19.1 SOLUTIONS 1307 CHAPTER NINETEEN - Valencia...

19.1 SOLUTIONS 1307

CHAPTER NINETEEN

Solutions for Section 19.1

Exercises

1. (a) The flux is positive, since ~F points in direction of positive x-axis, the same direction as the normal vector.(b) The flux is negative, since below the xy-plane ~F points toward negative x-axis, which is opposite the orientation of

the surface.(c) The flux is zero. Since ~F has only an x-component, there is no flow across the surface.(d) The flux is zero. Since ~F has only an x-component, there is no flow across the surface.(e) The flux is zero. Since ~F has only an x-component, there is no flow across the surface.

2. The vector field ~F = F1~i + F2~j + F3~k = −z~i + x~k is a field parallel to the xz-plane that suggests swirling aroundthe origin from the positive x-axis to the positive z-axis.

(a) The flux going through this surface is negative, because ~F · ~n = (−z~i + x~k ) ·~i = −z, z is positive here.(b) The flux going through this surface is positive, because ~F · ~n = −z, z is negative here.(c) The flux through this surface is negative, because ~F · ~n = (−z~i + x~k ) · (−~k ) = −x, x is positive.(d) The flux through this surface is negative, because ~F · ~n = −x, x is positive.(e) The flux through this surface is zero, because it is in the xz-plane, which is parallel to the vector field.

3. The vector field ~r is a field that always points away from the origin.(a) The flux through this surface is zero, because the plane is parallel to the field.(b) The flux through this surface is zero also, for the same reason.(c) The flux through this surface is zero also, for the same reason.(d) The flux through this surface is negative, because the field in that quadrant is going up and away from the origin, and

since the orientation is downward, the flux is negative.(e) The flux through this surface is zero also.

4. ~v · ~A = (2~i + 3~j + 5~k ) · ~k = 5.5. ~v · ~A = (2~i + 3~j + 5~k ) · 2~i = 4.6. The rectangle lies in the plane z+ 2y = 2. So a normal vector is 2~j +~k and a unit normal vector is 1√

5(2~j +~k ). Since

this points in the positive z-direction it is indeed an orientation for the rectangle. Since the area of this rectangle is√

5 wehave ~A = 2~j + ~k ,~v · ~A = (2~i + 3~j + 5~k ) · (2~j + ~k ) = 6 + 5 = 11.

7. The rectangle lies in the plane z + 2x = 2. So 2~i + ~k is a normal vector and 1√5(2~i + ~k ) is a unit normal vector. Since

this points in both the positive x-direction and the positive z-direction, it is an orientation for this surface. Since the areaof the rectangle is

√5, we have ~A = 2~i + ~k and ~v · ~A = (2~i + 3~j + 5~k ) · (2~i + ~k ) = 4 + 5 = 9.

8. The square has area 9, so its area vector is 9~i . Since ~F = −5~i on the square,

Flux = −5~i · 9~i = −45.

9. The square has area 16, so its area vector is 16~j . Since ~F = 5~j on the square,

Flux = 5~j · 16~j = 80.

10. Since the square, S, is in the plane y = 0 and oriented in the negative y-direction, d ~A = −~j dxdz and∫

S

~F · d ~A =∫

S

(0 + 3)~j · (−~j dxdz) = −3∫

S

dxdz = −3 · Area of square = −3(22) = −12.

1308 Chapter Nineteen /SOLUTIONS

11. The disk has area 25π, so its area vector is 25π~j . Thus

Flux = (2~i + 3~j ) · 25π~j = 75π.

12. Since ~F is a constant vector field, the flux through a closed surface is zero. (The flux that enters one side, exits the otherside.)

13. Since the square, S, is oriented upward, d ~A = ~k dxdy and

Flux =∫

S

~F · d ~A =∫

S

x~k · ~k dxdy =∫ 3

0

∫ 3

0

x dxdy =

∫ 3

0

x2

2

∣∣∣∣3

0

dy =9

2

∫ 3

0

dy =27

2.

14. The only contribution to the flux is from the ~k -component, and since the square, S, is oriented upward, we have

Flux =∫

S

(6~i + x2~j − ~k ) · d ~A =∫

S

−~k · d ~A = −Area of square = −4.

15. The only contribution to the flux is from the ~j -component, and since d ~A = ~j dx dz on the square, S, we have

Flux =∫

S

(6~i + x2~j − ~k ) · d ~A =∫ 2

−2

∫ 2

−2x2~j ·~j dx dz =

∫ 2

−2

x3

3

∣∣∣∣2

−2dz =

16

3· 4 = 64

3.

16. Since the vector field is constant, Flux = 0. The flux through opposite faces of the cube cancel.17. On the sphere of radius 3, the vector field has || ~F || = 21 and points outward everywhere. So

Flux =∫

S

~F · d ~A = ||~F || · Area of sphere = 21 · 4π32 = 756π.

18. We have d ~A =~i dA, and x = 4, so,∫

S

~F · d ~A =∫

S

((4 + 3)~i + (y + 5)~j + (z + 7)~k ) ·~i dA =∫

S

7 dA

= 7 · Area of rectangle = 7 · 6 = 42.

