Tuesday, December 8 ∗∗ More on Stokes’ Theorem1. Let F = 〈y2, x2, z2〉. Show that ∫

C1F ·dr =

∫C2

F ·drfor any two closed curves as shown lying on a cylinder about the z-axis.

980 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS

16. Let S be the portion of the plane z = x contained in the half-cylinder of radius R depicted in Figure 17. Use Stokes’ Theorem tocalculate the circulation of F = ⟨z, x, y + 2z⟩ around the boundary ofS (a half-ellipse) in the counterclockwise direction when viewed fromabove. Hint: Show that curl(F) is orthogonal to the normal vector tothe plane.

R

−R

FIGURE 17

17. Let I be the flux of F =〈ey, 2xex

2, z2

〉through the upper hemi-

sphere S of the unit sphere.(a) Let G =

〈ey, 2xex

2, 0

〉. Find a vector field A such that

curl(A) = G.(b) Use Stokes’ Theorem to show that the flux of G through S is zero.Hint: Calculate the circulation of A around ∂S .(c) Calculate I . Hint: Use (b) to show that I is equal to the flux of〈0, 0, z2

〉through S .

18. Let F = ⟨0, −z, 1⟩. Let S be the spherical cap x2 + y2 + z2 ≤ 1,where z ≥ 12 . Evaluate

∫∫

SF · dS directly as a surface integral. Then

verify that F = curl(A), where A = (0, x, xz) and evaluate the surfaceintegral again using Stokes’ Theorem.

19. Let A be the vector potential and B the magnetic field of the infinitesolenoid of radius R in Example 4. Use Stokes’ Theorem to compute:(a) The flux of B through a circle in the xy-plane of radius r < R(b) The circulation of A around the boundary C of a surface lyingoutside the solenoid

20. The magnetic field B due to a small current loop (which weplace at the origin) is called a magnetic dipole (Figure 18). Letρ = (x2 + y2 + z2)1/2. For ρ large, B = curl(A), where

A =〈− y

ρ3,

x

ρ3, 0

〉

(a) Let C be a horizontal circle of radius R with center (0, 0, c), wherec is large. Show that A is tangent to C.(b) Use Stokes’ Theorem to calculate the flux of B through C.

Rc

Current loop

A

z

y

x

FIGURE 18

21. Auniform magnetic field B has constant strengthb in the z-direction[i.e., B = ⟨0, 0, b⟩].

(a) Verify that A = 12 B × r is a vector potential for B, where r =⟨x, y, 0⟩.(b) Calculate the flux of B through the rectangle with vertices A, B,C, and D in Figure 19.

22. Let F =〈−x2y, x, 0

〉. Referring to Figure 19, let C be the closed

path ABCD. Use Stokes’ Theorem to evaluate∫

CF · dr in two ways.

First, regard C as the boundary of the rectangle with vertices A, B, C,and D. Then treat C as the boundary of the wedge-shaped box with anopen top.

FIGURE 19

23. Let F =〈y2, 2z + x, 2y2

〉. Use Stokes’ Theorem to find a plane

with equation ax + by + cz = 0 (where a, b, c are not all zero) suchthat

∮

CF · dr = 0 for every closed C lying in the plane. Hint: Choose

a, b, c so that curl(F) lies in the plane.

24. Let F =〈−z2, 2zx, 4y − x2

〉, and let C be a simple closed curve in

the plane x + y + z = 4 that encloses a region of area 16 (Figure 20).Calculate

∮

CF · dr, where C is oriented in the counterclockwise direc-

tion (when viewed from above the plane).

FIGURE 20

25. Let F =〈y2, x2, z2

〉. Show that

∫

C1F · dr =

∫

C2F · dr

for any two closed curves lying on a cylinder whose central axis is thez-axis (Figure 21).

FIGURE 21SOLUTION:Let A be the region of the cylinder bounded by C1 and C2, oriented via the outward pointing

normals. Thus ∂A =C1 −C2. Hence∫C1

F ·dr−∫

C2F ·dr =

∫∂A

F ·dr =∫ ∫

A(curlF) ·dS =

∫ ∫A〈0,0,2(x − y)〉 ·nd A = 0

since n has z-component 0. Thus∫

C1F ·dr = ∫C2 F ·dr

2. Consider the surface T which is the intersection of the plane x+2y+3z = 1 with the first octant.

(a) Draw a picture of T .

