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Dvostruki integrali
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Under the surface and above the triangle with vertices, , and
22. Enclosed by the paraboloid and the planes ,, ,
23. Bounded by the planes , , , and
24. Bounded by the planes , , , and
25. Enclosed by the cylinders , and the planes ,
26. Bounded by the cylinder and the planes , in the first octant
27. Bounded by the cylinder and the planes ,, in the first octant
28. Bounded by the cylinders and
; 29. Use a graphing calculator or computer to estimate the -coordinates of the points of intersection of the curves
and . If is the region bounded by these curves,estimate .
; 30. Find the approximate volume of the solid in the first octant that is bounded by the planes , , and and thecylinder . (Use a graphing device to estimate thepoints of intersection.)
31–32 |||| Find the volume of the solid by subtracting two volumes.
31. The solid enclosed by the parabolic cylinders ,and the planes ,
32. The solid enclosed by the parabolic cylinder and theplanes ,
33–36 |||| Use a computer algebra system to find the exact volumeof the solid.
33. Under the surface and above the regionbounded by the curves and for
34. Between the paraboloids and and inside the cylinder
35. Enclosed by
36. Enclosed by
37–42 |||| Sketch the region of integration and change the order ofintegration.
37. 38. y1
0 y
4
4x f �x, y� dy dxy
4
0 y
sx
0 f �x, y� dy dx
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
z � x 2 � y 2 and z � 2y
z � 1 � x 2 � y 2 and z � 0
x 2 � y 2 � 1z � 8 � x 2 � 2y 2z � 2x 2 � y 2
x � 0y � x 2 � xy � x 3 � xz � x 3y 4 � xy 2
CAS
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
z � 2 � yz � 3yy � x 2
2x � 2y � z � 10 � 0x � y � z � 2y � x 2 � 1
y � 1 � x 2
y � cos xz � xz � 0y � x
xxD x dADy � 3x � x 2
y � x 4x
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y 2 � z2 � r 2x 2 � y 2 � r 2
z � 0x � 0y � zx 2 � y 2 � 1
z � 0x � 0x � 2y,y 2 � z2 � 4
y � 4z � 0y � x 2z � x 2
z � 0x � y � 2y � xz � x
x � y � z � 1z � 0y � 0x � 0
z � 0y � xy � 1x � 0z � x 2 � 3y 2
�1, 2��4, 1��1, 1�z � xy21.1–6 |||| Evaluate the iterated integral.
1. 2.
3. 4.
6.
7–18 |||| Evaluate the double integral.
7.
8.
9.
10.
11. ,
12. ,
, is bounded by , ,
14. , is bounded by
15. ,
is the triangular region with vertices (0, 2), (1, 1), and
16.
is bounded by the circle with center the origin and radius 2
18. is the triangular region with vertices ,
, and
19–28 |||| Find the volume of the given solid.
19. Under the plane and above the regionbounded by and
20. Under the surface and above the region boundedby and x � y 3x � y 2
z � 2x � y 2
y � x 4y � xx � 2y � z � 0
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
�0, 3��1, 2�
�0, 0�yyD
2xy dA, D
D
yyD
�2x � y� dA,17.
yyD
xy 2 dA, D is enclosed by x � 0 and x � s1 � y 2
�3, 2�D
yyD
y 3 dA
y � sx and y � x 2DyyD
�x � y� dA
x � 1y � x 2y � 0DyyD
x cos y dA13.
D � ��x, y� � 0 � y � 1, 0 � x � y�yyD
xsy 2 � x 2 dA
D � ��x, y� � 1 � y � 2, y � x � y 3 �yyD
e x y dA
yyD
e y2 dA, D � ��x, y� � 0 � y � 1, 0 � x � y�
yyD
2y
x 2 � 1 dA, D � {�x, y� � 0 � x � 1, 0 � y � sx}
yyD
4y
x 3 � 2 dA, D � ��x, y� � 1 � x � 2, 0 � y � 2x�
yyD
x 3y 2 dA, D � ��x, y� � 0 � x � 2, �x � y � x�
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y1
0 y
v
0 s1 � v 2 du dvy
� 2
0 y
cos �
0 e sin � dr d�5.
y1
0 y
2�x
x �x 2 � y� dy dxy
1
0 y
e y
y sx dx dy
y2
1 y
2
y xy dx dyy
1
0 y
x 2
0 �x � 2y� dy dx
|||| 15.3 Exercises
1002 ❙ ❙ ❙ ❙ CHAPTER 15 MULTIPLE INTEGRALS
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙ ❙ ❙ ❙ 1003
51–52 |||| Use Property 11 to estimate the value of the integral.
51. ,
52. , is the disk with center the origin and radius
53. Prove Property 11.
In evaluating a double integral over a region , a sum of iterated integrals was obtained as follows:
Sketch the region and express the double integral as an iterated integral with reversed order of integration.
55. Evaluate , where
[Hint: Exploit the fact that is symmetric with respect to bothaxes.]
56. Use symmetry to evaluate , where is the region bounded by the square with vertices and .
57. Compute , where is the disk, by first identifying the integral as the volume
of a solid.
58. Graph the solid bounded by the plane and the paraboloid and find its exact volume.(Use your CAS to do the graphing, to find the equations of theboundary curves of the region of integration, and to evaluatethe double integral.)
z � 4 � x 2 � y 2x � y � z � 1CAS
x 2 � y 2 � 1DxxD s1 � x 2 � y 2 dA
�0, �5���5, 0�
DxxD �2 � 3x � 4y� dA
D
D � ��x, y� � x 2 � y 2 � 2�.xxD �x 2 tan x � y 3 � 4� dA
D
yyD
f �x, y� dA � y1
0 y
2y
0 f �x, y� dx dy � y
3
1 y
3�y
0 f �x, y� dx dy
D54.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
12Dyy
D
e x 2�y 2 dA
D � �0, 1� �0, 1�yyD
sx 3 � y 3 dA
39. 40.
42.
43–48 |||| Evaluate the integral by reversing the order of integration.
44.
45. 46.
47.
48.
49–50 |||| Express as a union of regions of type I or type II andevaluate the integral.
50.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
x0
y
y=_1
x=_1
x=1
x=¥
y=1+≈
Dx0
y
1
_1
_1 1
D(1, 1)
yyD
xy dAyyD
x 2 dA49.
D
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y8
0 y
2
sy3 e x4
dx dy
y1
0 y
� 2
arcsin y cos x s1 � cos2x dx dy
y1
0 y
1
x 2 x 3 sin�y 3 � dy dxy
3
0y
9
y 2 y cos�x 2 � dx dy
y1
0 y
1
sy sx 3 � 1 dx dyy
1
0 y
3
3y e x2
dx dy43.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y1
0 y
� 4
arctan x f �x, y� dy dxy
2
1 y
ln x
0 f �x, y� dy dx41.
y3
0 y
s9�y
0 f �x, y� dx dyy
3
0 y
s9�y 2 �s9�y 2
f �x, y� dx dy
|||| 15.4 D o u b l e I n t e g r a l s i n P o l a r C o o r d i n a t e s
Suppose that we want to evaluate a double integral , where is one of theregions shown in Figure 1. In either case the description of in terms of rectangular coor-dinates is rather complicated but is easily described using polar coordinates.