19. The vector field ~F and the area vector on the surface of the sphere are parallel, but in opposite directions. Since ~F ispointing inward and || ~F || = 6 on the surface

Flux =∫

S

~F · d ~A =∫

S

||~F || cosπ dA = −6 · Area of sphere = −6 · 4π22 = −96π.

20. We have d ~A =~i dA, so∫

S

~F · d ~A =∫

S

(2z~i + x~j + x~k ) ·~i dA =∫

S

2z dA

=

∫ 2

0

∫ 3

0

2z dzdy = 18.

21. Since the vector field is constant, if ~A is the area vector of the square

Flux = ~F · ~A .

An upward normal to the plane is~i + ~j + ~k , so a unit vector in this direction is (~i + ~j + ~k )√

3. The area vector hasmagnitude 4, so ~A = 4(~i +~j + ~k )/

√3. Thus

Flux = (~i + 2~j ) · 4(~i +~j + ~k )√

3=

4(1 + 2)√3

= 4√

3.

19.1 SOLUTIONS 1309

22. Since the disk is in the xy-plane and oriented upward, d ~A = ~k dxdy and∫

Disk

~F · d ~A =∫

Disk(x2 + y2)~k · ~k dxdy =

∫

Disk(x2 + y2) dxdy.

Using polar coordinates ∫

Disk

~F · d ~A =∫ 2π

0

∫ 3

0

r2 · r drdθ = 2π r4

4

∣∣∣∣3

0

=81π

2.

23. Since the disk is oriented in the positive x-direction, d ~A =~i dydz, so we have

Flux =∫

Disk

~F · d ~A =∫

Diskey

2+z2~i ·~i dydz =∫

Diskey

2+z2dydz.

To calculate this integral, we use polar coordinates with y = r cos θ and z = r sin θ. Then r2 = y2 + z2 and∫

Disk

~F · d ~A =∫ 2π

0

∫ 2

0

er2 · rdrdθ = 2π · e

r2

2

∣∣∣∣2

0

= π(e4 − 1).

24. Since the disk is horizontal and oriented upward, d ~A = ~k dxdy, so∫

Disk

~F · d ~A =∫

Diskcos(x2 + y2)~k · ~k dxdy =

∫

Diskcos(x2 + y2) dxdy.

Using polar coordinates, since the disk has radius 3, we have∫

Disk

~F · d ~A =∫

Diskcos(x2 + y2)dxdy =

∫ 2π

0

∫ 3

0

cos(r2)rdrd θ

=

∫ 2π

0

1

2sin(r2)

∣∣∣∣3

0

dθ = 2π · 12

sin(r2)

∣∣∣∣3

0

= 2π(

1

2sin(32)− 1

2sin(02)

)= π sin 9.

25. On the curved sides of the cylinder, the ~k component of ~F does not contribute to the flux. Since the~i and~j componentsare constant, these components contribute 0 to the flux on the entire cylinder. Therefore the only nonzero contribution tothe flux results from the ~k component through the top, where z = 2 and d ~A = ~k dA, and from the ~k component throughthe bottom, where z = −2 and d ~A = −~k dA:

Flux =∫

Top

~F · d ~A +∫

Bottom

~F · d ~A

=

∫

Top2~k · ~k dA+

∫

Bottom(−2~k ) · (−~k dA)

= 4

∫

TopdA = 4 · Area of top = 4 · π(32) = 36π.

26. On the curved side of the cylinder, only the components x~i + z~k contribute to the flux. Since x~i + z~k is perpendicularto the curved surface and ||x~i + z~k || = 2 there (because the cylinder has radius 2), we have

Flux through sides = 2 · Area of curved surface = 2 · 2π · 2 · 6 = 48π.On the flat ends, only y~j contributes to the flux. On one end, y = 3 and d ~A = ~j dA; on the other end, y = −3 andd ~A = −~j dA. Thus

Flux through ends = Flux through top + Flux through bottom

= 3~j ·~j π(22) + (−3~j ) · (−~j π(22)) = 24π.So,

Total flux = 48π + 24π = 72π.


Problems

27. (a) For a flat surface, flux through ~A is ~v · ~A . Therefore, the flux through each face of the cube is equal to (−~i + 2~j +~k ) · ( ~A of the face).

First we shall find the flux through the two faces parallel to the xy-plane, beginning with the one with negativez. The unit vector normal to this face and pointing outward is −~k . The area of the face equals 4, so ~A = −4~k . Theflux through the face with negative z equals

(−~i + 2~j + ~k ) · (−4~k ) = 0 + 0− 4 = −4For the face with positive z, the unit normal vector that points outward is ~k . Therefore ~A = 4~k . The flux throughthis face is given by

(−~i + 2~j + ~k ) · 4~k = 0 + 0 + 4 = 4Next, we will find the flux through the two faces parallel to the xz-plane, beginning with the one with negative

y. A unit vector normal to this face pointing outward is −~j . Therefore ~A = −4~j . The flux then equals(−~i + 2~j + ~k ) · (−4~j ) = 0− 8 + 0 = −8

