the meaning of this kind of line integral? Based on the interpretation that we gave to theline integral of a function, we can say that
1length(γ)
∫
γ
F · dr =1
length(γ)
∫
γ
(F · T) ds= averageγ
(F · T).
Recall from§2.8 that the dot productF ·T is the size of the projection ofF onto the direction ofT,taken with positive or negative sign according as the projection andT point in the same directionor in opposite directions. SinceT(t) is the direction of the curve at the point corresponding to thevaluet of the parameter, we conclude that
1length(γ)
∫
γ
F · dr
is the “average projection ofF in the direction of the curve”.121
Example 19.4.Evaluate∫
γ
(
ye1 + ze2 + xe3
)
· dr ,
whereγ is the helixr (t) = 〈2 cost,2 sint,3t〉, 0 ≤ t ≤ 2π.
Solution.By (19.7),
∫
γ
(
ye1 + ze2 + xe3
)
· dr =∫ 2π
0〈2 sint,3t,2 cost〉 · 〈(2 cost)′, (2 sint)′, (3t)′〉 dt
=
∫ 2π
0〈2 sint,3t,2 cost〉 · 〈−2 sint,2 cost,3〉 dt
=
∫ 2π
0
(
− 4 sin2 t + 6t cost + 6 cost)
dt
= −2∫ 2π
0(1− cos 2t) dt+ 6
∫ 2π
0(t + 1)d(sint)
=[
− 2t + sin 2t]2π
0+[
6(t + 1) sint]2π
0− 6
∫ 2π
0sint dt = −4π. �
In the above example, the parametrization ofγ was given explicitly. However, we often haveonly a geometric description of the curve and have to obtain the parametrization ourselves. In suchsituations, it is important to remember that (19.7) is sensitive to theorientationof the curve, thatis, the direction in whichr (t) traces the curve ast increases. Indeed, ifγ1 andγ2 are two parametriccurves that represent the same geometric curveγ but with opposite orientations, then
∫
γ1
F · dr = −∫
γ2
F · dr . (19.8)
That is, if we change the orientation of the curve, the line integral changes sign. This fact goesback to the property of definite integrals that
∫ a
bf (x) dx= −
∫ b
af (x) dx.
Because of (19.8), a geometric description of the curve must always specify its orientation.
Example 19.5.Evaluate∫
γ
(
x2e1 + y2e2
)
· dr ,
whereγ is the right half of the ellipsex2 + 4y2 = 4 oriented in the counterclockwise direction.
Solution.The point (a,b) is on the given ellipse exactly when the point (a,2b) is on the circlex2 + y2 = 4. Thus, we can parametrizeγ by
x = 2 cost, y = sint (−π/2 ≤ t ≤ π/2).122
Note that ast increases from−π/2 toπ/2, the point (x(t), y(t)) moves alongγ in the counterclock-wise direction. Applying (19.7) to the above parametrization, we get
∫
γ
(
x2e1 + y2e2
)
· dr =∫ π/2
−π/2
(
(2 cost)2e1 + (sint)2e2
)
·(
(2 cost)′e1 + (sint)′e2
)
dt
=
∫ π/2
−π/2
(
4 cos2 te1 + sin2 te2
)
·(
− 2 sinte1 + coste2
)
dt
=
∫ π/2
−π/2
(
− 8 cos2 t sint + sin2 t cost)
dt
= −8∫ π/2
−π/2cos2 t sint dt+
∫ π/2
−π/2sin2 t cost dt
= 0+ 2∫ π/2
0sin2 t cost dt = 2
∫ 1
0u2 du=
23
(u = sint). �
Suppose thatF(x, y, z) = f (x, y, z)e1. Then (19.7) yields∫
γ
F · dr =∫ b
a
(
f (r (t))e1
)
·(
x′(t)e1 + y′(t)e2 + z′(t)e3
)
dt =∫ b
af (r (t))x′(t) dt.
Another notation for this integral is∫
γf (x, y, z) dx, that is,
∫
γ
f (x, y, z) dx=∫
γ
( f e1) · dr =∫ b
af (r (t))x′(t) dt. (19.9)
Similarly, applying (19.7) to the vector fieldsF(x, y, z) = f (x, y, z)e2 andF(x, y, z) = f (x, y, z)e3,we obtain
∫
γ
f (x, y, z) dy=∫
γ
( f e2) · dr =∫ b
af (r (t))y′(t) dt (19.10)
and∫
γ
f (x, y, z) dz=∫
γ
( f e3) · dr =∫ b
af (r (t))z′(t) dt. (19.11)
These three integrals are called theline integrals of f with respect to x, y, andz, respectively. Sincethey are derived from (19.7), these integrals are also sensitive to the orientation of the curveγand change signs when we reverse the orientation of the curve. For example, ifγ1 andγ2 are twoparametric curves that represent the same geometric curve with opposite orientations, then
∫
γ1
f (x, y, z) dx= −∫
γ2
f (x, y, z) dx.
Note that ifF(x, y, z) = P(x, y, z)e1+Q(x, y, z)e2+R(x, y, z)e3, then we can use (19.9)–(19.11) towrite
∫
γ
F · dr =∫
γ
P(x, y, z) dx+∫
γ
Q(x, y, z) dy+∫
γ
R(x, y, z) dz.
In other words, the line integral of a vector field decomposesinto a sum of the line integrals of itscomponents with respect to the corresponding variables. The right side of the last identity is often
123
written as∫
γP dx+ Q dy+ R dz. With this notation, we can rewrite that identity in the form
∫
γ
(Pe1 + Qe2 + Re3) · dr =∫
γ
P dx+ Q dy+ R dz. (19.12)
Example 19.6.Evaluate∫
γ
y dx+ x dy+ xyz dz,
whereγ is the space curve
x = et + e−t, y = et − e−t, z= t2 (0 ≤ t ≤ 1).
Solution.By (19.9)–(19.11),∫
γ
y dx+ x dy+ xyz dz=∫ 1
0
[(
et − e−t)(
et + e−t)′+(
et + e−t)(
et − e−t)′+(
e2t − e−2t)
t2(2t)]
dt
=
∫ 1
0
[(
et − e−t)2+(
et + e−t)2+ 2t3
(
e2t − e−2t)]
dt
=
∫ 1
0
(
e2t − 2+ e−2t + e2t + 2+ e−2t)
dt+∫ 1
0t3 d
(
e2t + e−2t)
= 2∫ 1
0
(
e2t + e−2t)
dt+[
t3(
e2t + e−2t)]1
0−∫ 1
0
(
e2t + e−2t)
(3t2) dt
=[
e2t − e−2t]1
0+ e2 + e−2 − 3
2
∫ 1
0t2 d
(
e2t − e−2t)
= 2e2 − 32
[
t2(
e2t − e−2t)]1
0+
32
∫ 1
0
(
e2t − e−2t)
(2t) dt
=e2 + 3e−2
2+
32
∫ 1
0t d(
e2t + e−2t)
=e2 + 3e−2
2+
32
[
t(
e2t + e−2t)]1
0− 3
2
∫ 1
0
(
e2t + e−2t)
dt
= 2e2 + 3e−2 − 34
[
e2t − e−2t]1
0=
5e2 + 15e−2
4. �
Example 19.7.Evaluate∫
γ
(
y2 − xy)
dx,
whereγ is the curvey = ln x from (1,0) towards (e,1).
Solution.We can parametrizeγ as
x = t, y = ln t (1 ≤ t ≤ e).
Hence, (19.9) gives∫
γ
(
y2 − xy)
dx=∫ e
1
(
ln2 t − t ln t)
dt.
124
Using integration by parts, we find that∫ e
1ln2 t dt =
[
t ln2 t]e
1−∫ e
1t d(
ln2 t)
= e−∫ e
1t(
2t−1 ln t)
dt = e− 2∫ e
1ln t dt
= e− 2[
t ln t]e
1+ 2
∫ e
1t(
t−1)
dt = −e+ 2(e− 1) = e− 2;∫ e
1t ln t dt =
∫ e
1ln t d
(
12t2
)
=[
12t2 ln t
]e
1−∫ e
1
12t2
(
t−1)
dt = 12e2 − 1
2
∫ e
1t dt = 1
4
(
e2 + 1)
.
Thus, the given line integral equalse− 2− 14
(
e2 + 1)
≈ −1.379. �
Exercises
1. Evaluate the given line integral:
(a)∫
γ
x3y ds, γ is given byr (t) = te1 + t2e2, 0 ≤ t ≤ 1
(b)∫
γ
x2y ds, γ is the upper half of the circlex2 + y2 = 4
(c)∫
γ
(xy+ ln x) dy, γ is the arc of the curvey = x3/2 from (1,1) to (0,0)
(d)∫
γ
xy dx− x2 dy, γ is the arc of the circlex2 + y2 = 4 from (2,0) to (0,2)
(e)∫
γ
(
xy+ z2)
ds, γ is given byr (t) = cos 2te1 + sin 2te2 + 3te3, 0 ≤ t ≤ π
(f)∫
γ
z2 dx+ y dz, γ is given byr (t) = (t − sint)e1 + (1− cost)e2 + sin(t/2)e3, 0 ≤ t ≤ 2π
(g)∫
γ
yz dx+ 2xz dy− xy dz, γ consists of the line segments from (1,0,0) to (1,0,2) to (2,2,2)
(h)∫
γ
y2 dx+ x2 dy+ xyz dz, γ consists of the upper half of the circlex2 + y2 = 1 in thexy-plane taken
clockwise and the line segment from (1,0,0) to (0,1,1)
2. Evaluate the line integral∫
γ
F · dr for the given vector fieldF and the given curveγ:
(a)F(x, y, z) = x2ze2 − (x+ y)e3, γ is the ellipser (t) = 2e1 + 3 sinte2 + coste3, 0 ≤ t ≤ 2π(b) F(x, y, z) = sinxe1 + cosye2 − 2xyze3, γ is given byr (t) = t3e1 + 2t2e2 +
√te3, 0 ≤ t ≤ 1
(c) F(x, y) = xsinye1 + ye2, γ is the arc of the parabolax = y2 from (1,−1) to (4,2)
125
20. The fundamental theorem for line integrals
20.1. The main theorem. So far, we have been reducing line integrals to definite integrals. In thecase of conservative vector fields, there is another, fasterway to evaluate their line integrals. Thatalternative approach is based on the following theorem, known as thefundamental theorem for lineintegrals.
Theorem 20.1.Let γ be a piecewise smooth curve given by a vector functionr (t), a ≤ t ≤ b, andlet f(x, y, z) be a differentiable function, whose gradient vector field∇ f is continuous onγ. Then
∫
γ
∇ f · dr = f (r (b)) − f (r (a)). (20.1)
Proof. We have∫
γ
∇ f · dr =∫
γ
fx dx+ fy dy+ fz dz by (19.12)
=
∫ b
a
(
fx(r (t))x′(t) + fy(r (t))y′(t) + fz(r (t))z′(t))
dt by (19.9)–(19.11)
=
∫ b
a
ddt
[
f (r (t))]
dt by the chain rule
= f (r (b)) − f (r (a)) by the FTC �
Suppose thatF(x, y, z) is a conservative vector field with potential functionf (x, y, z), that is,F = ∇ f . Theorem 20.1 then gives
∫
γ
F · dr =∫
γ
∇ f · dr = f (r (b)) − f (r (a)).
This demonstrates the importance of conservative vector fields. If F is conservative and its potentialf is known, the evaluation of the line integral ofF is rather easy: all we need to do is calculate thenet change off along the curve. The next example illustrates this.
Example 20.1.Evaluate∫
γ
F · dr ,
whereF(x, y, z) = 2ye1 + 2xe2 + e3 andγ is the curve given byr (t) = 〈sin 2t, cos 2t, ln(t + 1)〉,0 ≤ t ≤ π.
Solution.We observe thatF = ∇g, whereg(x, y, z) = 2xy+ z. Hence, by (20.1),∫
γ
F · dr =∫
γ
∇g · dr = g(r (π)) − g(r (0)) = g(0,1, ln(π + 1))− g(0,1,0) = ln(π + 1).
Observe that the answer will stay the same, if we replaceγ by any other piecewise smooth curveconnecting the points (0,1,0) and (0,1, ln(π + 1))! For example, the line integral ofF along thecurve displayed on Figure 20.1 (I have no idea what its parametrization is!) will also equal ln(π +1). �
126
b
b
(0,1,0)
(0,1, ln(π + 1))
γ
Figure 20.1. A curve with∫
γ(2ye1 + 2xe2 + e3) · dr = ln(π + 1)
20.2. Path independence.We just saw that ifF(x, y, z) is a conservative vector field, Theorem20.1 implies that the line integral
∫
γF · dr depends only on the endpoints ofγ. Thus, ifγ1 andγ2
are two curves in the domain ofF that share the same endpoints, then∫
γ1
F · dr =∫
γ2
F · dr ,
that is,∫
γF · dr is path independent. The next theorem says that the converse of this observationis
also true: if the line integrals ofF are path independent, thenF is conservative.