FIGURE 1
x0
y
R
≈+¥=1
(a) R=s(r, ¨) | 0¯r¯1, 0¯¨¯2πd
x0
y
R
≈+¥=4
≈+¥=1
(b) R=s(r, ¨) | 1¯r¯2, 0¯¨¯πd
RR
RxxR f �x, y� dA
1004 ❙ ❙ ❙ ❙ CHAPTER 15 MULTIPLE INTEGRALS
Recall from Figure 2 that the polar coordinates of a point are related to the rect-angular coordinates by the equations
(See Section 10.3.)The regions in Figure 1 are special cases of a polar rectangle
which is shown in Figure 3. In order to compute the double integral , whereis a polar rectangle, we divide the interval into subintervals of equal
width and we divide the interval into subintervals ofequal width . Then the circles and the rays divide the polarrectangle R into the small polar rectangles shown in Figure 4.
The “center” of the polar subrectangle
has polar coordinates
We compute the area of using the fact that the area of a sector of a circle with radius and central angle is . Subtracting the areas of two such sectors, each of which hascentral angle , we find that the area of is
Although we have defined the double integral in terms of ordinary rect-angles, it can be shown that, for continuous functions , we always obtain the same answer using polar rectangles. The rectangular coordinates of the center of are
, so a typical Riemann sum is
�m
i�1 �
n
j�1 f �ri* cos � j*, ri* sin � j*� Ai � �
m
i�1 �
n
j�1 f �ri* cos � j*, ri* sin � j*� ri* r �1
�ri* cos � j*, ri* sin � j*�Rij
fxxR f �x, y� dA
� 12 �ri � ri�1 ��ri � ri�1 � � � r i* r �
Ai � 12 ri
2 � �12 ri�12 � � 1
2 �ri
2 � ri�12 � �
Rij� � � j � � j�1
12 r 2��
rRij
� j* � 12 ��j�1 � �j�ri* � 1
2 �ri�1 � ri �
Rij � ��r, �� � ri�1 � r � ri, � j�1 � � � � j�
FIGURE 3 Polar rectangle FIGURE 4 Dividing R into polar subrectangles
O
∫å
r=a ¨=å
¨=∫r=b
R
Ψ
¨=¨j
¨=¨j_1
(ri*, ¨j*)
r=ri_1
r=ri
Rij
O
� � � jr � ri� � �� � �� n��j�1, �j�n��, ��r � �b � a� m
�ri�1, ri�m�a, b�RxxR f �x, y� dA
R � ��r, �� � a � r � b, � � � � ��
y � r sin �x � r cos �r 2 � x 2 � y 2
�x, y��r, ��
O
y
x
¨
x
yr
P(r, ̈ )=P(x, y)
FIGURE 2
If we write , then the Riemann sum in Equation 1 can be writ-ten as
which is a Riemann sum for the double integral
Therefore, we have
Change to Polar Coordinates in a Double Integral If is continuous on a polar rect-angle given by , , where , then
The formula in (2) says that we convert from rectangular to polar coordinates in a double integral by writing and , using the appropriate limits of
| integration for and , and replacing by . Be careful not to forget the additionalfactor r on the right side of Formula 2. A classical method for remembering this is shownin Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rect-angle with dimensions and and therefore has “area”
EXAMPLE 1 Evaluate , where is the region in the upper half-planebounded by the circles and .
SOLUTION The region can be described as
It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by , 1 � r � 2
R � ��x, y� � y � 0, 1 � x 2 � y 2 � 4�
R
x 2 � y 2 � 4x 2 � y 2 � 1RxxR �3x � 4y 2 � dA
O
d¨
r d¨
dr
dA
r
FIGURE 5
dA � r dr d�.drr d�
r dr d�dA�ry � r sin �x � r cos �
yyR
f �x, y� dA � y�
� y
b
a f �r cos �, r sin �� r dr d�
0 � � � � � 2�� � � � �0 � a � r � bRf2
� y�
� y
b
a f �r cos �, r sin �� r dr d�
� lim m, n l
�m
i�1 �
n
j�1 t�ri*, � j*� r � � y
�
�y
b
a t�r, � � dr d�
yyR
f �x, y� dA � lim m, n l
�m
i�1 �
n
j�1 f �ri* cos � j*, ri* sin � j*� Ai
y�
� y
b
a t�r, �� dr d�
�m
i�1 �
n
j�1 t�ri*, � j*� r �
t�r, �� � rf �r cos �, r sin ��
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙ ❙ ❙ ❙ 1005
. Therefore, by Formula 2,
EXAMPLE 2 Find the volume of the solid bounded by the plane and the paraboloid.
SOLUTION If we put in the equation of the paraboloid, we get . Thismeans that the plane intersects the paraboloid in the circle , so the solid liesunder the paraboloid and above the circular disk given by [see Figures 6and 1(a)]. In polar coordinates is given by , . Since
, the volume is
If we had used rectangular coordinates instead of polar coordinates, then we would haveobtained
which is not easy to evaluate because it involves finding the following integrals:
What we have done so far can be extended to the more complicated type of regionshown in Figure 7. It’s similar to the type II rectangular regions considered in Section 15.3.In fact, by combining Formula 2 in this section with Formula 15.3.5, we obtain the fol-lowing formula.
If is continuous on a polar region of the form
then yyD
f �x, y� dA � y�
� y
h2���
h1��� f �r cos �, r sin �� r dr d�
D � ��r, �� � � � � � �, h1��� � r � h2����
f3
y �1 � x 2 �3 2 dxy x 2s1 � x 2 dxy s1 � x 2 dx
V � yyD
�1 � x 2 � y 2 � dA � y1
�1 y
s1�x2
�s1�x2 �1 � x 2 � y 2 � dy dx
� y2�
0 d� y
1
0 �r � r 3 � dr � 2� r 2
2�
r 4
4 0
1
��
2
V � yyD
�1 � x 2 � y 2 � dA � y2�
0 y
1
0 �1 � r 2 � r dr d�
1 � x 2 � y 2 � 1 � r 20 � � � 2�0 � r � 1D
x 2 � y 2 � 1Dx 2 � y 2 � 1
x 2 � y 2 � 1z � 0
z � 1 � x 2 � y 2z � 0
� 7 sin � �15�
2�
15
4 sin 2�
0
�
�15�
2
� y�
0 [7 cos � �
152 �1 � cos 2��] d�
� y�
0 [r 3 cos � � r 4 sin2�]r�1
r�2 d� � y
�
0 �7 cos � � 15 sin2� � d�
� y�
0 y
2
1 �3r 2 cos � � 4r 3 sin2�� dr d�
yyR
�3x � 4y 2 � dA � y�
0 y
2
1 �3r cos � � 4r 2 sin2�� r dr d�
0 � � � �
1006 ❙ ❙ ❙ ❙ CHAPTER 15 MULTIPLE INTEGRALS
FIGURE 6
y
0
(0, 0, 1)
Dx
z
O
∫å
r=h¡(¨)
¨=å
¨=∫ r=h™(¨)
D
FIGURE 7D=s(r, ¨) | 寨¯∫, h¡(̈ )¯r¯h™(̈ )d
|||| Here we use the trigonometric identity
See Section 7.2 for advice on integrating trigonometric functions.
sin2� � 12 �1 � cos 2��
In particular, taking , , and in this formula, we seethat the area of the region bounded by , , and is
and this agrees with Formula 10.4.3.
EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the four-leavedrose .
SOLUTION From the sketch of the curve in Figure 8 we see that a loop is given by theregion
So the area is
EXAMPLE 4 Find the volume of the solid that lies under the paraboloid ,above the -plane, and inside the cylinder .
SOLUTION The solid lies above the disk whose boundary circle has equationor, after completing the square,
(see Figures 9 and 10). In polar coordinates we have and , sothe boundary circle becomes , or . Thus, the disk is given by
and, by Formula 3, we have
� 2[32 � � sin 2� �
18 sin 4�]0
� 2� 2�3
2���
2 � �3�
2
� 2 y� 2
0 �1 � 2 cos 2� �
12 �1 � cos 4��� d�
� 8 y� 2
0 cos4� d� � 8 y
� 2
0 �1 � cos 2�
2 �2
d�
� y� 2
�� 2 r 4
4 0
2 cos �
d� � 4 y� 2
�� 2 cos4� d�
V � yyD
�x 2 � y 2 � dA � y� 2
�� 2 y
2 cos �
0 r 2 r dr d�
D � {�r, � � � �� 2 � � � � 2, 0 � r � 2 cos �}Dr � 2 cos �r 2 � 2r cos �
x � r cos �x 2 � y 2 � r 2
�x � 1�2 � y 2 � 1
x 2 � y 2 � 2xD
x 2 � y 2 � 2xxyz � x 2 � y 2
� 14 y
� 4
�� 4 �1 � cos 4�� d� � 1
4 [� �14 sin 4�]�� 4
� 4�
�
8
� y� 4
�� 4 [ 1
2 r 2]0cos 2� d� � 1
2 y� 4
�� 4 cos2 2� d�
A�D� � yyD
dA � y� 4
�� 4 y
cos 2�
0 r dr d�
D � {�r, �� � �� 4 � � � � 4, 0 � r � cos 2�}
r � cos 2�
� y�
�
r 2
2 0
h���
d� � y�
� 12 �h����2 d�
A�D� � yyD
1 dA � y�
� y
h���
0 r dr d�
r � h���� � �� � �Dh2��� � h���h1��� � 0f �x, y� � 1
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙ ❙ ❙ ❙ 1007
FIGURE 8
¨=π4
¨=_π4
FIGURE 9
0
y
x1 2
D
(x-1)@+¥=1 (or r=2 cos ¨)
FIGURE 10
y
x
z
12. ,where
, where D is the region bounded by thesemicircle and the y-axis
14. , where is the region in the first quadrant enclosedby the circle
15. ,where
16. , where is the region in the first quadrant that liesbetween the circles and
17–20 |||| Use a double integral to find the area of the region.
One loop of the rose
18. The region enclosed by the curve
19. The region within both of the circles and
20. The region inside the circle and outside the circle
21–27 |||| Use polar coordinates to find the volume of the givensolid.
Under the paraboloid and above the disk
22. Inside the sphere and outside the cylinder
23. A sphere of radius
24. Bounded by the paraboloid and the plane
Above the cone and below the sphere
26. Bounded by the paraboloids and
27. Inside both the cylinder and the ellipsoid
28. (a) A cylindrical drill with radius is used to bore a holethrough the center of a sphere of radius . Find the volumeof the ring-shaped solid that remains.
(b) Express the volume in part (a) in terms of the height ofthe ring. Notice that the volume depends only on , not on
or .
29–32 |||| Evaluate the iterated integral by converting to polar coordinates.
29. 30. ya
�a y
sa 2�y 2
0 �x 2 � y 2 �3�2 dx dyy
1
0 y
s1�x 2
0 e x 2�y 2 dy dx
r2r1
hh
r2
r1
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
4x 2 � 4y 2 � z2 � 64x 2 � y 2 � 4
z � 4 � x 2 � y 2z � 3x 2 � 3y 2
x 2 � y 2 � z2 � 1z � sx 2 � y 225.
z � 4z � 10 � 3x 2 � 3y 2
a
x 2 � y 2 � 4x 2 � y 2 � z 2 � 16
x 2 � y 2 � 9z � x 2 � y 221.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
r � 2r � 4 sin �
r � sin �r � cos �
r � 4 � 3 cos �
r � cos 3�17.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
x 2 � y 2 � 2xx 2 � y 2 � 4Dxx
D x dA
R � ��x, y� � 1 � x 2 � y 2 � 4, 0 � y � x�xx
R arctan� y�x� dA
x 2 � y 2 � 25Rxx
R yex dA
x � s4 � y 2
xxD e�x 2�y 2
dA13.
R � ��x, y� � x 2 � y 2 � 4, x � 0�xx
R s4 � x 2 � y 2 dA1–6 |||| A region is shown. Decide whether to use polar coordi-
nates or rectangular coordinates and write as an iter-ated integral, where is an arbitrary continuous function on .
1. 2.
3. 4.
6.
7–8 |||| Sketch the region whose area is given by the integral andevaluate the integral.
7. 8.
9–16 |||| Evaluate the given integral by changing to polar coordinates.
9. ,where is the disk with center the origin and radius 3
10. , where is the region that lies to the left ofthe -axis between the circles and
11. , where is the region that lies above the-axis within the circle x 2 � y 2 � 9x
RxxR cos�x 2 � y 2� dA
x 2 � y 2 � 4x 2 � y 2 � 1yRxx
R �x � y� dA
DxxD xy dA
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y��2
0 y
4 cos �
0 r dr d�y
2�
� y
7
4 r dr d�
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
0
2
y
x
R
2052
5
2
y
x
R
5.
0 31
3
1
y
x
R
0 2
2
y
x
R
0 2
2
y
x
R
0 2
2
y
x
R
RfxxR f �x, y� dA
R
|||| 15.4 Exercises
1008 ❙ ❙ ❙ ❙ CHAPTER 15 MULTIPLE INTEGRALS
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS ❙ ❙ ❙ ❙ 1009
32.
33. A swimming pool is circular with a 40-ft diameter. The depthis constant along east-west lines and increases linearly from2 ft at the south end to 7 ft at the north end. Find the volume ofwater in the pool.
34. An agricultural sprinkler distributes water in a circular patternof radius 100 ft. It supplies water to a depth of feet per hourat a distance of feet from the sprinkler.(a) What is the total amount of water supplied per hour to
the region inside the circle of radius centered at the sprinkler?
(b) Determine an expression for the average amount of waterper hour per square foot supplied to the region inside thecircle of radius .
Use polar coordinates to combine the sum
into one double integral. Then evaluate the double integral.