For the face with positive y, the unit normal vector pointing outward is ~j . Therefore ~A = 4~j . The flux then equals

(−~i + 2~j + ~k ) · (4~j ) = 0 + 8 + 0 = 8Next, we will find the flux through the two faces parallel to the yz plane, beginning with the one with negative

x. A unit vector normal to this plane pointing outward is −~i . Therefore ~A = −4~i . The flux then equals(−~i + 2~j + ~k ) · (−4~i ) = 4 + 0 + 0 = 4

For the face with positive x, the unit normal vector pointing outward is~i . Therefore ~A = 4~i . The flux then equals

(−~i + 2~j + ~k ) · (4~i ) = −4 + 0 + 0 = −4Adding up all of these fluxes to get the flux out of the entire cube, we get

Total flux = −4 + 4− 8 + 8 + 4− 4 = 0(b) For any constant vector field ~v = a~i + b~j + c~k , we can calculate the flux out of the cube by the same method.

First we shall find the flux out of the two faces parallel to the xy plane, beginning with the one with negative z.A unit vector normal to this plane, that points negative (because of the orientation of the face) is −~k . The area of theface equals 4, therefore ~A = −4~k . The flux through ~A then equals

(a~i + b~j + c~k ) · (−4~k ) = 0 + 0− 4c = −4cFor the face with positive z, the unit normal vector pointing outward is ~k . Therefore ~A = 4~k . The flux then equals

(a~i + b~j + c~k ) · (4~k ) = 0 + 0 + 4c = 4c.Next, we will find the flux through the two faces parallel to the xz plane, beginning with the one with negative

y. A unit vector normal to this plane pointing outward is −~j . Therefore ~A = −4~j . The flux then equals(a~i + b~j + c~k ) · (−4~j ) = 0− 4b+ 0 = −4b

For the face with positive y, the unit normal vector pointing outward is ~j . Therefore ~A == 4~j . The flux then equals

(a~i + b~j + c~k ) · (4~j ) = 0 + 4b+ 0 = 4bNext, we will find the flux through the two faces parallel to the yz plane, beginning with the one with negative

x. A unit vector normal to this plane pointing outward is −~i . Therefore ~A = −4~i . The flux then equals(a~i + b~j + c~k ) · (−4~i ) = −4a+ 0 + 0 = −4a

For the face in the positive x, the unit normal vector pointing outward is~i . Therefore ~A = 4~i . The flux then equals

(a~i + b~j + c~k ) · (4~i ) = 4a+ 0 + 0 = 4aAdding up all of these fluxes to get the flux out of the entire cube, we get

Total flux = −4c+ 4c− 4b+ 4b+ 4a− 4a = 0(c) The answers in parts (a) and (b) make sense because the vector field is constant, and so it does not change as it comes

in the one side of the cube, and exits the other side. Therefore the two fluxes cancel each other out, making the totalflux zero.

19.1 SOLUTIONS 1311

28. (a) Let ~A be the area vector of any face of the tetrahedron in Figure 19.1. The flux through the face equals ~v · ~A becausethe vector field is constant. Therefore, the flux through each face of the tetrahedron is equal to (−~i + 2~j + ~k ) · ~A ,where ~A is the area of that face.

First we shall find the flux out of the triangle in the xy plane. A unit vector normal to that plane, that pointsnegative (because of the orientation of the face), is equal to −~k . The area of the face equals 0.5, therefore ~A =−0.5~k . The flux through ~A then equals

(−~i + 2~j + ~k ) · (−0.5~k ) = 0 + 0− 0.5 = −0.5.

Next, we will find the flux out of the triangle in the xz plane. A unit vector normal to that plane, that pointsnegative, is equal to −~j . The area of the face equals 0.5, therefore ~A = −0.5~j . The flux through ~A then equals

(−~i + 2~j + ~k ) · (−0.5~j ) = 0− 1 + 0 = −1.

Next, we will find the flux out of the triangle in the yz plane. A unit vector normal to that plane, that pointsnegative, is equal to −~i . The area of the face equals 0.5, therefore ~A = −0.5~i . The flux through ~A then equals

(−~i + 2~j + ~k ) · (−0.5~i ) = 0.5 + 0 + 0 = 0.5.

Last, we will find the flux out of the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). A unit vector normal tothat plane, that points positive, is equal to 1√

3(~i +~j +~k ). The area of the face equals

√3/2, since it is an equilateral

triangle with side√

2. Therefore:

~A =1√3

(~i +~j + ~k )(√

3/2) = 0.5(~i +~j + ~k ).

The flux through ~A then equals

(−~i + 2~j + ~k ) · (0.5~i + 0.5~j + 0.5~k ) = −0.5 + 1 + 0.5 = 1.

The total flux out of the tetrahedron is −0.5− 1 + 0.5 + 1 = 0. Therefore the flux equals zero.

x

y

z

Figure 19.1

(b) For any constant vector field ~v = a~i + b~j + c~k , we can find the flux out of the tetrahedron.First we shall find the flux out of the triangle in the xy plane. A unit vector normal to that plane, that points

negative (because of the orientation of the face), is equal to −~k . The area of the face equals 0.5, therefore ~A =−0.5~k . The flux through ~A then equals

(a~i + b~j + c~k ) · (−0.5~k ) = 0 + 0− 0.5 = −0.5c.