Theorem 20.2.LetF be a vector field that is continuous in a open, connected region D. If the lineintegral
∫
γF · dr is path independent in D, thenF is conservative in D.
Note that unlike Theorem 18.2, this theorem requires less ofthe setD: only that it be open andconnected (as opposed to simply connected). On the other hand, the test that it provides is muchmore elusive: we must determine thatall (infinitely many) line integrals ofF are path independent.This hardly seems to be a practical test. And yet, if you look up the proof of Theorem 18.2 in anadvanced calculus text, you will discover that it is deducedfrom Theorem 20.2. Thus, every timewe show that some vector field is conservative by means of Theorem 18.2, we implicitly make useof Theorem 20.2 too.
We now return to the vector field
F(x, y) =−y
x2 + y2e1 +
xx2 + y2
e2,
which we studied in Example 18.6(c). Recall that this field is defined and differentiable everywherebut at the origin. Furthermore, we have
∂
∂y
(
−yx2 + y2
)
=∂
∂x
(
xx2 + y2
)
=y2 − x2
(x2 + y2)2,
also for all (x, y) , (0,0). Back in Lecture #18, we showed that this vector field is conservativeon the plane cut along a half-line passing through the origin, but were unable to decide whetherFis conservative in its entire domain. The next example showsthat, in fact, this vector field is notconservative in its full domain.
Example 20.2.The vector fieldF(x, y) =−y
x2 + y2e1 +
xx2 + y2
e2 is not conservative.
Solution.Let γ be the unit circle
x = cost, y = sint (0 ≤ t ≤ 2π).127
By (19.9) and (19.10),∫
γ
F · dr =∫
γ
−yx2 + y2
dx+x
x2 + y2dy
=
∫ 2π
0
(
(− sint)(cost)′
cos2 t + sin2 t+
(cost)(sint)′
cos2 t + sin2 t
)
dt
=
∫ 2π
0
(− sint)2 + (cost)2
cos2 t + sin2 tdt = 2π.
On the other hand, ifF was conservative, with potential function, sayf (x, y), Theorem 20.1 wouldgive
∫
γ
F · dr = f (cos 2π, sin 2π) − f (cos 0, sin 0)= f (1,0)− f (1,0) = 0.
Since this value differs from the value of the integral we got via direct calculation, F must benon-conservative. �
Example 20.3.Evaluate∫
γ
−yx2 + y2
dx+x
x2 + y2dy,
whereγ is the circle (x− 2)2 + (y+ 1)2 = 25, oriented counterclockwise.
First solution (not recommended).We just showed that the vector field
F(x, y) =−y
x2 + y2e1 +
xx2 + y2
e2
is conservative on the plane with a half-line removed and non-conservative on the plane with justthe origin removed. Thus,F is conservative on curves that don’t loop around the origin and non-conservative on curves that do. In particular,F is not conservative onγ, since the origin lies insideit. This means that we can’t use Theorem 20.1 to evaluate the given integral, so we go back to themethods from the last lecture.
A point (a,b) is on γ exactly when (a − 2,b + 1) is on the circlex2 + y2 = 25. Thus, we canderive the parametrization ofγ from that of the latter circle: the parametric equations ofγ are
x = 2+ 5 cost, y = −1+ 5 sint (0 ≤ t ≤ 2π).
Denote the given line integral byI . By (19.9) and (19.10),
I =∫ 2π
0
(
(1− 5 sint)(2+ 5 cost)′
(2+ 5 cost)2 + (−1+ 5 sint)2+
(2+ 5 cost)(−1+ 5 sint)′
(2+ 5 cost)2 + (−1+ 5 sint)2
)
dt
=
∫ 2π
0
(1− 5 sint)(−5 sint) + (2+ 5 cost)(5 cost)
4+ 20 cost + 25 cos2 t + 1− 10 sint + 25 sin2 tdt
=
∫ 2π
0
25− 5 sint + 10 cost30− 10 sint + 20 cost
dt =∫ 2π
0
5− sint + 2 cost6− 2 sint + 4 cost
dt.
Most likely, you have not seen integrals like the last in single-variable calculus. It can be re-duced to an integral of a rational function by the substitution u = tan(t/2), but we must be ex-tremely careful, because this substitution is not “well-behaved”. We must even be careful, if we
128
use mathematical software or an integral table. For example, Mathematicareports that∫
5− sint + 2 cost6− 2 sint + 4 cost
dt =t2+ arctan
(
21− tan(t/2)
)
+C,
which would suggest that∫ 2π
0
5− sint + 2 cost6− 2 sint + 4 cost
dt =
[
t2+ arctan
(
21− tan(t/2)
)]2π
0
= π,
... if it wasn’t for the fact thatMathematicaalso proudly proclaims that∫ 2π
0
5− sint + 2 cost6− 2 sint + 4 cost
dt = 2π.
In fact, both answers thatMathematicareturns are correct and our calculation between them isflawed, but this comes to show that we may want to avoid this approach... �
Second solution (recommended).Recall that in the solution of the last example, we showed that∫
γ0
−yx2 + y2
dx+x
x2 + y2dy= 2π,
whereγ0 is the unit circle taken in the counterclockwise direction.It turns out that, although wecannot appeal to Theorem 20.1 to evaluate the given line integral directly, we can use it to showthat
∫
γ
−yx2 + y2
dx+x
x2 + y2dy=
∫
γ0
−yx2 + y2
dx+x
x2 + y2dy.
Let γ+ be the part ofγ that lies above thex-axis and letτ+ be the curve that has the sameendpoints asγ+: (2±
√24,0), but consists of the line segment from (2+
√24,0) to (1,0), the upper
half γ+0 of the unit circleγ0, and the line segment from (−1,0) to (2−√
24,0) (see Figure 20.2).Also, letD+ be the plane cut along the negativey-axis. Sinceγ+ andτ+ lie D+ andF is conservativein D+, Theorem 20.2 gives
∫
γ+
−yx2 + y2
dx+x
x2 + y2dy=
∫
τ+
−yx2 + y2
dx+x
x2 + y2dy.
x
y
γ+
τ+
Figure 20.2. The curvesγ+ andτ+ in D+
x
y
γ−
τ−
Figure 20.3. The curvesγ− andτ− in D−
Next, letγ− be the part ofγ that lies below thex-axis and letτ− be the curve that consists ofthe line segment from (2−
√24,0) to (−1,0), the lower halfγ−0 of the unit circleγ0, and the line
129
segment from (1,0) to (2+√
24,0) (see Figure 20.3). Also, letD− be the plane cut along thepositivey-axis. Sinceγ− andτ− lie D− andF is conservative inD−, Theorem 20.2 gives
∫
γ−
−yx2 + y2
dx+x
x2 + y2dy=
∫
τ−
−yx2 + y2
dx+x
x2 + y2dy.
Combining this identity with the one before it, we get∫
γ
F · dr =∫
γ+F · dr +
∫
γ−F · dr =
∫
τ+F · dr +
∫
τ−F · dr .
We break the integrals overτ+ andτ− into six integrals overγ+0 , γ−0 , and the two horizontal edges.Each horizontal edge appears in the sum twice: once orientedfrom left to right and once orientedfrom right to left. Thus, those contributions cancel out (recall (19.8)) and we get
∫
τ+F · dr +
∫
τ−F · dr =
∫
γ+0
F · dr +∫
γ−0
F · dr =∫
γ0
F · dr .
It follows that, as promised,∫
γ
−yx2 + y2
dx+x
x2 + y2dy=
∫
γ0
−yx2 + y2
dx+x
x2 + y2dy= 2π. �
Exercises
1. Find a potential for the given vector field and then use the fundamental theorem to evaluate∫
γ
F · dr :
(a)F(x, y) = eye1 + xeye2, γ is a smooth curve connecting (1,0) and (2,−1)(b) F(x, y) = (4x+ siny)e1 + (siny+ xcosy)e2, γ is given byr (t) = (t − sint)e1 + (1− cost)e2, 0 ≤ t ≤ 2π(c) F(x, y) = ye1 + (x+ 3y)e2, γ follows the ellipsex2 + 2y2 = 5 from (1,2) to (−
√5,0) and then the line
segment from (−√
5,0) to (1,−2)(d) F(x, y, z) = (yz+ x2)e1 + (xz+ 4y)e2 + (xy− 3z2)e3, γ consists of the line segments from (1,2,3) to
(0,2,1) to (−1,3,0) to (3,2,1)(e)F(x, y, z) = y2 cosze1 + 2xycosze2 − xy2 sinze3, γ is given byr (t) = t2e1 + t3e2 − 2t2e3, 0 ≤ t ≤ 2
130
21. Green’s theorem
In this lecture we discuss the first of the three major theorems of vector calculus. It is known asGreen’s theoremand provides an alternative technique for evaluation of certain line integrals.
21.1. Some definitions.First, we need to fix some terminology. A curveγ, given byr (t), a ≤ t ≤b, is calledclosed, if r (a) = r (b). In other words, the curve is closed, if its initital point is alsoits terminal point (see Figure 21.1). A curveγ is calledsimple, if r (t1) , r (t2) whenevert1 , t2,
b r (a) = r (b)
γ
Figure 21.1. A closed, positively oriented curve
except that we allowr (a) = r (b). In other words, a simple curve does not self-intersect anddoesnot touch itself, but can be closed. For example, among the curves displayed on Figure 21.2, thosein (a) and (e) are simple and those in (b)–(d) are not: the curve in (b) self-intersects at pointA; thecurve in (c) passes through its initial point before reaching its terminal point; and the curve in (d)touches itself at pointB.
(a)
b
A
(b)b
r (a)
(c)
b B
(d) (e)
Figure 21.2. Simple and non-simple curves
It is intuitively clear (though quite difficult to prove) that a piecewise smooth, simple, closedplane curveγ divides the plane into two sets, which we can call intuitively the interior and theexteriorof γ, the interior being the bounded set and the exterior the unbounded. This fact is knownasJordan’s theorem, named after the French mathematician Camille Jordan for hiswork on planecurves, including this result. However, the first full proofof Jordan’s theorem was given by anAmerican: Oswald Veblen.
131
Let γ be a piecewise smooth, simple, closed curve. We say thatγ is positively oriented, if itsinterior lies always to the left of the pointr (t) as it traces the curve in the direction of increasing oft. For “round” curves, such as a circle or the curve on Figure 21.2(e), the positive orientation of thecurve is the counterclockwise orientation. However, this is not the case for all curves: the positiveorientation goes clockwise for the part of the curve on Figure 21.1 that follows immediately theinitial point r (a).
21.2. Green’s theorem.
Theorem 21.1(Green). Let P(x, y) and Q(x, y) have continuous first-order partials on an open setU. Letγ be a positively oriented, piecewise smooth, simple, closed plane curve such thatγ andallof its interior D are contained inside U. Then
∫
γ
P dx+ Q dy=∫∫
D
(
∂Q∂x− ∂P∂y
)
dA. (21.1)
In other words, Green’s theorem relates a line integral along a simple, closed plane curves toa double integral over the interior of the curve. Since we know how to evaluate double integralsthis yields another possible integration technique for line integrals along closed curves. However,there is a catch: we absolutelymust check the conditions of Green’s theorem before we use it. Forexample, we showed in Example 18.6(c) that the functions
P(x, y) =−y
x2 + y2, Q(x, y) =
xx2 + y2
are differentiable and satisfy the condition∂P/∂y = ∂Q/∂x everywhere except at the origin. Thus,if γ is a simple, closed curve whose interiorD does not contain the origin, Green’s theorem gives
∫
γ
P dx+ Q dy=∫∫
D
(
∂Q∂x− ∂P∂y
)
dA= 0.
Note that for such a curve the application of Green’s theoremis legitimate. On the other hand, ifthe origin is in the interior of the curveγ, the conditions of Green’s theorem are violated (at onepoint) and the theorem is not applicable. Were we to apply it anyways, the result would again bethat the line integral alongγ is 0. For example, such an illegitimate application of Green’s theoremyields
∫
γ0
−yx2 + y2
dx+x
x2 + y2dy= 0
whenγ0 is the unit circlex2 + y2 = 1. However, we know from Example 20.2 that the last integralequals 2π, not 0. We see that if the functionsP(x, y) andQ(x, y) or some of their derivatives arenot defined at even a single point in the interior of the curve,Green’s theorem is not applicable andshould not be used. Now, let us take a look at some of its legitimate uses.