36. (a) We define the improper integral (over the entire plane
� lima l �
yyDa
e��x 2�y 2 � dA
I � yy� 2
e��x 2�y 2 � dA � y�
�� y
�
�� e��x 2�y 2 � dy dx
� 2�
y1
1�s2 y
x
s1�x 2 xy dy dx � y
s2
1 y
x
0 xy dy dx � y
2
s2 y
s4�x 2
0 xy dy dx
35.
R
R
re�r
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y2
0 y
s2x�x 2
0 sx 2 � y 2 dy dxy
2
0 y
s4�y 2
�s4�y 2
x 2 y 2 dx dy31. where is the disk with radius and center the origin. Show that
(b) An equivalent definition of the improper integral in part (a)is
where is the square with vertices . Use this toshow that
(c) Deduce that
(d) By making the change of variable , show that
(This is a fundamental result for probability and statistics.)
37. Use the result of Exercise 36 part (c) to evaluate the followingintegrals.
(a) (b) y�
0 sx e�x dxy
�
0 x 2e�x 2 dx
y�
�� e�x 2�2 dx � s2�
t � s2 x
y�
�� e�x 2 dx � s�
y�
�� e�x 2 dx y
�
�� e�y 2 dy � �
�a, a�Sa
yy� 2
e��x 2�y 2 � dA � lima l �
yySa
e��x 2�y 2 � dA
y�
�� y
�
�� e��x 2�y 2 � dA � �
aDa
|||| 15.5 A p p l i c a t i o n s o f D o u b l e I n t e g r a l s
We have already seen one application of double integrals: computing volumes. Anothergeometric application is finding areas of surfaces and this will be done in the next section.In this section we explore physical applications such as computing mass, electric charge,center of mass, and moment of inertia. We will see that these physical ideas are also impor-tant when applied to probability density functions of two random variables.
D e n s i t y a n d M a s s
In Section 8.3 we were able to use single integrals to compute moments and the center ofmass of a thin plate or lamina with constant density. But now, equipped with the doubleintegral, we can consider a lamina with variable density. Suppose the lamina occupies aregion of the -plane and its density (in units of mass per unit area) at a point in
is given by , where is a continuous function on . This means that
where and are the mass and area of a small rectangle that contains and thelimit is taken as the dimensions of the rectangle approach 0. (See Figure 1.)
�x, y�Am
��x, y� � lim m
A
D���x, y�D�x, y�xyD
FIGURE 1
0 x
y
D
(x, y)
0
1
0
x2
(x+2y)dydx =0
1
xy+y2 y=x
2
y=0dx=
0
1
x(x2)+(x
2)2
0 0 dx
=0
1
(x3+x
4)dx=
14
x4+
15
x5 1
0=
920
1
2
y
2
xydxdy =1
212
x2y
x=2
x=ydy=
1
212
y(4 y2)dy=
12 1
2
(4y y3)dy
=12
2y2 1
4y
4 2
1=
12
8 4 2+14
=98
0
1
y
ey
x dxdy =0
123
x3/2 x=e
y
x=ydy=
23 0
1
(e3y/2
y3/2
)dy=23
23
e3y/2 2
5y
5/2 1
0
=23
23
e3/2 2
523
e0+0 =
49
e3/2 32
45
0
1
x
2 x
(x2
y)dydx =0
1
x2y
12
y2 y=2 x
y=xdx=
0
1
x2(2 x)
12
(2 x)2
x2(x)+
12
x2
dx
=0
1
( 2x3+2x
2+2x 2)dx=
12
x4+
23
x3+x
22x
1
0=
56
0
/2
0
cos
esin
dr d =0
/2
resin r=cos
r=0d =
0
/2
(cos )esin
d = esin /2
0
=esin ( /2)
e0=e 1
1.
2.
3.
4.
5.
6.
1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
0
1
0
v
1 v2
dudv =0
1
u 1 v2 u=v
u=0dv=
0
1
v 1 v2
dv=13
(1 v2)3/2 1
0
=13
(0 1)=13
Dx
3y
2dA =
0
2
x
x
x3y
2dydx=
0
213
x3y
3 y=x
y= xdx=
13 0
2
2x6dx
=23
17
x7 2
0=
221
27
0 =25621
D
4y
x3+2
dA =1
2
0
2x4y
x3+2
dydx=1
22y
2
x3+2
y=2x
y=0
dx=1
28x
2
x3+2
dx
=83
ln x3+2
2
1=
83
(ln 10 ln 3)=83
ln103
0
1
0
x2y
x2+1
dydx =0
1y
2
x2+1
y= x
y=0
dx=0
1x
x2+1
dx
=12
ln x2+1
1
0=
12
(ln 2 ln 1)=12
ln 2
0
1
0
y
ey
2
dxdy=0
1
xey
2 x=y
x=0dy=
0
1
yey
2
dy=12
ey
2 1
0=
12
(e 1)
1
2
y
y3
ex/y
dxdy=1
2
yex/y x= y
3
x= ydy=
1
2
yey
2
ey( )dy=12
ey
2 12
ey2
2
1=
12
(e4
4e)
0
1
0
y
x y2
x2
dxdy=0
113
(y2
x2)3/2 x=y
x=0dy=
13 0
1
y3dy=
13
14
y4 1
0=
112
7.
8.
9.
10.
11.
12.
2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
0
1
0
x2
xcos ydydx=0
1
xsin yy=x
2
y=0dx=
0
1
xsin x2dx=
12
cos x2 1
0=
12
(1 cos 1)
0
1
x2
x
(x+y)dydx =0
1
xy+12
y2 y= x
y=x2
dx
=0
1
x3/2
+12
x x3 1
2x
4dx
=25
x5/2
+14
x2 1
4x
4 110
x5 1
0=
310
1
2
2 y
2y 1
y3dxdy =
1
2
xy3 x=2y 1
x=2 ydy
=1
2
(2y 1) (2 y) y3dy
=1
2
(3y4
3y3)dy=
35
y5 3
4y
4 2
1
=965
1235
+34
=14720
13.
14.
15.
16.
3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
Dxy
2dA
=1
1
0
1 y2
xy2dxdy
=1
1
y2 1
2x
2 x= 1 y2
x=0dy=
12 1
1
y2(1 y
2)dy
=12 1
1
(y2
y4)dy=
12
13
y3 1
5y
5 1
1
=12
13
15
+13
15
=215
2
2
4 x2
4 x2
(2x y)dydx
=2
2
2xy12
y2 y= 4 x
2
y= 4 x2dx
=2
2
2x 4 x2 1
24 x
2( )+2x 4 x2
+12
4 x2( ) dx
=2
2
4x 4 x2
dx=43
4 x2( ) 3/2 2
2=0
4x 4 x2
17.
(Or, note that is an odd function, so
4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
2
2
4x 4 x2
dx=0
DxydA =
0
1
2x
3 x
2xydydx=0
1
xy2 y=3 x
y=2xdx
=0
1
x[(3 x)2
(2x)2]dx
=0
1
( 3x3
6x2+9x)dx
=34
x4
2x3+
92
x2 1
0=
34
2+92
=74
V =0
1
x4
x
(x+2y)dydx
=0
1
xy+y2 y=x
y=x4dx=
0
1
(2x2
x5
x8)dx
=23
x3 1
6x
6 19
x9 1
0=
23
16
19
=7
18
.)