Next, we will find the flux out of the triangle in the xz plane. A unit vector normal to that plane, that pointsnegative, is equal to −~j . The area of the face equals 0.5, therefore: ~A = −0.5~j . The flux through ~A then equals

(a~i + b~j + c~k ) · (−0.5~j ) = 0− 0.5b+ 0 = −0.5b.

Next, we will find the flux out of the triangle in the yz plane. A unit vector normal to that plane, that pointsnegative, is equal to (−~i ). The area of the face equals 0.5, therefore ~A = −0.5~i . The flux through ~A then equals

(a~i + b~j + c~k ) · (−0.5~i ) = −0.5a+ 0 + 0 = −0.5a.


Last, we will find the flux out of the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). A unit vector normal tothat plane, that points positive, is equal to 1√

3(~i +~j + ~k ). The area of the face equals

√3/2, therefore:

~A =1√3

(~i +~j + ~k )(√

3/2) = 0.5(~i +~j + ~k ).

The flux through ~A then equals

(a~i + b~j + c~k ) · (0.5~i + 0.5~j + 0.5~k ) = 0.5a+ 0.5b+ 0.5c.

The total flux out of the tetrahedron is −0.5c − 0.5b − 0.5a + 0.5a + 0.5b + 0.5c = 0. Therefore, the flux isequal to zero.

(c) The answers in (a) and (b) make sense because the vector field is constant, so it does not change as it enters throughone side of the tetrahedron, and exits the other side. Therefore the two cancel each other out,causing the flux to beequal to zero.

29. The square of side 2 in the plane x = 5, oriented in the positive x-direction, has area vector ~A = 4~i . Since the vectorfield is constant

Flux = (a~i + b~j + c~k ) · 4~i = 4a = 24.Thus, a = 6 and we cannot say anything about the values of b and c.

30. (a) The net electric flux through this surface is zero, because the surface is placed so that it is always parallel with theelectric field, and there is no flow through the surface.

(b) The net flux is zero, because the flow in through one half of the cylinder is canceled by the flow out through the otherhalf.

31. (a) In the first and third integrals, only the~i component contributes to the flux integral. The orientation is in the positive~i directions in both cases, the disks are the same size, and the (x2 +4) is larger on the surface S3, so the flux throughS3 is larger than the flux through S1. In the case of S2, only the ~j -component contributes. Since the disks in S2 andS3 are the same size, and S2 oriented in the positive y-direction, the fact that (x2 + 4) is larger on S3 than y is on S2tells us that the flux through S3 is largest.

(b) On S3, the vector field is ((−3)2 + 4)~i + y~j = 13~i + y~j and d ~A =~i dA, so∫

S3

((x2 + 4)~i + y~j ) · d ~A =∫

S3

(13~i + y~j ) ·~i dA =∫

S3

13 dA = 13 · Area of disk = 13π.

32. (a) At the north pole, the area vector of the plate is upward (away from the center of the earth), and so is in the oppositedirection to the magnetic field. Thus the magnetic flux is negative.

(b) At the south pole, the area vector of the plate is again away from the center of the earth (because that is upward in thesouthern hemisphere), and so is in the same direction as the magnetic field. Thus, the magnetic flux is positive.

(c) At the equator the magnetic field is parallel to the plate, so the flux is zero.

33. Suppose ~n is the normal vector at a point on the surface, and ~F is the vector field at the same point, then

~F · ~n = ‖ ~F ‖‖~n ‖ cos θ.

By the assumption that ~F is normal to S, the angle θ = 0 at all points on the surface. In addition ‖~n ‖ = 1. Thus,~F · ~n = ‖ ~F ‖ = constant.

So, ∫

S

~F · d ~A =∫

S

‖~F ‖dA = ‖ ~F ‖∫

S

dA = ‖ ~F ‖ · Area of S.

34. (a) Figure 19.2 shows the electric field ~E . Note that ~E points radially outward from the z-axis.

19.1 SOLUTIONS 1313

−2 2

−2

2

x

y

Figure 19.2: The electric field in the xy-plane due to a line of positive charge uniformly

distributed along the z-axis: ~E (x, y, 0) = 2λx~i + y~j

x2 + y2

(b) On the cylinder x2 + y2 = R2, the electric field ~E points in the same direction as the outward normal ~n , and

‖ ~E ‖ = 2λR2‖x~i + y~j ‖ = 2λ

R.

So ∫

S

~E · d ~A =∫

S

~E · ~n dA =∫

S

‖ ~E ‖ dA =∫

S

2λ

RdA

=2λ

R

∫

S

dA =2λ

R· Area of S = 2λ

R· 2πRh = 4πλh,

which is positive, as we expected.