Example 21.1.Evaluate∫
γ
(
e4x sinx+ 2y)
dx+(
x2 + arctany)
dy,
whereγ is the rectangle with vertices (1,2), (5,2), (5,4), (1,4).132
Solution.We haveP(x, y) = e4x sinx+2y andQ(x, y) = x2+arctany. The partials of these functionsare:
Px(x, y) = (4 sinx+ cosx)e4x, Py(x, y) = 2, Qx(x, y) = 2x, Qy(x, y) =1
1+ y2.
Since these are all continuous everywhere, the open setU in the statement of Green’s theorem canbe taken to beR2. We can therefore apply Green’s theorem. We get
∫
C
(
e4x sinx+ 2y)
dx+(
x2 + arctany)
dy=∫∫
D(2x− 2)dA=
∫ 5
1
∫ 4
2(2x− 2)dydx
=
∫ 5
12(2x− 2)dx= 2
[
x2 − 2x]5
1= 32.
Here,D denotes the rectangle [1,5] × [2,4]. �
Example 21.2.Evaluate∫
γ
(
e4x sinx+ 2y)
dx+(
x2 + arctany)
dy,
whereγ is the boundary of the semiannular region shown on Figure 21.3 (the radii of the twosemicircles are 2 and 3).
x
y(0,3)
(0,−2)
γ
Figure 21.3. The regionD
Solution.The integrand is the same as in the previous problem, but the curve is different. We canappeal to Green’s theorem for the same reasons as in the previous example. We get
∫
γ
(
e4x sinx+ 2y)
dx+(
x2 + arctany)
dy=∫∫
D(2x− 2)dA,
whereD is the semiannular region shown on Figure 21.3. Note that this region is convenientlydescribed in polar coordinates:
D ={
(r, θ) : 2 ≤ r ≤ 3, −π/2 ≤ θ ≤ π/2}
.133
Thus, by converting the double integral to polar coordinates, we obtain∫
γ
(
e4x sinx+ 2y)
dx+(
x2 + arctany)
dy=∫ 3
2
∫ π/2
−π/2(2r cosθ − 2)r dθdr
=
∫ 3
2
[
(2r sinθ − 2θ)r]π/2
−π/2 dr
=
∫ 3
2
(
4r2 − 2πr)
dr = 2513 − 5π. �
In the above examples, we used Green’s theorem to evaluate line integrals by means of doubleintegrals. However, Green’s theorem is a relation between the line integral and the double integralappearing in (21.1) that can be used in either direction. Sometimes, we want to use Green’s theoremto evaluate a double integral by means of a line integral. Forexample, under the assumption ofTheorem 21.1, we have
∫
γ
x dy=∫∫
D
(
∂x(x) − ∂y(0))
dA=∫∫
D1dA= area(D),
∫
γ
(−y) dx=∫∫
D
(
∂x(0)− ∂y(−y))
dA=∫∫
D1dA= area(D).
In other words, we have the following new formulas for the area enclosed by a simple, closedcurve.
Theorem 21.2.Let D be a plane region, bounded by a piecewise smooth, simple,closed curveγ.Then
area(D) =∫
γ
x dy=∫
γ
(−y) dx=12
∫
γ
x dy− y dx. (21.2)
Example 21.3.Find the area enclosed by the ellipsex2
a2+
y2
b2= 1.
Solution.We will use formula (21.2) and the parametric representation of the ellipse:
x = acost, y = bsint (0 ≤ θ ≤ 2π).
By (21.2),
area(D) =12
∫
Cx dy− y dx=
12
∫ 2π
0
[
(acost)(bcost) − (bsint)(−asint)]
dt
=12
∫ 2π
0ab(
cos2 t + sin2 t)
dt =12
∫ 2π
0ab dt= πab. �
21.3. Vector forms of Green’s theorem*. We can use the curl and divergence of a vector field togive alternative formulations of Green’s theorem. Supposethatγ is a positively oriented, piecewisesmooth, simple, closed curve with parametrization
r (t) = x(t)e1 + y(t)e2 + 0e3 (a ≤ t ≤ b),
that is,γ is a plane curve viewed as a space curve. Furhter, consider the vector field
F(x, y, z) = P(x, y)e1 + Q(x, y)e2 + 0e3.134
By (19.12),∫
γ
F · dr =∫
γ
P(x, y) dx+ Q(x, y) dy.
Also, by the definition of curl,
curlF =(
∂0∂y− ∂Q∂z
)
e1 +
(
∂P∂z− ∂0∂x
)
e2 +
(
∂Q∂x− ∂P∂y
)
e3 =
(
∂Q∂x− ∂P∂y
)
e3,
becauseP(x, y) andQ(x, y) do not depend onz. Thus, we can rewrite (21.1) as∫
γ
F · dr =∫∫
D(curlF) · e3 dA. (21.3)
Another formulation of Green’s theorem involves the vectorfield
G(x, y, z) = Q(x, y)e1 − P(x, y)e2 + 0e3,
which is orthogonal to the vector fieldF that we used in (21.3):
F ·G = P(x, y)Q(x, y) + Q(x, y)(−P(x, y)) + 0 = 0.
Note that
div F =∂Q∂x− ∂P∂y.
Also, by (19.7),∫
γ
F · r =∫
γ
(F · T) ds,
whereT is the unit tangent vector toγ. It is a nice exercise to try to use the properties of the dotproduct to convince yourself that
n(t) =y′(t)e1 − x′(t)e2√
x′(t)2 + y′(t)2
is a unit vector that is normal to the curveγ and points towards the exterior of the curve; it is calledtheoutward normalvector ofγ. SinceF · T = G · n, we get
∫
γ
F · r =∫
γ
(G · n) ds,
and we can restate (21.1) as∫
γ
(G · n) ds=∫∫
Ddiv G dA. (21.4)
Exercises
1. (a) Letγ be the square with vertices (1,1), (2,1), (2,2), and (1,2). Parametrizeγ and use the formulas fromLecture #19 to evaluate the integral
∫
γ
x2y dx+ xy2 dy.
(b) Evaluate the integral from part (a) by means of Green’s theorem.135
2. Use Green’s theorem to evaluate the given line integral (all the curves are assumed positively oriented):
(a)∫
γ
y3 dx− x3 dy, γ is the circlex2 + y2 = 4
(b)∫
γ
(xy+ tanx) dx+(
x2 + ecosy)
dy, γ is the boundary of the region bounded byy = x andy = x2
(c)∫
γ
xy3 dx+ 4x2y2 dy, γ is consists of the semicirclex =√
9− y2 and the line segment from (0,3) to
(0,−3)
(d)∫
γ
(
xy2+sinx2)
dx+(
(x2+y2)3/2+arctany)
dy, γ is the boundary of region bounded by the semicircle
y =√
4− x2 and the linesy = x andy = −√
3x
(e)∫
γ
ex siny dx+ ex cosy dy, γ is the ellipse (x+ 1)2 + 5(y− 2)2 = 7
3. Use Green’s theorem to evaluate the line integral∫
γ
F · dr for the given fieldF(x, y) and the given curveγ:
(a) F(x, y) =(
y3/2 + x)
e1 +(
x2 + y)
e2, γ consists of the graph ofy = sin2 x from (0,0) to (π,0) and theline segment from (π,0) to (0,0)
(b) F(x, y) =(
x3 + y)
e1 +(
2xy+ y4)
e2, γ is the cardioidr = 2+ cosθ, positively oriented(c) F(x, y) = xe1 +
(
x3 + 3xy2)
e2, γ consists of the graph ofy = 4− x2 from (−2,0) to (2,0) and the linesegment from (2,0) to (−2,0)
4. When the circlex2 + (y− 1)2 = 1 is rolled along thex-axis in the positive direction, the point whose originalposition is the origin traces a curve called acycloid (see Figure 21.4). This curve can be parametrized by thevector function
r (t) = (t − sint)e1 + (1− cost)e2 (t ≥ 0).
Use Theorem 21.2 to find the area bounded by thex-axis and the first arc of the cycloid (that is, the part ofthe cycloid with 0≤ t ≤ 2π).
Parameters: Plotting Area Axes Division Printing Format
x�Axis:
y�Axis:
[ 0.0 | 16.0 ]
[ 0.0 | 16.0 ]
1E = automatic
1E = automatic
1E = 0.5 cm
1E = 0.5 cm
Functions:
xf(t)=t�sin(t)
yf(t)=1�cos(t)
Figure 21.4. Cycloid
5. When a circle of radius 1 is rolled along the outside of the circle x2+y2 = 4 a fixed point on the smaller circletraces a curve called anepicycloid(see Figure 21.5). This curve can be parametrized by the vector function
r (t) = (3 cost − cos 3t)e1 + (3 sint − sin 3t)e2 (0 ≤ t ≤ 2π).
Use Theorem 21.2 to find the area enclosed by this epicycloid.
Parameters: Plotting Area Axes Division Printing Format
x�Axis:
y�Axis:
[ �8.0 | 8.0 ]
[ �8.0 | 8.0 ]
1E = automatic
1E = automatic
1E = 0.5 cm
1E = 0.5 cm
Functions:
xf(t)=3*cos(t)�cos(3t)
yf(t)=3*sin(t)�sin(3t)
Figure 21.5. Epicycloid
Parameters: Plotting Area Axes Division Printing Format
x�Axis:
y�Axis:
[ �8.0 | 8.0 ]
[ �8.0 | 8.0 ]
1E = automatic
1E = automatic
1E = 0.5 cm
1E = 0.5 cm
Functions:
xf(t)=2*cos(t)+cos(2t)
yf(t)=2*sin(t)�sin(2t)
Figure 21.6. Hypocycloid
Parameters: Plotting Area Axes Division Printing Format
x�Axis:
y�Axis:
[ �8.0 | 8.0 ]
[ �8.0 | 8.0 ]
1E = automatic
1E = automatic
1E = 0.5 cm
1E = 0.5 cm
Functions:
xf(t)=3cos(t)+cos(3t)
yf(t)=3sin(t)�sin(3t)
Figure 21.7. Astroid
136
6. When a circle of radius 1 is rolled along the inside of the circle x2 + y2 = 9 a fixed point on the smaller circletraces a curve called ahypocycloid(see Figure 21.6). This curve can be parametrized by the vector function
r (t) = (2 cost + cos 2t)e1 + (2 sint − sin 2t)e2 (0 ≤ t ≤ 2π).
Use Theorem 21.2 to find the area enclosed by this hypocycloid.
7. When a circle of radius 1 is rolled along the inside of the circle x2+y2 = 16 a fixed point on the smaller circletraces a different kind of hypocycloid called also anastroid(see Figure 21.7). This curve can be parametrizedby the vector function
r (t) = (3 cost + cos 3t)e1 + (3 sint − sin 3t)e2 (0 ≤ t ≤ 2π).
Use Theorem 21.2 to find the area enclosed by this astroid.
137
22. Surface integrals
22.1. Surface integrals of scalar functions.We start with a smooth parametric surfaceΣ, givenby the vector function
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D).
Recall that the smoothness ofΣmeans thatru(u, v) andr v(u, v) are continuous and (ru×r v)(u, v) , 0in D (see§10.3). Further, suppose thatf (x, y, z) is a continuous function on a setE in R3 thatcontains the curveΣ. In particular,f (x, y, z) is defined at each point (x, y, z) onΣ.
For simplicity, we shall assume thatD is a rectangle in theuv-plane:D = [a,b] × [c,d]. We cantreat the general case similarly, but there are more technical details to pay attention to and littleextra insight to gain. For eachn ≥ 1, leta = u0 < u1 < · · · < un = b be the points that divide theinterval [a,b] into n subintervals of equal lengths; letc = v0 < v1 < · · · < vn = d be the points thatdivide the interval [c,d] into n subintervals of equal lengths. The points
(un, v0) (un, v1) · · · (un, vj) · · · (un, vn),...
......
...(ui , v0) (ui , v1) · · · (ui , vj) · · · (ui , vn),...
......
...(u1, v0) (u1, v1) · · · (u1, vj) · · · (u1, vn),(u0, v0) (u0, v1) · · · (u0, vj) · · · (u0, vn),
define a partition of the rectangleD into n2 congruent subrectangles. The vector functionr mapseach of those small rectangles to a small “patch” onΣ and each grid point (ui , vj) to a pointPi j onthe surface. The patches form a partition ofΣ into n2 pieces, which we label asΣi j , 1 ≤ i, j ≤ n,so thatΣi j is the image of the rectangle that has (ui , vj) as its upper right corner point. Further, wepick n2 sample points (u∗i , v
∗j ) so that
ui−1 ≤ u∗i ≤ ui , vj−1 ≤ v∗j ≤ vj .