18.
19.
5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V =0
1
y3
y2
(2x+y2)dxdy
=0
1
x2+xy
2 x=y2
x=y3dy=
0
1
(2y4
y6
y5)dy
=25
y5 1
7y
7 16
y6 1
0=
19210
V =1
2
1
7 3y
xydxdy=1
212
x2y
x=7 3y
x=1dy
=12 1
2
(48y 42y2+9y
3)dy
=12
24y2
14y3+
94
y4 2
1=
318
20.
21.
22.
6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V =0
1
x
1
(x2+3y
2)dydx
=0
1
x2y+y
3 y=1
y=xdx=
0
1
(x2+1 2x
3)dx
=13
x3+x
12
x4 1
0=
56
V =0
1
0
1 x
1 x y( ) dydx
=0
1
y xyy
2
2
y=1 x
y=0dx
=0
1
(1 x)2 1
2(1 x)
2dx
=0
112
(1 x)2dx=
16
(1 x)3 1
0=
16
23.
24.
7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V =0
1
x
2 x
xdydx
=0
1
x yy=2 x
y=xdx=
0
1
(2x 2x2)dx
= x2 2
3x
3 1
0=
13
V =2
2
x2
4
x2dydx
=2
2
x2
yy=4
y=x2dx=
2
2
(4x2
x4)dx
=43
x3 1
5x
5 2
2=
323
325
+323
325
=12815
25.
26.
8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V =0
2
0
2y
4 y2
dxdy
=0
2
x 4 y2 x=2y
x=0dy=
0
2
2y 4 y2
dy
=23
4 y2( ) 3/2 2
0=0+
163
=163
V =0
1
0
1 x2
ydydx=0
1y
2
2
y= 1 x2
y=0dx
=0
11 x
2
2dx=
12
x13
x3 1
0=
13
V 8 V1
V1 =
0
r
0
r2
y2
r2
y2
dxdy
=0
r
x r2
y2 x= r
2y
2
x=0dy
27.
28.
By symmetry, the desired volume is times the volume in the first octant. Now
9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
r
(r2
y2)dy= r
2y
13
y3 r
0=
23
r3
V =163
r3
x=0 x 1.213
DxdA
0
1.213
x4
3x x2
xdydx=0
1.213
xyy=3x x
2
y=x4
dx
=0
1.213
(3x2
x3
x5)dx= x
3 14
x4 1
6x
6 1.213
0
0.713
y=cos xy=x x 0.7391
V0
0.7391
x
cos x
zdydx=0
0.7391
x
cos x
xdydx
Thus .
29.
From the graph, it appears that the two curves intersect at and at . Thus the desiredintegral is
30.
The desired solid is shown in the first graph. From the second graph, we estimate that intersects at . Therefore the volume of the solid is
10
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
0.7391
xyy=cos x
y=xdx=
0
0.7391
(xcos x x2)dx
= cos x+xsin x13
x3 0.7391
00.1024
y=0
V0
0.7391
0
x
xdydx+0.7391
/2
0
cos x
xdydx 0.4684
y=1 x2
y=x2
1 1,0( ) 1 x2
x2
1 1,1z=2x+2y+10 z=2 x y
V =1
1
x2
1
1 x2
(2x+2y+10)dydx1
1
x2
1
1 x2
(2 x y)dydx
=1
1
x2
1
1 x2
(2x+2y+10 (2 x y))dydx=1
1
x2
1
1 x2
(3x+3y+8)dydx
=1
1
3xy+32
y2+8y
y=1 x2
y=x2
1dx
=1
1
3x(1 x2)+
32
(1 x2)2+8(1 x
2) 3x(x
21)
32
(x2
1)2
8(x2
1) dx
=1
1
( 6x3
16x2+6x+16)dx=
32
x4 16
3x
3+3x
2+16x
1
1
=32
163
+3+16+32
163
3+16=643
y=1 z=3
y=x2
y=1 2+y 3y 0 y 1z=2+y z=3y
Note: There is a different solid which can also be construed to satisfy the conditions stated in theexercise. This is the solid bounded by all of the given surfaces, as well as the plane . In case youcalculated the volume of this solid and want to check your work, its volume is
.
31. The two bounding curves and intersect at with on .Within this region, the plane is above the plane , so
32. The two planes intersect in the line , , so the region of integration is the plane region
enclosed by the parabola and the line . We have for , so the solid region isbounded above by and bounded below by .
11
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
V =1
1
x2
1
(2+y)dydx1
1
x2
1
(3y)dydx=1
1
x2
1
(2+y 3y)dydx=1
1
x2
1
(2 2y)dydx
=1
1
2y y2 y=1
y=x2dx=
1
1
(1 2x2+x
4)dx= x
23
x3+
15
x5 1
1=
1615
y=x3
x y=x2+x x=2 x
2+x>x
3x
0,2( )
V =0
2
x3
x
x2+x
zdydx=0
2
x3
x
x2+x
(x3y
4+xy
2)dydx=
13,984,735,61614,549,535
x 1 y 1 2x2+y
2<8 x
22y
2
1 x 1 1 x2
y 1 x2
V =1
1
1 x2
1 x2
(8 x2
2y2) (2x
2+y
2) dydx=
132
x2+y
2=1 z=0 D
x2+y
21
D(1 x
2y
2)dA=
1
1
1 x2
1 x2
(1 x2
y2)dydx=
2
xy x2+y
2=2y
x2+y
22y=0 x
2+(y 1)
2=1 1 x 1
33. The two bounding curves and intersect at the origin and at , with on . Using a CAS, we find that the volume is
34. For and , . Also, the cylinder is described by the inequalities
, . So the volume is given by
[using a CAS]
35. The two surfaces intersect in the circle , and the region of integration is the disk :
. Using a CAS, the volume is .
36. The projection onto the plane of the intersection of the two surfaces is the circle
, so the region of integration is given by ,
12
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
1 1 x2
y 1+ 1 x2
2y x2+y
2
V =1
1
1 1 x2
1+ 1 x2
[2y (x2+y
2)]dydx=
2
D = (x,y) |0 y x ,0 x 4{ }= (x,y) | y
2x 4,0 y 2{ }
0
4
0
x
f (x,y)dydx=D
f (x,y)dA=0
2
y2
4
f (x,y)dxdy
D = x,y( ) |4x y 4,0 x 1{ }
= x,y( ) |0 xy4
,0 y 4{ }
0
1
4x
4
f (x,y)dydx =D
f (x,y)dA
. In this region, so, using a CAS, the volume is
.
37.
Because the region of integration is
we have
.
38.
Because the region of integration is
we have
13
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
4
0
y/4
f (x,y)dxdy
D = (x,y) | 9 y2
x 9 y2
,0 y 3{ }= (x,y) |0 y 9 x
2, 3 x 3{ }
0
3
9 y2
9 y2
f(x,y)dxdy=D
f(x,y)dA=3
3
0
9 x2
f(x,y)dydx
D = (x,y) |0 x 9 y ,0 y 3{ }= (x,y) |0 y 3,0 x 6{ } x,y( ) |0 y 9 x
2, 6 x 3{ }
0
3
0
9 y
f (x,y)dxdy=
Df (x,y)dA
39.