35. Since Pressure = Force/Area, we have

Force on a small patch with area ∆ ~A at point (x, y, z) ≈ P (x, y, z)‖∆ ~A ‖.This force is directed inward and normal to the surface, so the force is P (x, y, z)∆ ~A (if S is oriented with the inwardnormal). For buoyancy, take the upward component of this force, so

Buoyancy force = P (x, y, z)∆ ~A · ~k .Then:

Total buoyancy = lim‖∆ ~A ‖→0

∑

s

P (x, y, z)∆ ~A · ~k

=

∫

S

P (x, y, z)~k · d ~A

=

∫

S

~F · d ~A

36. (a) (i) The integral∫Wρ dV represents the total charge in the volume W .

(ii) The integral∫S~J · d ~A represents the total current flowing out of the surface S.

(b) The total current flowing out of the surface S is the rate at which the total charge inside the surface S (i.e., in thevolume W ) is decreasing. In other words,

Rate current flowing out of S = − ∂∂t

(charge in W ),

so ∫

S

~J · d ~A = − ∂∂t

(∫

W

ρ dV

).


37. (a) If we examine the equation for ~v , we see that when r = 0, that is, at the center of the pipe, ~v (0) becomes u~i . So u isthe speed at the center of the pipe; it is also the maximum speed since u(1− r2/a2) reaches its maximum at r = 0.

(b) The flow rate at the wall of the pipe (where r = a) is

~v (a) = u(1− a2/a2)~i = ~0 .

(c) To find the flux through a circular cross-sectional area, we use polar coordinates in the plane perpendicular to thevelocity. In these coordinates, an infinitesimal area, d ~A becomes r dr dθ~i . So the flux is given by

Flux =∫

S

~v · d ~A =∫

S

u(1− r2/a2)~i · r dr dθ~i =∫ 2π

0

∫ a

0

u(1− r2/a2)r dr dθ

= 2πu

∫ a

0

(r − r3

a2) dr = 2πu

(a2

2− a

2

4

)=πua2

2.

38. (a) From Newton’s law of cooling, we know that the temperature gradient will be proportional to the heat flow. If theconstant of proportionality is k then we have the equation ~F = k gradT . Since gradT points in the direction ofincreasing T , but heat flows toward lower temperatures, the constant k must be negative.

(b) This form of Newton’s law of cooling is saying that heat will be flowing in the direction in which temperature is de-creasing most rapidly, in other words, in the direction exactly opposite to gradT . This agrees with our intuition whichtells us that a difference in temperature causes heat to flow from the higher temperature to the lower temperature, andthe rate at which it flows depends on the temperature gradient.

(c) The rate of heat loss from W is given by the flux of the heat flow vector field through the surface of the body. Thus,

Rate of heatloss from W

=Flux of ~Fout of S

=

∫

S

~F · d ~A = k∫

S

(gradT ) · d ~A


Exercises

1. Using z = f(x, y) = x+ y, we have d ~A = (−~i −~j + ~k ) dx dy. As S is oriented upward, we have∫

S

~F · d ~A =∫ 3

0

∫ 2

0

((x− y)~i + (x+ y)~j + 3x~k ) · (−~i −~j + ~k ) dxdy

=

∫ 3

0

∫ 2

0

(−x+ y − x− y + 3x) dxdy =∫ 3

0

∫ 2

0

x dxdy = 6.

2. Using z = 1 − x − y, the upward pointing area element is d ~A = (~i + ~j + ~k ) dx dy, so the downward one isd ~A = (−~i −~j − ~k ) dx dy. Since S is oriented downward, we have

∫

S

~F · d ~A =∫

S

(x~i + y~j + z~k ) · d ~A

=

∫ 3

0

∫ 2

0

(x~i + y~j + (1− x− y)~k ) · (−~i −~j − ~k ) dxdy

=

∫ 3

0

∫ 2

0

(−x− y − 1 + x+ y) dxdy = −6.

19.2 SOLUTIONS 1315

3. Using z = x2 + y2, we find that the upward pointing area element is d ~A = (−2x~i − 2y~j + ~k ) dx dy. Since S isoriented downward, we have d ~A = (2x~i + 2y~j − ~k ) dx dy, so

∫

S

~F · d ~A =∫

S

(x~i + y~j + z~k ) · d ~A

=

∫

Disk(x~i + y~j + (x2 + y2)~k ) · (2x~i + 2y~j − ~k ) dx dy

=

∫

Disk(2x2 + 2y2 − x2 − y2) dx dy =

∫

Disk(x2 + y2) dx dy

=

∫ 2π

0

∫ 1

0

r2r dr dθ = 2π · r4

4

∣∣∣∣1

0

=π

2.

4. Writing the surface S as z = f(x, y) = −y + 1, we have

d ~A = (−fx~i − fy~j + ~k )dxdy.

Thus,∫

S

~F · d ~A =∫

R

~F (x, y, f(x, y)) · (−fx~i − fy~j + ~k ) dxdy

=

∫ 1

0

∫ 1

0

(2x~j + y~k ) · (~j + ~k ) dxdy

=

∫ 1

0

∫ 1

0

(2x+ y) dxdy =

∫ 1

0

(x2 + xy)

∣∣∣∣1

0

dy

=

∫ 1

0

(1 + y) dy = (y +y2

2)

∣∣∣∣1

0

=3

2.