These yieldn2 sample points P∗i j on the surface, with coordinatesx∗i j = x(u∗i , v∗j ), y∗i j = y(u∗i , v
∗j ),
z∗i j = z(u∗i , v∗j ). Then thesurface integral of f overΣ is defined by∫∫
Σ
f (x, y, z) dS = limn→∞
n∑
i=1
n∑
j=1
f (x∗i j , y∗i j , z∗i j )∆Si j , ∆Si j = area(Σi j ), (22.1)
provided that the limit exists.Recall that, by Theorem 17.1,
∆Si j =
∫∫
Di j
‖ru × r v‖dA,
whereDi j is the rectangle in theuv-plane whose image isΣi j . Using this formula, the abovedefinition, the properties of continuous functions, and thedefinition of the double integral, it canbe shown that for continuous functionsf the limit (22.1) always exists and the surface integral canbe expressed as a double integral.
Theorem 22.1.LetΣ be a smooth parametric surface, given by the vector function
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D).138
If f (x, y, z) is continuous onΣ, then∫∫
Σ
f (x, y, z) dS =∫∫
Df (r (u, v))‖ru × r v‖dA. (22.2)
Similarly to line integrals, we can extend the above definition of the surface integral to piecewisesmooth surfaces. IfΣ is a piecewise smooth surface that is the union of the smooth surfacesΣ1,Σ2, . . . ,Σn, Σ = Σ1 ∪ Σ2 ∪ · · · ∪ Σn, then we define
∫∫
Σ
f (x, y, z) dS =∫∫
Σ1
f (x, y, z) dS+∫∫
Σ2
f (x, y, z) dS+ · · · +∫∫
Σn
f (x, y, z) dS.
Remark. Like all other types of integrals we have encountered so far,the surface integral can beinterpreted in terms of averages. In this case, we are dealing with the average value of the functionon the surface:
average(x,y,z)∈Σ
f (x, y, z) =1
area(Σ)
∫∫
Σ
f (x, y, z) dS. (22.3)
Example 22.1.Evaluate∫∫
Σ
xy dS,
whereΣ is the graph ofz= 23
(
x3/2 + y3/2)
over the rectangle 0≤ x, y ≤ 1.
Solution.The surface can be parametrized by the vector function
r (u, v) = ue1 + ve2 +23
(
u3/2 + v3/2)
e3 (0 ≤ u, v ≤ 1).
Hence,
ru(u, v) = e1 + 0e2 + u1/2e3, r v(u, v) = 0e1 + e2 + v1/2e3,
and
ru × r v =
∣
∣
∣
∣
∣
∣
e1 e2 e3
1 0 u1/2
0 1 v1/2
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
0 u1/2
1 v1/2
∣
∣
∣
∣
e1 −∣
∣
∣
∣
1 u1/2
0 v1/2
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
1 00 1
∣
∣
∣
∣
e3 = −u1/2e1 − v1/2e2 + e3.
Applying (22.2), we get∫∫
Sxy dS=
∫∫
0≤u,v≤1
uv√
u+ v+ 1dA=∫ 1
0
[
v∫ 1
0u√
u+ v+ 1du
]
dv. (∗)
Using the integration by parts, we find that∫ 1
0u√
u+ v+ 1du= 23
∫ 1
0u d(u+ v+ 1)3/2
=[
23u(u+ v+ 1)w3/2
]1
0− 2
3
∫ 1
0(u+ v+ 1)3/2 du
= 23(v+ 2)3/2 − 2
3
[
25(u+ v+ 1)5/2
]1
0
= 23(v+ 2)3/2 − 4
15(v+ 2)5/2 + 415(v+ 1)5/2.
139
Inserting the result back into (∗) and using another integration by parts, we obtain∫∫
Sxy dS=
∫ 1
0v[
23(v+ 2)3/2 − 4
15(v+ 2)5/2 + 415(v+ 1)5/2
]
dv
=
∫ 1
0vd(
415(v+ 2)5/2 − 8
105(v+ 2)7/2 + 8105(v+ 1)7/2
)
=[
v(
415(v+ 2)5/2 − 8
105(v+ 2)7/2 + 8105(v+ 1)7/2
)]1
0
−∫ 1
0
(
415(v+ 2)5/2 − 8
105(v+ 2)7/2 + 8105(v+ 1)7/2
)
dv
= 415
(
35/2 − 2737/2 + 2
727/2)
− 415
[
27(v+ 2)7/2 − 4
63(v+ 2)9/2 + 463(v+ 1)9/2
]1
0
= 4105
(
9√
3+ 16√
2)
− 8105
19
(
81√
3− 8√
2− 2)
≈ 0.372. �
22.2. Surface integrals of vector fields.
Definition. Let F be a continuous vector field defined on an oriented surfaceΣ with unit normalvectorn. Then thesurface integral ofF overΣ is
∫∫
Σ
F · dS=∫∫
Σ
(F · n) dS. (22.4)
Remark. SinceF · n is the size of the projection ofF in the direction ofn, we can interpret thesurface integral ofF as the average size of its projection onto the direction of orientation ofΣ.
Suppose that the surface is given by the vector function
r (u, v) = x(u, v)e1 + y(u, v)e2 + z(u, v)e3 ((u, v) ∈ D),
and that it is oriented using the unit normal vector
n(u, v) =ru × r v
‖ru × r v‖.
Combining (22.4) and (22.2), we get∫∫
Σ
F · dS=∫∫
Σ
(F · n) dS =∫∫
Σ
(
F · ru × r v
‖ru × r v‖
)
dS
=
∫∫
D
(
F · ru × r v
‖ru × r v‖
)
‖ru × r v‖dA=∫∫
D
(
F · (ru × r v))
dA. (22.5)
In particular, whenΣ is the graphz = f (x, y), (x, y) ∈ D, with the upward orientation (recallExample 10.7), we obtain
∫∫
Σ
(Pe1 + Qe2 + Re3) · dS=∫∫
D
(
− P(x, y) fx(x, y) − Q(x, y) fy(x, y) + R(x, y))
dA. (22.6)
Example 22.2.Evaluate∫∫
Σ
(xe1 − 2ye2 + ze3) · dS,
whereΣ is the elliptic paraboloidz= 4− 2x2 − 2y2, z≥ 0, with the upward orientation.140
Solution.Conditionz≥ 0 restrictsx, y to the circle
4− 2x2 − 2y2 ≥ 0 ⇐⇒ x2 + y2 ≤ 2.
Since we are using the upward orientation, we can appeal to (22.6). We have
∂x
(
4− 2x2 − 2y2)
= −4x, ∂y
(
4− 2x2 − 2y2)
= −4y,
so (22.6) gives∫∫
Σ
(xe1 − 2ye2 + ze3) · dS=∫∫
x2+y2≤2
(
4x2 − 8y2 + z(x, y))
dA
=
∫∫
x2+y2≤2
(
4x2 − 8y2 + 4− 2x2 − 2y2)
dA
=
∫∫
x2+y2≤2
(
4+ 2x2 − 10y2)
dA.
Converting the double integral to polar coordinates, we obtain∫∫
Σ
(xe1 − 2ye2 + ze3) · dS=∫ 2π
0
∫
√2
0
(
4+ 2r2 cos2 θ − 10r2 sin2 θ)
r drdθ
=
∫ 2π
0
[
2r2 + 12r4 cos2 θ − 5
2r4 sin2 θ]
√2
0dθ
=
∫ 2π
0
(
4+ 2 cos2 θ − 10 sin2 θ)
dθ
=
∫ 2π
0
(
4+ (1+ cos 2θ) − 5(1− cos 2θ))
dθ = 0. �
Example 22.3.Evaluate∫∫
Σ
(xze1 + xe2 + ye3) · dS,
whereΣ is the hemispherex2 + y2 + z2 = 25,y ≥ 0, oriented in the direction of the positivey-axis.
Solution. It’s best to represent the hemisphere in spherical coordinates:
r (θ,φ) = 5 cosθ sinφe1 + 5 sinθ sinφe2 + 5 cosφe3 (0 ≤ θ,φ ≤ π).Then
r θ = −5 sinθ sinφe1 + 5 cosθ sinφe2,
rφ = 5 cosθ cosφe1 + 5 sinθ cosφe2 − 5 sinφe3;
r θ × rφ = −25 sin2 θ sinφ cosφ(e1 × e2) + 25 sinθ sin2φ(e1 × e3)
+ 25 cos2 θ sinφ cosφ(e2 × e1) − 25 cosθ sin2φ(e2 × e3)
= −25 cosθ sin2φe1 − 25 sinθ sin2φe2 − 25 sinφ cosφe3
= −25 sinφ(cosθ sinφe1 + sinθ sinφe2 + cosφe3).
141
Note that the vectorr θ × rφ has a negative second component for allθ,φ with 0 ≤ θ,φ ≤ π.Therefore, we can orientΣ in the direction of the positivey-axis by taking
n =−(r θ × rφ)‖r θ × rφ‖
=rφ × r θ‖rφ × r θ‖
.
With this choice, we have
F(r (θ,φ)) = 25 cosθ sinφ cosφe1 + 5 cosθ sinφe2 + 5 sinθ sinφe3
= 5 sinφ(5 cosθ cosφe1 + cosθe2 + sinθe3),
F(r (θ,φ)) ·(
rφ × r θ)
= 125 sin2φ(
5 cos2 θ sinφ cosφ + sinθ cosθ sinφ + sinθ cosφ)
.
Thus, (22.5) yields∫∫
Σ
F · dS=∫∫
0≤θ,φ≤π
125 sin2φ(5 cos2 θ sinφ cosφ + sinθ cosθ sinφ + sinθ cosφ) dA
= 125∫ π
0
∫ π
0
(
5 cos2 θ sin3φ cosφ + sinθ cosθ sin3φ + sinθ sin2φ cosφ)
dθdφ.
We now use that∫ π
0cos2 θdθ =
π
2,
∫ π
0sinθ cosθdθ = 0,
∫ π
0sinθdθ = 2.
Substituting these values in the iterated integral above, we obtain∫∫
SF · dS= 125
∫ π
0(2.5π sin3φ cosφ + 2 sin2φ cosφ) dφ
= 125∫ 0
0(2.5πu3 + 2u2) du= 0. �
Figure 22.1. Two views of the helicoid
Example 22.4.The surfaceΣ with vector equation
r (u, v) = ucosve1 + usinve2 + ve3,
where 0≤ u ≤ 1, 0≤ v ≤ π, is calledhelicoid (see Figure 22.1). Evaluate the surface integral of
F(x, y, z) = x√
1+ x2 + y2e1 + y√
1+ x2 + y2e2 + (x+ y)ze3142
overΣ with upward orientation.
Solution.We have
ru = cosve1 + sinve2, r v = −usinve1 + ucosve2 + e3;
ru × r v = ucos2 v(e1 × e2) + cosv(e1 × e3) − usin2 v(e2 × e1) + sinv(e2 × e3)
= sinve1 − cosve2 + ue3.
Note that the vectorru × r v has a positive third component for allu with 0 ≤ u ≤ 1. Therefore, wecan orientΣ in the upward direction by the unit normal vector
n =ru × r v
‖ru × r v‖.
With this choice, we have
F(r (u, v)) = ucosv√
1+ u2e1 + usinv√
1+ u2e2 + uv(cosv+ sinv)e3
F(r (u, v)) · (ru × r v) = u2v(cosv+ sinv).
Thus, (22.5) yields∫∫
Σ
F · dS=∫ π
0
∫ 1
0u2v(cosv+ sinv) dudv
=13
∫ π
0v(cosv+ sinv) dv=
13
∫ π
0v d(sinv− cosv)
=13
[
v(sinv− cosv)]π
0− 1
3
∫ π
0(sinv− cosv) dv=
π
3− 2
3. �
Exercises
1. Evaluate the surface integral∫∫
Σx2yz dS, whereΣ is the surface
z= 1+ 2x+ 3y, 0 ≤ x ≤ 3, 0 ≤ y ≤ 2,
that is, the part of the planez= 1+ 2x+ 3y that lies above the rectangle [0,3] × [0,2].
2. Evaluate the surface integral∫∫
Σx2z2 dS, whereΣ is the surface
z=√
x2 + y2, 1 ≤ x2 + y2 ≤ 4,
that is, the part of the conez2 = x2 + y2 that lies between the planesz= 1 andz= 2.
3. Evaluate the surface integral∫∫
Σ
(
x2z+ y2z)
dS, whereΣ is the hemispherex2 + y2 + z2 = 4, z ≥ 0, that is,the parametric surface given by
r (u, v) = 2 cosusinve1 + 2 sinusinve2 + 2 cosve3, 0 ≤ u ≤ 2π, 0 ≤ v ≤ π/2.