Because the region of integration is
we have
40.
To reverse the order, we must break the region into two separate type I regions. Because the region ofintegration is
we have
14
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
6
0
3
f (x,y)dydx+6
3
0
9 x2
f (x,y)dydx
D = (x,y) |0 y ln x,1 x 2{ }
= (x,y) |ey
x 2,0 y ln 2{ }
1
2
0
ln x
f (x,y)dydx =D
f (x,y)dA
=0
ln 2
ey
2
f (x,y)dxdy
D = (x,y) |arctanx y4
,0 x 1{ }= (x,y) |0 x tan y,0 y
4{ }
0
1
arctanx
/4
f (x,y)dydx =D
f (x,y)dA
=0
/4
0
tan y
f (x,y)dxdy
41.
Because the region of integration is
we have
42.
Because the region of integration is
we have
15
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
0
1
3y
3
ex
2
dxdy =0
3
0
x/3
ex
2
dydx
=0
3
ex
2
yy=x/3
y=0dx=
0
3x3
ex
2
dx
=16
ex
2 3
0=
e9
16
0
1
y
1
x3+1 dxdy =
0
1
0
x2
x3+1 dydx
=0
1
x3+1 y
y=x2
y=0dx=
0
1
x2
x3+1 dx
=29
(x3+1)
3/2 1
0=
29
23/2
1( )
0
3
y2
9
ycos x2dxdy =
0
9
0
x
ycos x2dydx
43.
44.
45.
16
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
9
cos x2 y
2
2
y= x
y=0dx=
0
912
xcos x2dx
=14
sin x2 9
0=
14
sin 81
0
1
x2
1
x3sin (y
3)dydx =
0
1
0
y
x3sin (y
3)dxdy
=0
1x
4
4sin (y
3)
x= y
x=0dy
=0
114
y2sin (y
3)dy
=112
cos (y3)
1
0=
112
(1 cos 1)
0
1
arcsiny
/2
cos x 1+cos2x dxdy
=0
/2
0
sin x
cos x 1+cos2x dydx
=0
/2
cos x 1+cos2x y
y=sin x
y=0dx
46.
47.
17
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
=0
/2
cos x 1+cos2x sin xdx
Let u=cos x,du= sin xdx,dx=du/( sin x)
=1
0
u 1+u2
du=13
1+u2( ) 3/2 0
1
=13
8 1( )=13
2 2 1( )
0
8
3y
2
ex
4
dxdy =0
2
0
x3
ex
4
dydx
=0
2
ex
4
yy=x
3
y=0dx=
0
2
x3e
x4
dx
=14
ex
4 2
0=
14
(e16
1)
D = (x,y) |0 x 1, x+1 y 1{ } (x,y) | 1 x 0,x+1 y 1{ }(x,y) |0 x 1, 1 y x 1{ } (x,y) | 1 x 0, 1 y x 1{ }
Dx
2dA =
0
1
1 x
1
x2dydx+
1
0
x+1
1
x2dydx+
0
1
1
x 1
x2dydx+
1
0
1
x 1
x2dydx
= 40
1
1 x
1
x2dydx[ by symmetry of the regions and because f (x,y)=x
20]
= 40
1
x3dx=4
14
x4 1
0=1
48.
49.
,
all type I.
18
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
D = (x,y) | 1 x 0, 1 y 1+x2{ } (x,y) |0 x 1, x y 1+x
2{ }(x,y) |0 x 1, 1 y x{ }
DxydA =
1
0
1
1+x2
xydydx+0
1
x
1+x2
xydydx+0
1
1
x
xydydx
=1
012
xy2 y=1+x
2
y= 1dx+
0
112
xy2 y=1+x
2
y= xdx+
0
112
xy2 y= x
y= 1dx
=1
0
x3+
12
x5
dx+0
112
(x5+2x
3x
2+x)dx+
0
112
(x2
x)dx
=14
x4+
112
x6 0
1+
12
16
x6+
12
x4 1
3x
3+
12
x2 1
0+
12
13
x3 1
2x
2 1
0
=13
+5
12112
=0
D= 0,1 0,1 0 x3+y
32 A(D)=1 0
Dx
3+y
3dA 2
D= (x,y) |x2+y
2 14{ } 1=e
0e
x2+y
2
e1/4
A(D)=4 4 D
ex
2+y
2
dA (e1/4
)4
m f (x,y) MD
mdAD
f (x,y)dAD
MdA
mD
1dAD
f (x,y)dA MD
1dA mA(D)D
f (x,y)dA MA(D)
50.
,
all type I.
51. For , and , so .
52. Since , and , so
.
53. Since , by (8)
by (7) by (10).
54.
19
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
Df (x,y)dA =
0
1
0
2y
f (x,y)dxdy+1
3
0
3 y
f (x,y)dxdy
=0
2
x/2
3 x
f (x,y)dydx
D(x
2tan x+y
3+4) dA=
Dx
2tan xdA+
Dy
3dA+
D4dA x
2tan x x D
yD
x2tan xdA=0 y
3y D
xD
y3dA=0
D(x
2tan x+y
3+4)dA=4
DdA=4(area of D)=4 2( ) 2
=8
D y
3x xD
3xdA=0 4y y D
xD
4ydA=0
D2 3x+4y( ) dA =
D2dA=2
DdA
= 2(area of D)=2(50)= 100
1 x2
y2
0D
1 x2
y2
dA
55. . But is an odd function of and is
symmetric with respect to the axis, so . Similarly, is an odd function of and
is symmetric with respect to the axis, so . Thus
56. First, 0in0.19in ] The region , shown in the figure, is symmetric with respect to the axis and
is an odd function of , so . Similarly, is an odd function of and is symmetric
with respect to the axis, so . Then
57. Since , we can interpret as the volume of the solid that lies below
20
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
z= 1 x2
y2
D xy z= 1 x2
y2
x2+y
2+z
2=1 z 0 xy x
2+y
2=1 D
112
43
(1)3
=23
z
y=1 13+4x 4x
2
2
13+4x 4x2=0
x=1 14
2
V =1 14( ) /2
1+ 14( ) /2
1 13+4x 4x2( ) /2
1+ 13+4x 4x2( ) /2
[(4 x2
y2) (1 x y)]dydx=
498
the graph of and above the region in the plane. is equivalent to
, which meets the plane in the circle , the boundary of . Thus, the
solid is an upper hemisphere of radius which has volume .
58. To find the equations of the boundary curves, we require that the values of the two surfaces bethe same. In Maple, we use the command solve(4 x^2 y^2=1 x y,y);and in Mathematica, we use Solve[4 x^2 y^ 2==1 x y,y] . We find that the curves have equations
.
To find the two points of intersection of these curves, we use the CAS to solve , finding
that . So, using the CAS to evaluate the integral, the volume of intersection is
.