5. y

x

R

1

2

y = −2x+ 2

Figure 19.3

Writing the surface S as z = f(x, y) = −2x− 4y + 1, we have


With R as shown in Figure 19.3, we have∫

S

~F · d ~A =∫

R


=

∫

R

(3x~i + y~j + (−2x− 4y + 1)~k ) · (2~i + 4~j + ~k ) dxdy


=

∫

R

(4x+ 1) dxdy =

∫ 1

0

∫ −2x+2

0

(4x+ 1) dydx

=

∫ 1

0

(4x+ 1)(−2x+ 2) dx

=

∫ 1

0

(−8x2 + 6x+ 2) dx = (−8x3

3+ 3x2 + 2x)

∣∣∣∣1

0

=7

3.

6. Writing the surface S as z = f(x, y) = 25− x2 − y2, we have


Thus,∫

S

~F · d ~A =∫

R


=

∫

R

(x~i + y~j ) · (2x~i + 2y~j + ~k ) dxdy

=

∫

R

2(x2 + y2) dxdy =

∫ 2π

0

∫ 5

0

2r2r drdθ

=

∫ 2π

0

r4

2

∣∣∣∣5

0

dθ =625

2(2π) = 625π.

7. Writing the surface S as z = f(x, y) = 25− x2 − y2, we have

d ~A = (−fx~i − fy~j + ~k )dxdy = (2x~i + 2y~j + ~k )dxdy.

Thus∫

S

~F · d ~A =∫

R

cos(x2 + y2)~k · (2x~i + 2y~j + ~k ) dxdy

=

∫

R

cos(x2 + y2) dxdy =

∫ 2π

0

∫ 5

0

cos r2 · r drdθ

=

∫ 2π

0

sin r2

2

∣∣∣∣5

0

dθ = π sin 25.

8. Writing the surface S as z = f(x, y) = y2 + 5, we have

d ~A = (−fx~i +−fy~j + ~k )dxdy.

Thus,∫

S

~F · d ~A =∫

R


=

∫

R

(−y~j + (y2 + 5)~k ) · (−2y~j + ~k ) dxdy

=

∫ 1

0

∫ 1

−2(3y2 + 5) dxdy =

∫ 1

0

(9y2 + 15) dy

= (3y3 + 15y)

∣∣∣∣1

0

= 18.

19.2 SOLUTIONS 1317

9. We have 0 ≤ z ≤ 6 so 0 ≤ x2 + y2 ≤ 36. Let R be the disk of radius 6 in the xy-plane centered at the origin. Becauseof the cone’s point, the flux integral is improper; however, it does converge. We have

∫

S

~F · d ~A =∫

R


=

∫

R

(−x√x2 + y2~i − y

√x2 + y2~j + (x2 + y2)~k )

·(− x√

x2 + y2~i − y√

x2 + y2~j + ~k

)dxdy

=

∫

R

2(x2 + y2) dxdy

= 2

∫ 6

0

∫ 2π

0

r3 dθdr

= 4π

∫ 6

0

r3 dr = 1296π.

10. Since y = f(x, z) = x2 + z2, we have

d ~A = (−fx~i +~j − fz~k ) dxdz = (−2x~i +~j − 2z~k ) dxdz.Thus, substituting y = x2 + z2 into ~F , we have

∫

S

~F · d ~A =∫

x2+z2≤1((x2 + z2)~i +~j − xz~k ) · (−2x~i +~j − 2z~k ) dxdz

=

∫

x2+z2≤1(−2x3 − 2xz2 + 1 + 2xz2) dxdz

=

∫ 1

−1

∫ √1−z2

−√

1−z2(1− 2x3) dxdz

=

∫ 1

−1

∫ √1−z2

−√

1−z2dxdz −

∫ 1

−1

∫ √1−z2

−√

1−z22x3 dxdz

= Area of disk−∫ 1

−1

x

4

2

∣∣∣∣

√1−z2

−√

1−z2

dz = π − 0 = π

11. Here z =√

9− x2 − y2, sozx = − x√

9− x2 − y2zy = − y√

9− x2 − y2.

The flux integral is given by∫

S

~F · d ~A =∫

S

(x√

9− x2 − y2~i + y~k)·(

x√9− x2 − y2

~i +y√

9− x2 − y2~j + ~k

)dxdy

=

∫ 3

−3

∫ √9−x2

−√

9−x2(x2 + y) dydx

Changing to polar coordinates gives∫

S

~F · d ~A =∫ 2π

0

∫ 3

0

(r2 cos2 θ + r sin θ) rdrdθ

=

∫ 2π

0

(81

4cos2 θ +

27

3sin θ

)dθ =

81

4π.