SOLUTION:

(b) Use Stokes’ Theorem to evaluate∫∂T

F ·dr for F = 〈y, −2z, 4x〉.SOLUTION:A normal vector v to the plane is 〈1,2,3〉 so a unit normal vector for T is 1p

14〈1,2,3〉.

By Stokes’, we need to evaluate∫ ∫T

(curlF) ·nd A =∫ ∫

T〈2,−4,−1〉 ·nd A =

∫ ∫T

−9p14

d A = −9p14

Area(T ).

Area(T ) = 12 |a × b| where a = 〈−1,1/2,0〉 and b = 〈0,−1/2,1/3〉, so Area(T ) = 12√

718 . So∫ ∫

T (curlF) ·nd A = −92p14 ·p

73p

2=−34 .

Alternative approach: Parametrize T by r (u, v) = (u, v, 13 (1−u −2v) with domain

D = {0 ≤ u ≤ 1 and 0 ≤ v ≤ 12− u

2}.

ru × rv =

∣∣∣∣∣∣∣i j k1 0 −1/30 1 −2/3

∣∣∣∣∣∣∣= 〈1/3,2/3,1〉So flux = ∫ ∫D〈2,−4,−1〉 · 〈1/3,2/3,1〉d vdu = ∫ ∫D −3d A =−3 Area(D) =−3/4.

3. Carefully explain how Green’s Theorem is actually a special case of Stokes’ Theorem.

SOLUTION:

Green’s theorem is just Stokes’ theorem for a surface and vector field that lie totally inside the

xy-plane. Let’s consider the situation of Green’s theorem — we start with a vector field in the xy-

plane, which looks like 〈P (x, y),Q(x, y)〉, and a region D inside the plane with some boundarycurves ∂D . Let’s think about these living inside of the xy-plane in 3-space, R3, just extending in

the most obvious way, so that we can see what Stokes’ theorem says. Let us now have the vector

field F on R3 defined byF(x, y, z) = 〈P (x, y),Q(x, y),0〉

The region D is now a surface which happens to lie entirely inside the plane z = 0. Computingthe curl of F gives

curl(F ) =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q 0

∣∣∣∣∣∣∣= 〈−∂Q

∂z,∂P

∂z,∂Q

∂x− ∂P∂y

〉 = 〈0,0, ∂Q∂x

− ∂P∂y

〉

as the functions P and Q only depend on x and y . Looking at the flux of the curl of this vector

field F through the surface with upwards normal n = 〈0,0,1〉, we have the following expression:

Flux through D of curl(F) =Ï

Dcurl(F) ·ndS =

ÏD〈0,0, ∂Q

∂x− ∂P∂y

〉 · 〈0,0,1〉dS =Ï

D

∂Q

∂x− ∂P∂y

dS

Technically, we should call the surface living inside of R3 something like D ′, and parametrizeby r (x, y) = (x, y,0), where the domain for (x, y) is D . In this case, the surface integral abovebecomes exactly a standard double integral in the plane:Ï

D

∂Q

∂x− ∂P∂y

d A

On the other hand, Stokes’ theorem gives us a second expression which computes the flux of

curl(F): the line integral of F along the boundary of D . Note that with upward normal, the rulefor the direction is exactly the same as the one for Green’s theorem – left arm points in walking

along the direction where your head points up in the direction of the normal. So, we have

exactly the statement of Green’s theorem:

Flux through D of curl(F) =Ï

D

∂Q

∂x− ∂P∂y

d A

and

Flux through D of curl(F) =∫∂D

F ·dr

4. Work the following problem.

980 C H A P T E R 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS

16. Let S be the portion of the plane z = x contained in the half-cylinder of radius R depicted in Figure 17. Use Stokes’ Theorem tocalculate the circulation of F = ⟨z, x, y + 2z⟩ around the boundary ofS (a half-ellipse) in the counterclockwise direction when viewed fromabove. Hint: Show that curl(F) is orthogonal to the normal vector tothe plane.

R

−R

FIGURE 17

17. Let I be the flux of F =〈ey, 2xex

2, z2

〉through the upper hemi-

sphere S of the unit sphere.(a) Let G =

〈ey, 2xex

2, 0

〉. Find a vector field A such that

curl(A) = G.(b) Use Stokes’ Theorem to show that the flux of G through S is zero.Hint: Calculate the circulation of A around ∂S .(c) Calculate I . Hint: Use (b) to show that I is equal to the flux of〈0, 0, z2

〉through S.