21
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
R R= (r, ) |0 r 2,0 2{ }
Rf (x,y)dA=
0
2
0
2
f (rcos ,rsin )r dr d
R R= (x,y) |0 x 2,0 y 2 x{ }
Rf (x,y)dA=
0
2
0
2 x
f (x,y)dydx
R R= (x,y) | 2 x 2,x y 2{ }
Rf (x,y)dA=
2
2
x
2
f (x,y)dydx
R R= (r, ) |1 r 3,02{ }
Rf (x,y)dA=
0
/2
1
3
f (rcos ,rsin )r dr d
R R= (r, ) |2 r 5,0 2{ }
Rf (x,y)dA=
0
2
2
5
f (rcos ,rsin )r dr d
R
R= (r, ) |0 r 2 2,4
54{ }
Rf (x,y)dA=
/4
5 /4
0
2 2
f (rcos ,rsin )r dr d
2
4
7
r dr d R= (r, ) |4 r 7, 2{ }
2
4
7
r dr d =2
d4
7
r dr
=2 1
2r
2 7
4=
12
49 16( )=33
2
1. The region is more easily described by polar coordinates: . Thus
.
2. The region is more easily described by rectangular coordinates: .
Thus .
3. The region is more easily described by rectangular coordinates: .
Thus .
4. The region is more easily described by polar coordinates: .
Thus .
5. The region is more easily described by polar coordinates: . Thus
.
6. The region is more easily described by polar coordinates:
. Thus .
7. The integral represents the area of the region , the lower
half of a ring.
1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
0
/2
0
4cos
r dr d R= (r, ) |0 r 4cos ,0 /2{ }
r=4cos r2=4rcos x
2+y
2=4x (x 2)
2+y
2=4 R
2,0( )
0
/2
0
4cos
r dr d =0
/212
r2 r=4cos
r=0d =
0
/2
8cos2
d
=0
/2
4(1+cos 2 )d =4 +12
sin 2/2
0=2
D D= (r, ) |0 r 3,0 2{ }
DxydA =
0
2
0
3
(rcos )(rsin )r dr d =0
2
sin cos d0
3
r3dr
=12
sin2 2
0
14
r4 3
0=0
R(x+y)dA =
/2
3 /2
1
2
(rcos +rsin )r dr d =/2
3 /2
1
2
r2(cos +sin dr d
=/2
3 /2
(cos +sin d1
2
r2dr = sin cos
3 /2
/2
13
r3 2
1
=( 1 0 1+0)83
13
=143
8. The integral represents the area of the region .
Since , is the portion in the first quadrant of acircle of radius 2 with center .
9. The disk can be described in polar coordinates as . Then
10.
2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
Rcos (x
2+y
2)dA =
0 0
3
cos (r2)r dr d =
0d
0
3
rcos (r2)dr
=0
12
sin (r2)
3
0=
12
(sin 9 sin 0)=2
sin 9
R4 x
2y
2dA =
/2
/2
0
2
4 r2
r dr d =/2
/2
d0
2
r 4 r2
dr
=/2
/2
12
23
(4 r2)3/2 2
0=
2+
213
(0 43/2
) =83
De
x2
y2
dA =/2
/2
0
2
er
2
r dr d =/2
/2
d0
2
rer
2
dr
=/2
/2
12
er
2 2
0=
12
(e4
e0)=
2(1 e
4)
Rye
xdA=
0
/2
0
5
(rsin )ercos
r dr d =0
5
0
/2
r2sin e
rcosd dr
0
/2
r2sin e
rcosd u=rcos du= rsin d
0
/2
r2sin e
rcosd =
u=r
u=0
r eudu= r[e
0e
r]=re
rr
0
5
0
/2
r2sin e
rcosd dr=
0
5
(rer
r)dr= rer
er 1
2r
2 5
0=4e
5 232
R R= (r, ) |0 /4,1 r 2{ }
Rarctan(y/x)dA=
0
/4
1
2
arctan(tan )r dr d
y/x=tan (tan )= 0 /4
11.
12.
13.
14. . First we integrate
: Let , and
. Then
, where we integrated by parts in the
first term.
15. is the region shown in the figure, and can be described by .
Thus
since . Also, arctan for ,so the integral becomes
3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
0
/4
1
2
r dr d =0
/4
d1
2
r dr=12
2 /4
0
12
r2 2
1=
2
3232
=3
642
DxdA
=
x2+ y
24
x 0,y 0
xdA
x 1( )2+ y
21
y 0
xdA
=0
/2
0
2
r2cos dr d
0
/2
0
2cos
r2cos dr d
=0
/213
(8cos )d0
/213
(8cos4
)d
=83
812
cos3
sin +32
( +sin cos )/2
0
=83
23
0+32 2
=16 3
6
D= (r, ) /6 /6,0 r 3{ }
DdA =
/6
/6
0
cos 3
r dr d =/6
/612
r2 r=cos 3
r=0d
=/6
/612
cos23 d =2
0
/612
1+cos 62
d
.
16.
17. One loop is given by the region , so the area is
4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
=12
+16
sin 6/6
0=
12
D= (r, ) |0 2 ,0 r 4+3cos{ }
A(D) =D
dA=0
2
0
4+3cos
r dr d =0
212
r2 r=4+3cos
r=0d =
12 0
2
(4+3cos )2d
=12 0
2
(16+24cos +9cos2
)d =12 0
2
16+24cos +91+cos 2
2d
=12
16 +24sin +92
+94
sin 22
0=
412
A =20
/4
0
sin
r dr d =20
/412
r2 r=sin
r=0d
=0
/4
sin2
d =0
/412
(1 cos 2 )d
=12
12
sin 2/4
0
=12 4
12
sin2
0+12
sin 0 =18
2( )
18. , so
19. By symmetry,
5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
2=4sin =6
56
A =/6
5 /6
2
4sin
r dr d =/6
5 /612
r2 r=4sin
r=2d =
/6
5 /6
(8sin2
2)d
=/6
5 /6
[4(1 cos 2 ) 2]d = 2 2sin 25 /6
/6=
43
+2 3
V =
x2+ y
29
(x2+y
2)dA=
0
2
0
3
(r2)r dr d =
0
2
d0
3
r3dr=
2
0
14
r4 3
0=2
814
=81
2
x2+y
2+z
2=16 xy x
2+y
2=16
V = 2
4 x2+y
216
16 x2
y2
dA [by symmetry]
= 20
2
2
4
16 r2
r dr d =20
2
d2
4
r(16 r2)1/2
dr
= 22
0
13
(16 r2)3/2 4
2=
23
(2 )(0 123/2
)=43
12 12( )=32 3
V =2
x2+ y
2a
2a
2x
2y
2dA=2
0
2
0
a
a2
r2
r dr d =20
2
d0
a
r a2
r2
dr
=22
0
13
(a2
r2)3/2 a
0=2(2 ) 0+
13
a3
=43
a3
z=10 3x2
3y2
z=4 4=10 3x2
3y2
x2+y
2=2
20. implies that or , so
.
21.