4. Evaluate the surface integral∫∫
Σ
(
x2y+z2)
dS, whereΣ is the part of the cylinderx2+y2 = 9 that lies betweenthe planesz= 0 andz= 2, that is, the parametric surface given by
r (u, v) = 3 cosue1 + 3 sinue2 + ve3, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2.
5. Evaluate the surface integral∫∫
ΣF ·dS, whereF(x, y, z) = xye1+yze2+zxe3 andΣ is the part of the paraboloid
z= 4− x2 − y2 that lies above the square 0≤ x, y ≤ 1 and is oriented upward.
6. Evaluate the surface integral∫∫
ΣF · dS, whereF(x, y, z) = xzeye1 − xzeye2 + ze3 andΣ is the part of the plane
x+ y+ z= 1 in the first octant with the upward orientation.143
7. Evaluate the surface integral∫∫
ΣF · dS, whereF(x, y, z) = xze1 + 2yze3 andΣ is the paraboloidy = x2 + z2,
0 ≤ y ≤ 1, oriented in the direction of the positivey-axis.
8. Evaluate the surface integral∫∫
ΣF · dS, whereF(x, y, z) = xe1 + ye2 + xyze3 andΣ is the helicoid given by
r (u, v) = ucosve1 + usinve2 + ve3, 0 ≤ u ≤ 1, 0 ≤ v ≤ π,and oriented byn = ru × r v.
144
23. Stokes’ theorem
In the last two lectures, we discuss the other two major integral theorems of vector calculus:Stokes’ theorem and Gauss’ divergence theorem. Stokes’ theorem is a generalization of Green’stheorem to closed curves on surfaces in space. Before we stateit, we introduce the notion oforientation of curves on an oriented surface.
Definition. Suppose thatΣ is an oriented surface with unit normal vectorn. We say that a closedcurveγ onΣ is positively oriented, if the vectorT × n points towards the interior of the curve; hereT is the unit tangent vector toγ.
Theorem 23.1(Stokes). LetF be a vector field that has continuous partial derivatives on an openset U inR3. LetΣ be an oriented, piecewise smooth surface that is contained inU and is boundedby a simple, closed, piecewise smooth, positively oriented curveγ. Then
∫
γ
F · dr =∫∫
Σ
curlF · dS. (23.1)
We said above that Stokes’ theorem is a generalization of Green’s. It would be nice to corrobo-rate that. Consider the parametric surfaceΣ given by
r (x, y) = xe1 + ye2 + 0e3 ((x, y) ∈ D),
that is,Σ is the regionD in the xy-plane turned into a “three-dimensional surface”. If we give thissurface an upward orientation, then the unit normal vectorn is e3, and Stokes’ theorem gives
∫
γ
F · dr =∫∫
Σ
curlF · dS=∫∫
Σ
(curlF) · e3 dS.
Converting the latter surface integral into a double integral, we find
r x(x, y) = e1, r y(x, y) = e2, r x × r y = e3,
and hence,∫
γ
F · dr =∫∫
Σ
(curlF) · e3 dS =∫∫
D(curlF) · e3 dA.
This is exactly one of the vector forms of Green’s theorem (recall (21.3)).We now move to describe the main applications of Stokes’ theorem. They are similar to the
applications of Green’s theorem. First, we can apply Stokes’s theorem to convert a line integralover a closed space curveγ to a surface integral over a surfaceΣ that hasγ as its boundary.
Example 23.1.Evaluate∫
γ
[(
ex3+x + z2)
e1 +(
arctany+ x2)
e2 +(
ln(z2 + 1)+ y2)
e3
]
· dr ,
whereγ is the positively oriented boundary of the part of the spherex2+ y2+ z2 = 9 that lies in thefirst octant, with downward orientation.
Solution.All the partials ofF will be continuous everywhere, so we can apply Stokes’ theorem.We get
∫
γ
F · dr =∫∫
Σ
curlF · dS,
145
whereΣ is the given surface and
F(x, y, z) =(
ex3+x + z2)
e1 +(
arctany+ x2)
e2 +(
ln(z2 + 1)+ y2)
e3.
As we usually do when dealing with spheres, we parametrizeΣ using spherical coordinates:
r (θ,φ) = 3 cosθ sinφe1 + 3 sinθ sinφe2 + 3 cosφe3 (0 ≤ θ,φ ≤ π/2).
Then, similarly to Example 22.3,
r θ = −3 sinθ sinφe1 + 3 cosθ sinφe2, rφ = 3 cosθ cosφe1 + 3 sinθ cosφe2 − 3 sinφe3;
r θ × rφ = −9 sinφ(cosθ sinφe1 + sinθ sinφe2 + cosφe3).
Since−9 sinφ cosφ ≤ 0 for all φ, 0 ≤ φ ≤ π/2, the downward orientation ofΣ is given by
n =r θ × rφ‖r θ × rφ‖
.
Furthermore, we have
curlF =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
ex3+x + z2 arctany+ x2 ln(z2 + 1)+ y2
∣
∣
∣
∣
∣
∣
= (2y− 0)e1 − (0− 2z)e2 + (2x− 0)e3 = 2ye1 + 2ze2 + 2xe1;
curlF(r (θ,φ)) = 6 sinθ sinφe1 + 6 cosφe2 + 6 cosθ sinφe3,(
curlF(r (θ,φ)))
· (r θ × rφ) = −54 sinφ(
sinθ cosθ sin2φ + sinθ sinφ cosφ + cosθ sinφ cosφ)
= −54 sin2φ(
sinθ cosθ sinφ + (sinθ + cosθ) cosφ)
.
We conclude that∫
γ
F · dr =∫∫
0≤θ,φ≤π/2
(
curlF(r (θ,φ)))
· (r θ × rφ) dA
= −54∫ π/2
0
∫ π/2
0sin2φ
(
sinθ cosθ sinφ + (sinθ + cosθ) cosφ)
dθdφ
= −54∫ π/2
0sin2φ
[
12 sin2 θ sinφ + (− cosθ + sinθ) cosφ
]π/2
0dφ
= −54∫ π/2
0sin2φ
(
12 sinφ + 2 cosφ
)
dφ
= −27∫ π/2
0
(
1− cos2φ)
sinφdφ − 108∫ π/2
0sin2φ cosφdφ
= −27∫ 1
0
(
1− u2)
du− 108∫ 1
0u2 du= −54. �
Example 23.2.Evaluate∫
γ
(yze1 + 2xze2 + 3xye3) · dr ,
whereγ is the intersection curve of the ellipsoidx2+ 2y2+ z2 = 4 and the planex+ z= 2, orientedpositively relative to the outward orientation of the ellipsoid.
146
Figure 23.1. The intersection curve ofx2 + 2y2 + z2 = 4 andx+ z= 2
Solution.The intersection curveγ of the ellipsoid and the plane is displayed on Figure 23.1. Itcanbe seen from the figure that the positive orientation ofγ relative to the outward orientation of theellipsoid is in the counterclockwise direction. Note that this is also the positive orientation ofγrelative to the upward orientation of the planex+ z= 2. The equations ofγ are
x2 + 2y2 + z2 = 4, x+ z= 2 ⇐⇒ x2 + 2y2 + (2− x)2 = 4, z= 2− x.
Some basic algebra shows that
x2 + 2y2 + (2− x)2 = 4 ⇐⇒ x2 + 2y2 + 4− 4x+ x2 = 4 ⇐⇒ 2x2 − 4x+ 2y2 = 0
⇐⇒ x2 − 2x+ y2 = 0 ⇐⇒ (x− 1)2 + y2 = 1,
soγ is described by the equations
(x− 1)2 + y2 = 1, z= 2− x.
That is,γ contains the points (x, y, z) on the planez = 2 − x for which the point (x, y) lies on thecircle (x−1)2+ y2 = 1 in thexy-plane. It is possible to parametrizeγ and to evaluate the given lineintegral directly (see the exercises), but we want to use Stokes’ theorem.
Let Σ be the portion of the planez= 2− x that lies withinγ, with the upward orientation. Then,by Stokes’ theorem,
∫
γ
(yze1 + 2xze2 + 3xye3) · dr =∫∫
Σ
curl(yze1 + 2xze2 + 3xye3) · dS.
We have
curl(yze1 + 2xze2 + 3xye3) =
∣
∣
∣
∣
∣
∣
e1 e2 e3
∂x ∂y ∂z
yz 2xz 3xy
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∂y ∂z
2xz 3xy
∣
∣
∣
∣
e1 −∣
∣
∣
∣
∂x ∂z
yz 3xy
∣
∣
∣
∣
e2 +
∣
∣
∣
∣
∂x ∂y
yz 2xz
∣
∣
∣
∣
e3
= (3x− 2x)e1 − (3y− y)e2 + (2z− z)e3 = xe1 − 2ye2 + ze3.
147
Further,Σ is the graph ofz = 2− x for (x, y) in the diskD with equation (x− 1)2 + y2 ≤ 1. Thus,(22.6) gives
∫∫
Σ
curl(yze1 + 2xze2 + 3xye3) · dS=∫∫
Σ
curl(xe1 − 2ye2 + ze3) · dS
=
∫∫
D(−x∂x(2− x) − (−2y)∂y(2− x) + z(x, y)) dA
=
∫∫
D(x+ 0+ (2− x)) dA=
∫∫
D2dA.
The last double integral represents twice the area ofD, which is a disk of radius 1. Consequently,∫
γ
(yze1 + 2xze2 + 3xye3) · dr =∫∫
D2dA= 2area(D) = 2π. �
In the last two examples, we used Stokes’ theorem to reduce a line integral of a vector fieldto a surface integral of its curl. In the next example, we describe the other main application ofStokes’ theorem, which is to reduce a surface integral of a vector field to a line integral along theboundary curve of the surface. This is a much more practical application, because it reduces a two-dimensional integral to a single-dimensional one. However, this application of Stokes’ theoremcomes with one caveat, and it is a big one: we need to recognizethe given vector field as the curlof some other vector field, and this is a difficult problem.
Example 23.3.Evaluate∫∫
Σ
(xe1 − 2ye2 + ze3) · dS,
whereΣ is the part of the paraboloidz= 4−2x2−2y2 that lies above thexy-plane, oriented upward.
We evaluated this integral in Example 22.2 by writing it as a double integral. Now, we give analternative solution, which uses Stokes’ theorem in reverse.
Figure 23.2. The paraboloidz= 4− 2x2 − 2y2, z≥ 0 and its boundary
Solution.We recall from the previous example that
xe1 − 2ye2 + ze3 = curlF, F = yze1 + 2xze2 + 3xye3.
Hence, when we apply Stokes’ theorem toF onΣ, we get∫∫
Σ
(xe1 − 2ye2 + ze3) · dS=∫∫
Σ
curlF · dS=∫
γ
F · dr ,
148
whereγ is the boundary of the given surface, positively oriented. The curveγ is the intersectioncurve of the paraboloid and thexy-plane, that is, the circlex2 + y2 = 2, z= 0. We can parametrizeγ using
r (t) =√
2 coste1 +√
2 sinte2 + 0e3 (0 ≤ t ≤ 2π).Note that with this parametrization,γ is positively oriented relative to the upward orientation of Σ(see Figure 23.2). Thus,
∫
γ
F · dr =∫ 2π
0F(r (t)) · r ′(t) dt
=
∫ 2π
0
(
0e1 + 0e2 + 6 cost sinte3
)
·(
−√
2 sinte1 +√
2 coste2
)
dt = 0. �
Remark. In the last example, the vector fieldF resulted from a “happy coincidence”. Had we notjust worked out Example 23.2, we would have started the solution of Example 23.3 by “observing”that
xe1 − 2ye2 + ze3 = curlF, F = yze1 + 2xze2 + 3xye3.
Such an “observation” falls just short of reliance upon “divine intervention”. In general, given asurface integral
∫∫
ΣG · dS, in order to argue similarly to Example 23.3, we need to find a vector
field F such that curlF = G is some given vector field. This is a difficult problem, which limits theapplicability of the method. Still, the above example is nota complete waste of time: sometimes,one has to calculate the surface integral of the curl of a given field, and in those situations one canargue as above.