18. Let F = ⟨0, −z, 1⟩. Let S be the spherical cap x2 + y2 + z2 ≤ 1,where z ≥ 12 . Evaluate

∫∫

SF · dS directly as a surface integral. Then

verify that F = curl(A), where A = (0, x, xz) and evaluate the surfaceintegral again using Stokes’ Theorem.

19. Let A be the vector potential and B the magnetic field of the infinitesolenoid of radius R in Example 4. Use Stokes’ Theorem to compute:(a) The flux of B through a circle in the xy-plane of radius r < R(b) The circulation of A around the boundary C of a surface lyingoutside the solenoid

20. The magnetic field B due to a small current loop (which weplace at the origin) is called a magnetic dipole (Figure 18). Letρ = (x2 + y2 + z2)1/2. For ρ large, B = curl(A), where

A =〈− y

ρ3,

x

ρ3, 0

〉

(a) Let C be a horizontal circle of radius R with center (0, 0, c), wherec is large. Show that A is tangent to C.(b) Use Stokes’ Theorem to calculate the flux of B through C.

Rc

Current loop

A

z

y

x

FIGURE 18

21. Auniform magnetic field B has constant strengthb in the z-direction[i.e., B = ⟨0, 0, b⟩].

(a) Verify that A = 12 B × r is a vector potential for B, where r =⟨x, y, 0⟩.(b) Calculate the flux of B through the rectangle with vertices A, B,C, and D in Figure 19.

22. Let F =〈−x2y, x, 0

〉. Referring to Figure 19, let C be the closed

path ABCD. Use Stokes’ Theorem to evaluate∫

CF · dr in two ways.

First, regard C as the boundary of the rectangle with vertices A, B, C,and D. Then treat C as the boundary of the wedge-shaped box with anopen top.

FIGURE 19

23. Let F =〈y2, 2z + x, 2y2

〉. Use Stokes’ Theorem to find a plane

with equation ax + by + cz = 0 (where a, b, c are not all zero) suchthat

∮

CF · dr = 0 for every closed C lying in the plane. Hint: Choose

a, b, c so that curl(F) lies in the plane.

24. Let F =〈−z2, 2zx, 4y − x2

〉, and let C be a simple closed curve in

the plane x + y + z = 4 that encloses a region of area 16 (Figure 20).Calculate

∮

CF · dr, where C is oriented in the counterclockwise direc-

tion (when viewed from above the plane).

FIGURE 20

25. Let F =〈y2, x2, z2

〉. Show that

∫

C1F · dr =

∫

C2F · dr

for any two closed curves lying on a cylinder whose central axis is thez-axis (Figure 21).

FIGURE 21SOLUTION:

(a) Parametrize the circle at height c and radius R centered on the z axis by r(t ) = (R cos(t ),R sin(t ),c).Then, the tangent is given by

r′(t ) = (−R sin(t ),R cos(t ),0)

Substituting x(t ) = R cos(t ) and so on into the vector field A, we get what the vector field isat the corresponding point:

A(r(t )) = 〈−R sin(t )ρ3

,R cos(t )

ρ3,0〉

It doesn’t matter what ρ is exactly — we just want to know if the two vectors are multiples

of each other at r(t ), which they are:

r′(t ) · 1ρ3

= A(r(t ))

(b) Let D be the horizontal disk enclosed by C . We’ll compute the flux with upward normal.

By Stokes’ theorem, ÏD

B ·ndS =Ï

Dcurl(A) ·ndS =

∫C

A ·dr

Here C is traversed counterclockwise as seen from above to match the upward normal.

Then, using our parametrization from before:∫C

A ·dr =∫ 2π

0A(r(t )) · r′(t )d t =

∫ 2π0

r′(t )ρ3

· r′(t )d t =∫ 2π

0

R2

(R2 + c2) 32d t

as r′(t ) · r′(t ) = R2 sin(t )2 +R2 cos(t )2 +02 and ρ3 =(√

(R cos(t ))2 + (R sin(t ))2 + c2)3

Because R and c are fixed, we get a final answer immediately ofÏD

B ·ndS = 2πR2

(R2 + c2) 32