12. The plane through the points (1, 0, 0), (0, 1, 0), and (0, 0, 1) is given by x + y + z = 1, so S is the part of the graph ofz = f(x, y) = 1− x− y above the region R in the xy-plane where x ≥ 0, y ≥ 0, and x+ y ≤ 1. Thus

∫

S

~F · d ~A =∫

R

~F (x, y, f(x, y)) · (−fx~i − fy~j + ~k ) dx dy

=

∫

R

(x2~i + y2~j + (1− x− y)2~k ) · (~i +~j + ~k ) dx dy

=

∫

R

(x2 + y2 + (1− x− y)2) dx dy

=

∫ 1

0

∫ 1−x

0

(1 + 2x2 + 2y2 − 2x− 2y + 2xy) dy dx

=

∫ 1

0

[(1− x) + 2x2(1− x) + 23

(1− x)3 − 2x(1− x)

−(1− x)2 + x(1− x)2] dx = 14.

13. Since the radius of the cylinder is 1, using cylindrical coordinates we have

d ~A = (cos θ~i + sin θ~j )dθdz.

Thus,∫

S

~F · d ~A =∫ 6

0

∫ 2π

0

(cos θ~i + sin θ~j ) · (cos θ~i + sin θ~j ) dθ dz

=

∫ 6

0

∫ 2π

0

1 dθ dz = 12π.

14. Since the radius of the cylinder is 1, using cylindrical coordinates we have d ~A = (cos θ~i + sin θ~j )dθdz. Thus,∫

S

~F · d ~A =∫ 6

0

∫ 2π

0

(z cos θ~i + z sin θ~j + z3~k ) · (cos θ~i + sin θ~j ) dθ dz

=

∫ 6

0

∫ 2π

0

z dθ dz = 2π

(z2

2

)∣∣∣∣6

0

= 36π.

15. Since the radius of the sphere is 5, using spherical coordinates we have

d ~A = (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )25 sinφ dθ dφ.

Thus,∫

S

~F · d ~A =∫ π

2

0

∫ 2π

0

(25 cos2 φ~k ) · (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )25 sinφ dθ dφ

= 625

∫ π2

0

∫ 2π

0

cos3 φ sinφ dθdφ

= −1250π (cosφ)4

4

∣∣∣∣π2

0

=625

2π.

16. Since the radius of the sphere is a, using spherical coordinates we have

d ~A = (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )a2 sinφ dφ dθ.

19.2 SOLUTIONS 1319

Thus,∫

S

~F · d ~A =∫ 2π

0

∫ π

0

(a sinφ cos θ~i + a sinφ sin θ~j + a cosφ~k ) ·

(sinφ cos θ~i + sinφ sin θ~j + cosφ~k )a2 sinφ dφ dθ

= a3∫ 2π

0

∫ π

0

sinφ dφ dθ

= 2πa3∫ π

0

sinφ dφ = (2πa3)(2) = 4πa3.

Problems

17. On the disk, z = 0 and d ~A = ~k dx dy, so∫

S

~F · d ~A =∫

x2+y2≤1(xzeyz~i + x~j + (5 + x2 + y2)~k ) · ~k dx dy

=

∫

x2+y2≤1(5 + x2 + y2) dx dy =

∫ 2π

0

∫ 1

0

(5 + r2)r dr dθ

= 2π

(5r2

2+r4

4

)∣∣∣∣1

0

=11π

2.

18. The plane is x− z = 0 over region 0 ≤ x ≤√

2, 0 ≤ y ≤ 2. See Figure 19.4.

x

z

√2

2

2

y

Figure 19.4

Flux =

∫ 2

0

∫ √2

0

((exy + 3z + 5)~i + (exy + 5z + 3)~j + (3z + exy)~k

)· (~i − ~k ) dx dy

=

∫ 2

0

∫ √2

0

(exy + 3z + 5− 3z − exy) dx dy = 5(2)(√

2) = 10√

2

Alternatively, since a unit normal to the surface is ~n /√

2 = (~i −~j )/√

2, writing dA = ||d ~A ||, we have

Flux =

∫

S

~H · d ~A =∫

~H ·~i − ~k√

2dA =

∫5√2dA

=5√2

(Area of slanted square) =5√2

4 = 10√

2.


19. (a) The charge is contained in a sphere of radius a centered at the origin, and uniformly distributed through the regionenclosed by the sphere.

(b) Since ~e ρ is the unit vector outward normal to the sphere of radius ρ, we have ~E = E(ρ)~e ρ. Let S be a sphere offixed radius ρ, centered at the origin. Then

∫

W

δ dV =

{43πρ3δ0 ρ ≤ a

43πa3δ0 ρ > a.

On the other hand, since on the sphere d ~A = ~e ρdA, we have∫

S

~E · d ~A =∫

S

E(ρ)~e ρ · ~e ρdA = E(ρ)∫

S

dA = E(ρ)4πρ2.

Therefore, by Gauss’s Law,

E(ρ)4πρ2 =

{k 4

3πρ3δ0 ρ ≤ a

k 43πa3δ0 ρ > a.

Since ~E = E(ρ)~e ρ, simplifying gives

~E =

kδ03ρ~e ρ ρ ≤ a

kδ0a

3

3r3~e ρ ρ > a.

20. (a) The charge is contained in a cylinder of radius a centered at the origin, and uniformly distributed through the regionenclosed by the cylinder.