Exercises
1. Use Stokes’ theorem to evaluate∫
γ
F · dr for the given fieldF(x, y, z) and the given curveγ:
(a)F(x, y, z) = 2ze1 + xe2 + 3ye3, γ is the ellipse that is the intersection of the planez= x and the cylinderx2 + y2 = 4, oriented clockwise
(b) F(x, y, z) = (z− y)e1+ (z− x)e2+ (x+ y)e3, γ is the triangle with vertices (0,1,0), (0,2,2) and (1,2,2),oriented counterclockwise
(c) F(x, y, z) = x2e1 − 2xye2 + yz2e3, γ is the boundary of the surfacez = x2y, 0 ≤ x, y ≤ 1, orientedcounterclockwise
(d) F(x, y, z) = x2ye1+13 x3e2+ xye3, γ is the curve of intersection of the hyperbolic paraboloidz= y2− x2
and the cylinderx2 + y2 = 1, oriented counterclockwise as viewed from above
2. Use Stokes’ theorem to evaluate∫∫
Σ
curlF · dS for the given fieldF(x, y, z) and the given surfaceΣ:
(a)F(x, y, z) = 3ye1− xze2−yz2e3, Σ is the part of the elliptic paraboloid 2z= x2+y2 belowz= 2, orientedupward
(b) F(x, y, z) = yze1 + xze2 + xye3, Σ is the part of the paraboloidz = 9 − x2 − y2 abovez = 8, orienteddownward
(c) F(x, y, z) = x2e1 − 2xye2 + yz2e3, Σ is the the surfacez= x2y, 0 ≤ x, y ≤ 1, oriented upward(d) F(x, y, z) = ye1 + xze2 + 2xe3, Σ is the hemispherex2 + y2 + z2 = 1, y ≥ 0, oriented in the direction of
the positivey-axis
149
24. Gauss’ divergence theorem
We now turn to another generalization of Green’s theorem, known asGauss’ divergence theo-rem. It relates the surface integral over the boundary of a simply-connected solid to a triple integralover the entire solid, similarly to how Green’s theorem relates a line integral over the boundary ofa simply-connected plane region to the double integral overthe entire region.
Like Green’s theorem, the divergence theorem requires thatwe pay attention to terminology.Recall the definition of a simply-connected region from§18.3. By asimply-connected solid, wemean a simply-connected region in space that is closed (contains all its boundary points). Forsuch a solid, itsboundaryconsists of the points on its “outer” (visible) surface. Theboundaryof a simply-connected solid is the rightful three-dimensional analog of the simple, closed curveappearing in Green’s theorem. We also define thepositive orientationof the boundary of a solid(assuming that it is orientable) to be the outward orientation.
Theorem 24.1(Gauss’ divergence theorem). Let E be a bounded, simply-connected solid inR3,whose boundary is the piecewise smooth, positively oriented surfaceΣ. Let F be a vector fieldwhose first-order partials are continuous on an open set U containing E. Then
∫∫
Σ
F · dS=∫∫∫
Ediv F dV. (24.1)
Example 24.1.Evaluate∫∫
Σ
(
(x3 + ysinz)e1 + (y3 + zsinx)e2 + 3ze3
)
· dS,
whereΣ is the positively oriented boundary of the solidE bounded by thexy-plane and by thehemispheres
z=√
1− x2 − y2, z=√
4− x2 − y2.
Figure 24.1. The solid 1≤ x2 + y2 + z2 ≤ 4, z≥ 0
Solution.We writeF = (x3 + ysinz)e1 + (y3 + zsinx)e2 + 3ze3. Then, by Gauss’ theorem,∫∫
Σ
F · dS=∫∫∫
Ediv F dV =
∫∫∫
E
(
3x2 + 3y2 + 3)
dV.
The solidE is displayed on Figure 24.1. It is described by the inequalities
1 ≤ x2 + y2 + z2 ≤ 4, z≥ 0.
These inequalities are translated in spherical coordinates as
1 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2.150
Hence, converting the triple integral overE to spherical coordinates, we obtain
∫∫
Σ
F · dS=∫ 2
1
∫ π/2
0
∫ 2π
0(3ρ2 sin2φ + 3)ρ2 sinφdθdφdρ
= 6π∫ 2
1
∫ π/2
0(ρ4 sin2φ + ρ2) sinφdφdρ
= 6π∫ 2
1
[∫ π/2
0
(
ρ4(1− cos2φ) + ρ2)
sinφdφ
]
dρ
= 6π∫ 2
1
[∫ 1
0
(
ρ4(1− u2) + ρ2)
du
]
dρ
= 6π∫ 2
1
(
23ρ
4 + ρ2)
dρ = 1945 π. �
Example 24.2.Evaluate∫∫
Σ
(
x2e1 + y2e2 + z2e3
)
· dS,
whereΣ is the positively oriented surface of the solidE shown on Figure 24.2 (a cube of side length2 with a corner cube of side length 1 removed).
(0,2,2)(2,0,2)
(2,2,0)
Figure 24.2. The solidE
Solution. It is possible to evaluate the surface integral using doubleintegrals. However, were we todo that, we would have to parametrize each face separately, and there are nine faces. Furthermore,we would have to split the three double integrals corresponding to the L-shaped faces in two, so wewould have to evaluate twelve double integrals altogether.On the other hand,E is the difference oftwo cubes, so we can calculate a triple integral overE by writing it as the difference of two tripleintegrals over those cubes. This suggests that we use the divergence theorem.
By Gauss’ theorem,
∫∫
Σ
(
x2e1 + y2e2 + z2e3
)
· dS=∫∫∫
E(2x+ 2y+ 2z) dV.
151
SinceE is the difference of the cubes [0,2]3 and [1,2]3, it is easier to calculate the triple integraloverE as the difference of the triple integrals over those two cubes:∫∫∫
E(2x+ 2y+ 2z) dV =
∫∫∫
[0,2]3
(2x+ 2y+ 2z) dV−∫∫∫
[1,2]3
(2x+ 2y+ 2z) dV
=
∫ 2
0
∫ 2
0
∫ 2
0(2x+ 2y+ 2z) dzdydx−
∫ 2
1
∫ 2
1
∫ 2
1(2x+ 2y+ 2z) dzdydx
=
∫ 2
0
∫ 2
0
[
2xz+ 2yz+ z2]2
0dydx−
∫ 2
1
∫ 2
1
[
2xz+ 2yz+ z2]2
1dydx
=
∫ 2
0
∫ 2
0(4x+ 4y+ 4)dydx−
∫ 2
1
∫ 2
1(2x+ 2y+ 3)dydx
=
∫ 2
0
[
4xy+ 2y2 + 4y]2
0dx−
∫ 2
1
[
2xy+ y2 + 3y]2
1dx
=
∫ 2
0(8x+ 16)dx−
∫ 2
1(2x+ 6)dx=
[
4x2 + 16x]2
0−[
x2 + 6x]2
1= 39. �
With a little bit of extra effort, we can also use the divergence theorem to evaluate surfaceintegrals of scalar functions.
Example 24.3.Evaluate∫∫
Σ
(
x2z+ y2z)
dS,
whereΣ is the spherex2 + y2 + z2 = 4.
Solution.Note that we have to deal with the surface integral of a scalarfunction and not of a vectorfield. Thus, Gauss’ theorem is not directly applicable. However, we know that the sphereΣ can beparametrized using spherical coordinates by the vector function
r (θ,φ) = 2 cosθ sinφe1 + 2 sinθ sinφe2 + 2 cosφe3,
where 0≤ θ ≤ 2π and 0≤ φ ≤ π. We also know (we went through this calculation several timesalready) that the (outward) unit normal vector is then
n = rφ × r θ = cosθ sinφe1 + sinθ sinφe2 + cosφe3.
Note that in Cartesian coordinatesn = 12
(
xe1+ ye2+ ze3
)
. We can use this observation to representthe integrand in the given integral in the formF · n, with F some nice vector field; we can thenappeal to Gauss’ theorem. For example, if we chooseF = 2(x2 + y2)e3, we obtain
F · n = 0(
12x)
+ 0(
12y)
+ 2(x2 + y2)(
12z)
= x2z+ y2z,
whence∫∫
Σ
(
x2z+ y2z)
dS =∫∫
Σ
(F · n) dS =∫∫
Σ
F · dS=∫∫∫
Bdiv F dV = 0.
Here,B is the ballx2 + y2 + z2 ≤ 4.Note that the above choice ofF is not unique. We could have chosen also
F = 2xze1 + 2yze2, F = xze1 + yze2 + (x2 + y2)e3,152
or any other vector fieldF that satisfies
F · n = x2z+ y2z.
However, in most cases, those choices would yield triple integrals that equal 0 for less obviousreasons than the one we encountered. For example, the choiceF = 2xze1 + 2yze2 yields
∫∫
Σ
(
x2z+ y2z)
dS =∫∫∫
Bdiv F dV =
∫∫∫
B4z dV.
The last integral is still 0, but this requires some thought (try to convince yourself by interpretingit in terms of the average value off (x, y, z) = 4z). �
We can also use the divergence theorem to calculate volumes by means of surface integralssimilarly to how we earlier used Green’s theorem to calculate areas by means of line integrals. Forexample, ifE is a simply-connected solid andΣ is its boundary, then
∫∫
Σ
xe1 · dS=∫∫∫
Ediv(xe1) dV =
∫∫∫
E1dV = vol.(E).
Similarly, we obtain formulas for the volume ofE that use the vector fieldsye2 andze3. Here aresome of the surface integrals representing vol.(E):
vol.(E) =∫∫
Σ
xe1 · dS=∫∫
Σ
ye2 · dS=∫∫
Σ
ze3 · dS=13
∫∫
Σ
(xe1 + ye2 + ze3) · dS. (24.2)
Example 24.4.Find the volume of the spiral pipe shown on Figure 24.3. The outer surface of thepipe is parametrized by the vector function
r (u, v) = (2+ sinv) cos 2ue1 + (2+ sinv) sin 2ue2 + (u+ cosv)e3 (0 ≤ u, v ≤ 2π),
and its two openings are the disks (x− 2)2 + z2 ≤ 1 and (x− 2)2 + (z− 2π)2 ≤ 1 in thexz-plane.
Figure 24.3. The surfacer (u, v) = (2+ sinv) cos 2ue1 + (2+ sinv) sin 2ue2 + (u+ cosv)e3
Solution.Let Σ be the positively oriented, piecewise smooth surface consisting of the parametricsurfaceΣ0 that is the outer surface of the pipe and the two disks representing its two openings. Wedenote the disksΣ1 andΣ2. ThenΣ is the boundary of the simply-connected solidE whose volumewe want to find. By (24.2),
vol.(E) =∫∫
Σ
ze3 · dS=(∫∫
Σ0
+
∫∫
Σ1
+
∫∫
Σ2
)
ze3 · dS.
153
The unit normal vectors toΣ1 aree2 and−e2, and of those two, it is−e2 that points out of the solidE. Thus, by (22.4),
∫∫
Σ1
ze3 · dS=∫∫
Σ1
(
(ze3) · (−e2))
dS = 0.
Similarly,∫∫
Σ2
ze3 · dS=∫∫
Σ2
(
(ze3) · (e2))
dS = 0.
Hence,
vol.(E) =∫∫
Σ0
ze3 · dS.
We evaluate the last surface integral using the orientation
n =ru × r v
‖ru × r v‖.
Thus, by (22.5),∫∫
Σ0
ze3 · dS=∫∫
0≤u,v≤2π
(
(z(u, v)e3) · (ru × r v))
dA.
Note that the dot product depends only on the third componentof ru × r v. We have
ru(u, v) = −2(2+ sinv) sin 2ue1 + 2(2+ sinv) cos 2ue2 + e3,
r v(u, v) = cosvcos 2ue1 + cosvsin 2ue2 − sinve3.
Thus, the cross productru × r v is the sum of six nonzero terms:
ru × r v = −2 cosv(2+ sinv) sin2 2u(e1 × e2) + 2 cosv(2+ sinv) cos2 2u(e2 × e1) + other terms
= −2 cosv(2+ sinv)(
sin2 2u+ cos2 2u)
e3 + other terms
= −2 cosv(2+ sinv)e3 + other terms.
The omitted terms are cross products involvinge1× e3, e2× e3, etc., which do not contribute to thecoefficient ofe3 in ru × r v. Thus,
(z(u, v)e3) · (ru × r v) = ((u+ cosv)e3) · (ru × r v) = −2 cosv(u+ cosv)(2+ sinv).
We conclude that∫∫
Σ0
ze3 · dS=∫∫
0≤u,v≤2π
(
− 2 cosv(u+ cosv)(2+ sinv))
dA
=
∫ 2π
0
[∫ 2π
0
(
− 2ucosv(2+ sinv) − 2 cos2 vsinv− 4 cos2 v)
dv
]
du.
We have∫ 2π
0(2+ sinv) cosv dv= 0,
∫ 2π
0cos2 vsinv dv= 0,
∫ 2π
0cos2 v dv= π,
so∫∫
Σ0
ze3 · dS=∫ 2π
0
(
− 2u(0)− 2(0)− 4π)
du= −8π2.