22. The sphere intersects the plane in the circle , so
23. By symmetry,
24. The paraboloid intersects the plane when or . So
6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
V =x
2+ y
22
(10 3x2
3y2) 4 dA=
0
2
0
2
(6 3r2)r dr d
=0
2
d0
2
(6r 3r3)dr=
2
03r
2 34
r4 2
0=6
z= x2+y
2x
2+y
2+z
2=1 x
2+y
2+ x
2+y
2( ) 2
=1 x2+y
2=
12
V =x
2+ y
21/2
1 x2
y2
x2+y
2( )dA=0
2
0
1/ 2
1 r2
r( ) r dr d
=0
2
d0
1/ 2
r 1 r2
r2( )dr=
2
0
13
(1 r2)3/2 1
3r
3 1/ 2
0
= 213
1
21 =
32 2( )
3x2+3y
2=4 x
2y
2x
2+y
2=1
V =x
2+ y
21
[(4 x2
y2) 3(x
2+y
2)]dA=
0
2
0
1
4(1 r2)r dr d
=0
2
d0
1
(4r 4r3)dr=
2
02r
2r
4 1
0=2
x2+y
2=4 z= 64 4x
24y
2
z= 64 4x2
4y2
V =x
2+ y
24
64 4x2
4y2
64 4x2
4y2( ) dA
=x
2+y
24
2 64 4x2
4y2
dA=40
2
0
2
16 r2
r dr d
25. The cone intersects the sphere when or .
So
26. The two paraboloids intersect when or . So
27. The given solid is the region inside the cylinder between the surfaces
and . So
7
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
= 40
2
d0
2
r 16 r2
dr=42
0
13
(16 r2)3/2 2
0
= 813
(123/2
162/3
)=83
64 24 3( )
xy r2
1x
2+y
2r
2
2xy
V = 2
r2
1x
2+ y
2r
2
2
r2
2x
2y
2dA=2
0
2
r1
r2
r2
2r
2r dr d
= 20
2
dr
1
r2
r2
2r
2r dr=
43
(r2
2r
2)3/2 r
2
r1
=43
(r2
2r
2
1)3/2
r2
2=
12
h2+r
2
1
14
h2=r
2
2r
2
1
h V =43
14
h2 3/2
=6
h3
0
1
0
1 x2
ex
2+y
2
dydx =0
/2
0
1
er
2
r dr d
28. (a) Here the region in the plane is the annular region and the desired volume is
twice that above the plane. Hence
(b) A cross sectional cut is shown in the figure. So or .
Thus the volume in terms of is .
29.
8
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
=0
/2
d0
1
rer
2
dr
=/2
0
12
er
2 1
0=
14
(e 1)
/2
/2
0
a
(r2)3/2
r dr d =/2
/2
d0
a
r4dr
=/2
/2
15
r5 a
0
=15
a5
0 0
2
(rcos )2(rsin )
2r dr d =
0(sin cos )
2d
0
2
r5dr
=0
12
sin 22d
0
2
r5dr
=14
12
18
sin 40
16
r6 2
0
=14 2
646
=43
30.
31.
32.
9
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
0
/2
0
2cos
r2dr d =
0
/213
r3 r=2cos
r=0d
=0
/283
cos3
d
=83
sin13
sin3 /2
0=
169
D 20 DD f (x,y)
x,y( )
D= (x,y) |x2+y
2400{ } f (x,y) y
xy f (0, 20)=2 f (0,20)=7 yz
(0, 20,2) (0,20,7)7 2
20 20( ) =18
z 7=18
(y 20) z=18
y+92
f (x,y) x f (x,y)=18
y+92
Df (x,y)dA
D= (r, ) |0 r 20,0 2{ } x=rcos y=rsin
0
2
0
2018
rsin +92
r dr d =0
2124
r3sin +
94
r2 r =20
r =0d =
0
21000
3sin +900 d
=1000
3cos +900
2
0=1800
1800 56553
R
V =0
2
0
R
err dr d =
0
2
d0
R
rerdr=
2
0re
re
r R
0
= 2 [ ReR
eR+0+1]=2 (1 Re
Re
R)ft
3
33. The surface of the water in the pool is a circular disk with radius ft. If we place oncoordinate axes with the origin at the center of and define to be the depth of the water at
, then the volume of water in the pool is the volume of the solid that lies above
and below the graph of . We can associate north with the positive direction, so we are given that the depth is constant in the direction and the depth increases linearlyin the direction from to . The trace in the plane is a line segment from
to . The slope of this line is , so an equation of the line is
. Since is independent of , . Thus the volume is
given by , which is most conveniently evaluated using polar coordinates. Then
and substituting , the integral becomes
Thus the pool contains ft of water.
34. (a) The total amount of water supplied each hour to the region within feet of the sprinkler is
10
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
R
Vareaofregion
=V
R2
=2 1 Re
Re
R( )R
2
3
1/ 2
1
1 x2
x
xydydx+1
2
0
x
xydydx+2
2
0
4 x2
xydydx
=0
/4
1
2
r3cos sin dr d =
0
/4r
4
4cos sin
r =2
r =1d
=154 0
/4
sin cos d =154
sin2
2
/4
0=
1516
Da
e(x
2+y
2)dA=
0
2
0
a
rer
2
dr d =212
er
2 a
0= 1 e
a2( ) a
lima
1 ea
2( )= ea
2
0 a e(x
2+y
2)dA=
Sa
e(x
2+y
2)dA=
a
a
a
a
ex
2
ey
2
dxdy=a
a
ex
2
dxa
a
ey
2
dy a
=
R2
(x2+y
2)dA
=lima S
a
e(x
2+y
2)dA=lim
a a
a
ex
2
dxa
a
ey
2
dy = ex
2
dx ey
2
dy .
lima a
a
ex
2
dxa
a
ey
2
dy
(b) The average amount of water per hour per square foot supplied to the region within feet of the
sprinkler is ft (per hour per square foot). See the definition
of the average value of a function on page 1022 [ET 986].
35.
36. (a) for each . Then
since as . Hence .
(b) for each .
Then, from (a), , so
To evaluate , we are using the fact that these integrals are bounded.
11
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates
1,1 0<ex
2
1 , 1( ) 0<ex
2
ex
1,( ) 0<ex
2
<ex
0 ex
2
dx1
exdx+
1
1
dx+1
exdx=2(e
1+1)
ex
2
dx ey
2
dy = y x ex
2
dx
2
=
ex
2
dx= ex
2
0 x ex
2
dx=
t= 2 x ex
2
dx=1
2e
t2/2( )dt =
1
2e
t2/2
dt et2/2
dt= 2
u=x dv=xex
2
dx du=dx v=12
ex
2
0x
2e
x2
dx = limt 0
t
x2e
x2
dx=limt
12
xex
2 t
0+
0
t12
ex
2
dx
= limt
12
tet2
+12 0
ex
2
dx=0+12 0
ex
2
dx [by l’Hospital’s Rule]
=14
ex
2
dx [since ex
2
is an even function]
=14
[by Exercise 36(c)]
u= x u2=x dx=2udu
0x e
xdx = lim
t 0
t
x exdx=lim
t 0
t
ueu
2
2udu=20
u2e
u2
du
= 214
[by part(a)]=12
This is true since on , while on , and on , .
Hence .
(c) Since and can be replaced by , implies that
. But for all , so .
(d) Letting , , so that or .
37. (a) We integrate by parts with and . Then and , so
(b) Let . Then
12
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.4 Double Integrals in Polar Coordinates