154
We obtained a negative volume, because we did not check whether the unit vectorn gives theinward or the outward orientation of the surface. It turns out that it gives the inward (negative)orientation; hence, the negative volume. Consequently, thevolume of the pipe is 8π2. �
Exercises
1. Use Gauss’ theorem to evaluate∫∫
Σ
F · dS for the given fieldF(x, y, z) and the given surfaceΣ oriented
outward:(a)F(x, y, z) = y3e1+ z3e2+ x3e3, Σ is the boundary of the solidE defined byx2+ y2+ z2 ≤ 3, x ≥ 0, y ≥ 0(b) F(x, y, z) = x2z3e1+x2ye2+y2ze3, Σ is the boundary of the solidE enclosed by the spherex2+y2+z2 = 4
inside the cylinderx2 + y2 = 1(c) F(x, y, z) = xy2e1 + yz2e2 + zx2e3, Σ is the boundary of the solid shown on Figure 24.4, that is, thepart
of the solid spherex2 + y2 + z2 ≤ 4 outside the first octant
Figure 24.4. Seven eights of the spherex2 + y2 + z2 ≤ 4
(d) F(x, y, z) = y3eze1+y√
4− x2e2+ey cosxe3, Σ is the boundary of the solidE that lies below the surfacez= 2− x3 − y3, −1 ≤ x, y ≤ 1, and above thexy-plane
2. Use Gauss’ theorem to evaluate∫∫
Σ
(
x2 + y2 + z3)
dS,
whereΣ is the spherex2 + y2 + z2 = 4.
3. Use the volume formulas (24.2) to calculate the volume of the given solid:(a) the solid enclosed by the ellipsoidx2 + 2y2 + 4z2 = 4, that is, the parametric surface given by
r (θ,φ) = 2 cosθ sinφe1 +√
2 sinθ sinφe2 + cosφe3 (0 ≤ θ ≤ 2π,0 ≤ φ ≤ π)(b) the solid bounded by the torus
r (u, v) = cosu(4+ 2 cosv)e1 + sinu(4+ 2 cosv)e2 + 2 sinve3 (0 ≤ u, v ≤ 2π)
155
Appendix A. Determinants
An n× n matrix is an array of real numbers withn rows andn columns. For example,A belowis a 2× 2 matrix andB is a 3× 3 matrix:
A =
[
1 23 4
]
, B =
1 −3 10 2 −11 1 −2
.
Thedeterminant of a2× 2 matrix
A =
[
a11 a12
a21 a22
]
is the following number (the product of its diagonal entriesminus the product of its off-diagonalentries):
det(A) =
∣
∣
∣
∣
a11 a12
a21 a22
∣
∣
∣
∣
= a11a22− a12a21.
Thedeterminant of a3× 3 matrix
A =
a11 a12 a13
a21 a22 a23
a31 a32 a33
is given byexpansion along the first row, a process which requires the computation of three 2× 2determinants:
det(A) =
∣
∣
∣
∣
∣
∣
a11 a12 a13
a21 a22 a23
a31 a32 a33
∣
∣
∣
∣
∣
∣
= a11
∣
∣
∣
∣
a22 a23
a32 a33
∣
∣
∣
∣
− a12
∣
∣
∣
∣
a21 a23
a31 a33
∣
∣
∣
∣
+ a13
∣
∣
∣
∣
a21 a22
a31 a32
∣
∣
∣
∣
.
Example A.1. Compute the determinant of the matrixB above.
Solution.We have
det(B) = 1
∣
∣
∣
∣
2 −11 −2
∣
∣
∣
∣
− (−3)
∣
∣
∣
∣
0 −11 −2
∣
∣
∣
∣
+ 1
∣
∣
∣
∣
0 21 1
∣
∣
∣
∣
= (2 · (−2)− 1 · (−1))+ 3(0 · (−2)− 1 · (−1))+ (0 · 1− 2 · 1) = −2. �
156
Appendix B. Definite integrals of single-variable functions
B.1. Definition of the definite integral.
Definition. Let f (x) be a bounded function on [a,b]. For eachn = 1,2, . . . , set∆n = (b− a)/n anddefine the numbers
x0 = a, x1 = a+ ∆n, x2 = a+ 2∆n, . . . , xn = a+ n∆n = b.
Also, choosesample points x∗1, x∗2, . . . , x
∗n, with
x0 ≤ x∗1 ≤ x1, x1 ≤ x∗2 ≤ x2, . . . , xn−1 ≤ x∗n ≤ xn.
Thedefinite integral of f from a to bis∫ b
af (x) dx= lim
n→∞
n∑
i=1
f (x∗i )∆n, (B.1)
provided that the limit exists. The sums on the right side of (B.1) are calledRiemann sums of f on[a,b].
Remarks. 1. In the above definition, the limit is supposed to exist and be independent of thechoice of the sample points. This is not true for any old function, but it does hold for functionsthat are continuous on [a,b] or have only a finite number ofjump discontinuities inside [a,b]. Forsuch functions, we may evaluate the integral using any choice of the sample points (e.g., we cantake eachx∗i to be the midpoint of the respective subinterval).
a b
y = f (x)
R
Figure B.1. The definite integral as area
2. In single-variable calculus, the definite integral of a continuous function is usually motivatedby thearea problem: If f (x) is a continuous function on [a,b], with f (x) ≥ 0, calculate the area ofthe region (see Figure B.1)
R={
(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f (x)}
.
The answer turns out to be the definite integral off (x):
area(R) =∫ b
af (x) dx.
3. A different interpretation of the definite integral arises, if we divide both sides of (B.1) byb− a. Since∆n = (b− a)/n, we have
1b− a
n∑
i=1
f (x∗i )∆n =∆n
b− a
n∑
i=1
f (x∗i ) =1n
n∑
i=1
f (x∗i ).
157
Hence, dividing both sides of (B.1) byb− a, we derive
1b− a
∫ b
af (x) dx= lim
n→∞
1n
n∑
i=1
f (x∗i ). (B.2)
Note that the expression1n∑
i f (x∗i ) is the average of the sample valuesf (x∗1), f (x∗2), . . . , f (x∗n),which are about equally spaced. Thus, the limit on the right side of (B.2) represents the average of“infinitely many, equally spaced” sample values off (x). Consequently, we may argue that the leftside of (B.2) is the average value off (x):
averagex∈[a,b]
f (x) =1
b− a
∫ b
af (x) dx. (B.3)
This alternative interpretation of the definite integral israrely emphasized in single-variablecalculus, but is much more appropriate for the purpose of these notes, because we want to drawparallels between the properties of definite integrals and the properties of other types of integrals(double, triple, line, surface) whose geometric meaning may be less obvious.
B.2. Properties of the definite integral. The following is a list of some general properties ofdefinite integrals:
1.∫ b
a[ f (x) + g(x)] dx=
∫ b
af (x) dx+
∫ b
ag(x) dx
2.∫ b
ac f(x) dx= c
∫ b
af (x) dx
3.∫ c
af (x) dx+
∫ b
cf (x) dx=
∫ b
af (x) dx
4. If f (x) ≤ g(x), then∫ b
af (x) dx≤
∫ b
ag(x) dx
5. If m≤ f (x) ≤ M, thenm(b− a) ≤∫ b
af (x) dx≤ M(b− a).
6.∫ a
bf (x) dx= −
∫ b
af (x) dx,
∫ a
af (x) dx= 0.
Of course, the most important property of definite integralsis the Fundamental Theorem ofCalculus, also known as the Newton–Leibnitz theorem.
Theorem B.1(Fundamental theorem of calculus). Suppose that f(x) is a continuous function on[a,b].
i) We haveddx
∫ x
af (t) dt = f (x).
that is,∫ x
a f (t) dt is an antiderivative of f(x).ii) If F (x) is any antiderivative of f(x), then
∫ b
af (x) dx= F(b) − F(a).
158
Appendix C. Polar coordinates in the plane
C.1. Definition. Consider a pointP in the plane. Recall that itsCartesian coordinates(x, y) arethe signed distances fromP to two perpendicular axes (thex- andy-axes). In this appendix, werecall another set of coordinates: thepolar coordinatesof P. Thepolar coordinate systemconsistsof a pointO called thepole (or theorigin) and a ray starting atO called thepolar axis. The polarcoordinates of a pointP , O are a pair of real numbers (r, θ), wherer is the distance fromO to Pandθ ∈ [0,2π) is the oriented angle between the polar axis and the lineOP (see Figure C.1); thepolar coordinates of the poleO are (0, θ), whereθ is any number in [0,2π).
b
bP(r, θ)
θr
O
Figure C.1. The polar coordi-nates of a point
b
bP(x, y)
Q(x,0)θ
r =√ x2
+y2
O
y
x
Figure C.2. Polar and Cartesian coordinates
C.2. Relations between polar and Cartesian coordinates.We want to be able to convert polarcoordinates to Cartesian and vice versa. In order to do so, we need to set a Cartesian and a polarcoordinate systems in the same copy of the plane. We choose the two coordinate systems so thatthey share the same origin and the polar axis coincides with the positivex-axis. Since the distancebetween the origin andP(x, y) is the same no matter what coordinate system we use, we have
r = |OP| =√
x2 + y2.
The relation betweenθ and (x, y) is more technical, but not much more complicated. First, considera pointP(x, y) in the first quadrant. From△OPQon Figure C.2,
x = r cosθ, y = r sinθ, tanθ =yx.
The first two of these equations hold for all pointsP(x, y), even if they do not lie in the firstquadrant, that is, the Cartesian coordinates of a pointP(r, θ) are obtained from the formulas
x = r cosθ, y = r sinθ. (C.1)
On the other hand, the relation tanθ = y/x extends only to the points withx , 0 and determinesθuniquely only on the half-planesx > 0 or x < 0. Thus, when we write the formulas for convertingCartesian into polar coordinates
r =√
x2 + y2, tanθ =yx, (C.2)
we must remember that the latter formula must be applied cautiously.159
Example C.1. The pointP whose polar coordinates are (2, 23π) has Cartesian coordinates
x = 2 cos(
23π)
= −1, y = 2 sin(
23π)
=√
3.
�
Example C.2. The Cartesian pointP(2,2) has polar coordinates
r =√
22 + 22 = 2√
2, θ = arctan
(
22
)
=π
4.
Here, we used that sinceP lies in the first quadrant, the solution of the second equation in (C.2) isgiven byθ = arctan(y/x). �
Example C.3.The Cartesian pointP(2,−2) has polar coordinates (r, θ), wherer =√
22 + (−2)2 =2√
2 andθ is such thattanθ = −1, θ ∈
(
32π,2π
)
,
that is,P(2√
2, 74π). Notice that in this case the solution of the second equation in (C.2) isθ =
arctan(y/x)+2π, because arctan(y/x) lies in (−12π,
12π) whereas the polar angleθ lies in [0,2π). �
Example C.4. The Cartesian pointP(−3,4) has polar coordinates(r, θ), wherer =√
(−3)2 + 42 =
5 andθ is such that
tanθ = −43, θ ∈
(
12π, π
)
,
that is,P(5,arctan(−43) + π). �
C.3. Polar curves. A polar curveis the set of points in the plane whose (polar) coordinates (r, θ)satisfy an equationr = f (θ), where f is some function. Note that using equations (C.1), we canwrite every polar curve in parametric form:
x = r cosθ = f (θ) cosθ, y = r sinθ = f (θ) sinθ.
Example C.5. The curver = 3 is the set of points at distance 3 from the origin, that is, a circle ofradius 3 centered atO (whose Cartesian equation isx2 + y2 = 9, of course). Notice that we reachthe same conclusion if we use (C.2) to convert the given polar equation into a Cartesian equation:
r = 3 ⇐⇒√
x2 + y2 = 3 ⇐⇒ x2 + y2 = 9.
�
Example C.6. The curveθ = 34π is the ray starting at the origin and intersecting the polar axis at
an angle of 135◦. �
Example C.7. The curver = −2 cosθ is a circle. To see this, we pass to Cartesian coordinates:
r = −2 cosθ ⇐⇒ r2 = −2r cosθ ⇐⇒ x2 + y2 = −2x.
The last equation is the Cartesian equation of a circle of radius 1 centered at (−1,0):
x2 + y2 = −2x ⇐⇒ x2 + 2x+ 1+ y2 = 1 ⇐⇒ (x+ 1)2 + y2 = 1.
�
Example C.8. The curver = 1+ cosθ is known as thecardioid, because its graph is shaped likea heart (see Figure C.3). Unlike the curves in the previous examples, the cardioid does not have asimple Cartesian equation which we can use to sketch its graph. �
160
O
Figure C.3. The cardioid
C.4. Area and arc length in polar coordinates.
Theorem C.1. Let 0 ≤ a < b ≤ 2π and let f be a positive continuous function defined on[a,b].Then the area A of the region in the plane bounded by the curves
θ = a, θ = b, r = f (θ),
is given by
area(R) =12
∫ b
a[ f (θ)]2 dθ.
Example C.9. Find the area enclosed by the cardioid (see Example C.8).
Solution.By Theorem C.1, the area is given by the integral
12
∫ 2π
0(1+ cosθ)2 dθ =
12
∫ 2π
0
(
1+ 2 cosθ + cos2 θ)
dθ
=12
∫ 2π
0
[
1+ 2 cosθ + 12(1+ cos 2θ)
]
dθ
=12
[
θ + 2 sinθ + 12θ +
14 sin 2θ
]2π
0=
3π2. �
Theorem C.2. Let 0 ≤ a < b ≤ 2π and let f be a positive continuous function defined on[a,b].Then the length L of the curve r= f (θ) is given by
L =∫ b
a
√
f (θ)2 + f ′(θ)2 dθ.
Example C.10.Find the lengthL of the cardioid.
Solution.By Theorem C.2,
L =∫ 2π
0
√
(1+ cosθ)2 + (− sinθ)2 dθ =∫ 2π
0
√2+ 2 cosθ dθ
=
∫ 2π
0
√
4 cos2(θ/2)dθ =∫ 2π
02| cos(θ/2)|dθ = 4
∫ π
0| cosu|du
= 4∫ π/2
0cosu du− 4
∫ π
π/2cosu du= 8. �
161
Appendix D. Answers to selected exercises
§1. 1. (a)Q; (b) none. 3.√
14,√
14,√
14; equilateral. 4. (x− 1)2 + (y− 1)2 + (z− 1)2 = 10.6. (a) a plane parallel to theyz-plane and 3 units behind it; (c) all points on or between the planesy = −1 andy = 2; (e) all points between the spheres centered atO that have radii 2 and 3.7. (b) x2 + y2 + z2 < 4, z > 0. 8. (b) (32
√2, 3
2
√2,−3). 9. (a) (
√2,7π/4,4); (c) (2, π/6,−2).
10. (c) (−32
√3,−3
2,0). 11. (a) (√
6, π/4,arccos(2/√
6)); (b) (2,3π/4,3π/4).12. (a)r2 = 2z, ρ2 sin2φ = 2ρ cosφ; (c) z= r, φ = π/4. 13. (a)z= −2, a horizontal plane;(c) x2+ y2+ z2 = 4, a sphere with radius 2 and centerO; (e) x2+ y2−2z= 0. 14. (b) The spherewith radius 2 and centerO; (d) all points above thexy-plane that lie on or between the sphereswith centerO and radii 2 and 4; (f) all points below thexy-plane that lie on or inside the infinitecylinder with radius 1 and axis thez-axis.
§2. 2. (a) 2, 〈−5,−2〉 ,2√
2, 〈4,−7〉; (c)√
3, 〈5,−3,−4〉 , 12
√11, 〈3,−2,−1〉. 4. (b) 0; (d) 3.
5. (b)π/3. 6. π/2, π/3, π/6. 7. (b) neither. 8. (a)〈−1,−2,2〉. 9. (a)〈−7,−3,5〉. 10. (b)2√
1329; (d) 80; (f) undefined. 12.12√
110. 14. sin2φ cosθe1+sin2φ sinθe2+sinφ cosφe3.
§3. 2. x = t, y = −2+ 3t, z= 3+ 2t (t ∈ R); x = y+23 =
z−32 . 3. Skew. 5. 3x− 2z+ 3 = 0.
7. 4x+ 5y− 3z+ 1 = 0. 8. x−2.42 =
y+0.43 = z
5. 10. (b)π/2. 11. 89.
§4. 1. (b) 〈1,2,1〉. 2. (a)⟨
2tt2+1,
2√1−4t2,et
⟩
; (c)⟨
3t2,0,−6e2t⟩
. 3. (b)⟨
0, 12
(
e2 − e−2)
,0⟩
.
§5. 1. (a) Yes,t = 0; (c) No. 2. Yes,r1(0) = r2(2) = 〈0,0,0〉. 4.⟨
3√10,0, 1√
10
⟩
. 6. x = −5,z= 2. 7. (0,±5,−2). 8. (a) smooth; (b) piecewise smooth. 10. (b) 1+ ln 2.11. (a)s(a) = 0, s(b) = L; (e) r (s) =
⟨(
1 + 3s5
)
sin(
43 ln(1 + 3s
5 ))
,(
1 + 3s5
)
cos(
43 ln(1 + 3s
5 ))
,5⟩
,0 ≤ s≤ 5.
§6. 1. (a) Domain: all points (x, y); range: [0,∞); (c) Domain: the points (x, y, z) for which (x, y)is not on the hyperbolax2 − y2 = 1; range: all real numbers. 2. (b) The points on and above theline y = −x/2, except for those that lie on the circlex2 + y2 = 1. 3. (a) levels−1 and 0: no levelcurves; level 1: the liney = 0; level 2: the linesy = 1 andy = −1; level 3: the linesy =
√2 and
y = −√
2; (c) level−1: no level curve; level 0: the level curve consists of a single point, (0,0); fork = 1,2,3, thek-level curve is the circlex2 + y2 = k. 4. (a)−29; (c) 0; (d) D.N.E.; (f) 2. 5.(a) All (x, y); (b) the points withx2 + y2 + 4z2 ≤ 4; (c) all (x, y) , (0,0).
§7. 1. (a) Parabolas, parabolas, hyperbolas, hyperbolic paraboloid; (c) Ellipses, ellipses, ellipses,ellipsoid. 2. (b) (x− 2)2 + 4y2 + (z+ 1)2 = 9, ellipsoid centered at (2,0,−1).3. r (t) = 〈3 cost,3 sint,3 cost + 3 sint + 2〉, 0 ≤ t ≤ 2π. 4. (a) The elliptic paraboloidz= x2 + y2.6. r (y, z) =
⟨
−2√
y2 + z2 + 4, y, z⟩
,−∞ < y, z< ∞. 8. (a)r (x, y) =⟨
x, y,4−x2−y2⟩
, x2+y2 ≤ 4;(b) r (u, v) =
⟨
ucosv,usinv,4− u2⟩
, 0 ≤ u ≤ 2,0 ≤ v ≤ 2π.9. r (u, v) = 〈(b+ acosv) cosu, (b+ acosv) sinu,asinv〉, 0 ≤ u, v ≤ 2π.
§8. 1. (a) 4x3 − 8xy3/2, −6x2y1/2; (c) 3(3x+ 7yz)−1, 7z(3x+ 7yz)−1, 7y(3x+ 7yz)−1;(d)
(
2x2 + xy)(
x2 + xy+ y2)x−1+(
x2 + xy+ y2)x
ln(
x2 + xy+ y2)
,(
2x2 + xy)(
x2 + xy+ y2)x−1
.2. (a) 12x2 − 8y3/2, −12xy1/2, −3x2y−1/2; (b) −(2x+ y)2 cos
(
x2 + xy+ y2)
− 2 sin(
x2 + xy+ y2)
,−(2x+y)(x+2y) cos
(
x2+xy+y2)
−sin(
x2+xy+y2)
,−(x+2y)2 cos(
x2+xy+y2)
−2 sin(
x2+xy+y2)
.3. (a)−12y1/2; (d)
(
6z2 + 4yz3)
e2yz,(
12xz+ 24xyz2 + 8xy2z3)
e2yz.162
4. (a) 2y(2t + 3)+ (2x+ 3y2)(2t); (d)2(y− 3z)et − 3(x+ 2z)e−t + 2(2y− 3y)e2t
xy+ 2yz− 3xz.
5. (a) (4xy− y2)6s+ (2x2 − 2xy)3t, (4xy− y2)2t + (2x2 − 2xy)3s. 6. (b)−6e−24.9. (a) ln 7+ 3
7(x− 2)+ 57(y− 1). 10.ǫ1(x, y) = (y− 2), ǫ2(x, y) = 0. 11. (a) 0, 0.
§9. 1. (a) 2√5e2xy(x + 2y + x2y + 2xy2 + y3 + 2x3). 3. 1
60
√10. 4. (b) 7, in the direction of
〈−3,2,−6〉.
§10. 1. (a) 3x + 2y + z − 2 = 0 andx− 2
3=
y+ 12= z + 2; (c) x + y − 2z − 5 = 0 and
x − 2 = y + 1 =z+ 2−2
. 2. (a)x = x0 + tx0, y = y0 + ty0, z = z0 + tz0. 4.(
316,−
12,
18
)
. 5.
The tangent plane to both isx− 2y+ z = 2. 6. (a)−x+ π2y− z = 0; (c)−2y+ z+ 1 = 0. 7.(a) smooth; (c) piecewise smooth; (d) not even piecewise smooth: it is non-smooth on each curvexy= −π2 + 2kπ.
§11. 1. (a) saddle point at (12,−2); (c) local maxima at (±1√
2, ±1√
2), saddle point at (0,0); (f) local
maxima at (±1√2, ±1√
2), local minima at (±1√
2, ∓1√
2), saddle point at (0,0). 2. (a) 8, 20; (c)−4
√2
3√
3, 4√
23√
3.
3.√
2/3. 5. The cube of side length 5.
§12. 1. (a)− 827. 2. (b) 2.5. 3. (a) 2023; (c) 8(ln 3)−2. 4. (b) 3
4 ln 3− ln 2; (d) 258 π; (f) −1.
§13. 1. (a)−2113; (c) sin 1− cos 1. 2. (a) 257.2; (c) 1
2 ln 2; (e) 12(1− sin 1); (g) 7.
3. (a)∫ 1
0
∫ y2
0 f (x, y) dxdy; (c)∫ 2
0
∫
√2y
0 f (x, y) dxdy+∫ 4
2
∫ 4−y0 f (x, y) dxdy; (e)
∫ ln 20
∫ 2ex f (x, y) dydx.
4. (a) 14
(
1− e−1)
; (b) 14 ln 17. 5. 6. 7.32π +
5615. 9. 16
3 . 10. 23π.
§14. 1. (b) (e− 1)π2; (c)(
15√
15− 7√
7)
π6. 2. 8π. 4.
(
36− 203
√5)
π. 6. 83π −
329 .
§15. 1. (b) 12
(
1− e−16)
. 2. (a) 0.7; (c) 6415; (e) 1
2(1− sin 1). 3. 163 .
§16. 1. (a) 3415π; (c) 2π
3
(
2√
2− 1)
. 2. (a) 2π2; (c) 8π. 3. (a) 2π35. 4. (a) 81
2 π; (c) 143 π.
§17. 1. (a) 94
√6; (c) 8π
(
2−√
2)
; (e)π(
2√
6− 83
)
. 2. (b) 649 , (6
5,0). 3. (a) 32,(
76,
76,
76
)
.
4. In =12
32 · · ·
2n−12
√π
2 .
§18. 2. (a)⟨
ye2y cosxy, xe2y cosxy+ 2e2y sinxy⟩
; (c)
⟨
2xx2 + yz
,z
x2 + yz,
yx2 + yz
⟩
. 3. (a) open
and simply-connected; (b) simply-connected, but not open;(c) open and connected, but notsimply-connected; (e) open, but not connected. 4. (a) no; (c) yes, f (x, y) = xey + c; (d) yes,f (x, y) = y2 arctanx+ c. 5. (a) curlF =
⟨
−4xz2, x2y+ 4yz2,−x2z⟩
, divF = −6xyz; (c) curlF =⟨
2xze2xy,−2yze2xy− xsinxz,2ycosxy⟩
, divF = e2xy+2xcosxy− zsinxz. 6. (b) yes,f (x, y, z) =xy2 cosz+ c; (c) no.
§19. 1. (a) 1840
(
125√
5− 1)
; (c) 724; (e) 3π3
√13; (g) 14. 2. (a) 15π; (b) 2
3 − cos 1+ sin 2.
§20. For the potentials see Exercises 18.4 and 18.6. The values ofthe integrals are: (a)−1+ 2e−1;(c) −4; (e) 256 cos 8.
§21. 2. (a)−24π; (c) 162; (e) 0. 3. (a)43 −12π
2; (c) 9137. 4. 3π. 6. 2π.
163
§22. 1. 171√
14. 3. 16π. 5. 713/180. 7.−23π.
§23. 1. (a) 8π; (c) −1312. 2. (a) 20π; (c) −13
12.
§24. 1. (a) 0; (d)83π + 4
√3. 3. (a) 8
√2π; (b) 32π2.
164
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