(b) Since ~e r is the unit outward pointing normal to the cylinder of radius r, we have ~E = E(r)~e r . Let S be a cylinderof fixed radius r, height 1, centered along the z-axis. If r ≤ a,

∫

W

δ dV = πr2δ0,

and if r > a ∫

W

δ dV = πa2δ0.

On the other hand, since d ~A = ~e rdA on S, we can write∫

S

~E · d ~A =∫

S

E(r)(~e r · ~e r)dA = E(r)∫

S

dA = E(r)2πr.

(The flux across the top and bottom of the cylinder is zero.) So, by Gauss’s Law

E(r)2πr =

kπr2δ0 if r ≤ a

kπa2δ0 if r > a.

Since ~E = E(r)~e r , simplifying gives

~E =

{12kδ0r~e r if r ≤ a

12kδ0

a2

r~e r if r > a.

19.3 SOLUTIONS 1321


Exercises

1. Since S is given by~r (s, t) = (s+ t)~i + (s− t)~j + (s2 + t2)~k ,

we have∂~r

∂s=~i +~j + 2s~k and

∂~r

∂t=~i −~j + 2t~k ,

and

∂~r

∂s× ∂~r∂t

=

∣∣∣∣∣∣

~i ~j ~k

1 1 2s

1 −1 2t

∣∣∣∣∣∣= (2s+ 2t)~i + (2s− 2t)~j − 2~k .

Since the~i component of this vector is positive for 0 < s < 1, 0 < t < 1, it points away from the z-axis, and so has theopposite orientation to the one specified. Thus, we use

d ~A = −∂~r∂s× ∂~r∂t

ds dt,

and so we have∫

S

~F · d ~A = −∫ 1

0

∫ 1

0

(s2 + t2)~k ·((2s+ 2t)~i + (2s− 2t)~j − 2~k

)ds dt

= 2

∫ 1

0

∫ 1

0

(s2 + t2) ds dt = 2

∫ 1

0

(s3

3+ st2

)∣∣∣∣s=1

s=0

dt

= 2

∫ 1

0

(1

3+ t2) dt = 2(

1

3t+

t3

3)

∣∣∣∣1

0

= 2(1

3+

1

3) =

4

3.

2. Since S is parameterized by~r (s, t) = 3 sin s~i + 3 cos s~j + (t+ 1)~k ,

we have∂~r

∂s= 3 cos s~i − 3 sin s~j and ∂~r

∂t= ~k .

So

∂~r

∂s× ∂~r∂t

=

∣∣∣∣∣∣

~i ~j ~k

3 cos s −3 sin s 00 0 1

∣∣∣∣∣∣= −3 sin s~i − 3 cos s~j ,

which points toward the z-axis and thus opposite to the orientation we were given. Hence, we use

d ~A = −∂~r∂s× ∂~r∂t

ds dt,

and so we have∫

S

~F · d ~A = −∫ 1

0

∫ π

0

(3 cos s~i + 3 sin s~j ) · (−3 sin s~i − 3 cos s~j ) ds dt

= 9

∫ 1

0

∫ π

0

2 sin s cos s ds dt = 9

∫ 1

0

∫ π

0

sin 2s ds dt

= 9

∫ 1

0

(−cos 2s

2

∣∣∣∣s=π

s=0

)dt = 0.


3. The cross product ∂~r /∂s× ∂~r /∂t is given by

∂~r

∂s× ∂~r∂t

=

∣∣∣∣∣∣

~i ~j ~k

2s 2 0

0 2t 5

∣∣∣∣∣∣= 10~i − 10s~j + 4st~k .

Since the z-component, 4st, of the vector ∂~r /∂s × ∂~r /∂t is positive for 0 < s ≤ 1, 1 ≤ t ≤ 3, we see that ∂~r /∂s ×∂~r /∂t points upward, in the direction of the orientation of S we were given. Thus, we use

d ~A =

(∂~r

∂s× ∂~r∂t

)ds dt,

and so we have∫

S

~F · d ~A =∫ 1

0

∫ 3

1

(5t~i + s2~j ) · (10~i − 10s~j + 4st~k ) dt ds

=

∫ 1

0

∫ 3

1

(50t− 10s3) dt ds =∫ 1

0

(25t2 − 10s3t)∣∣∣∣t=3

t=1

ds

=

∫ 1

0

(200− 20s3) ds = (200s− 5s4)∣∣∣∣1

0

= 200− 5 = 195.

4. Since~r (a, θ) = a cos θ~i + a sin θ~j + sin a2~k ,

we have

∂~r

∂a× ∂~r∂θ

=

∣∣∣∣∣∣

~i ~j ~k

cos θ sin θ 2a cos a2

−a sin θ a cos θ 0

∣∣∣∣∣∣= −2a2 cos θ cos a2~i − 2a2 sin θ cos a2~j + a~k .

The z-component, a, of the vector ∂~r /∂a× ∂~r /∂θ is positive for 1 ≤ a ≤ 3, 0 ≤ θ ≤ π, so ∂~r /∂a× ∂~r /∂θ pointsupward, in the direction of the orientation of S we were given. Thus, we use d ~A = (∂~r /∂a× ∂~r /∂θ) da dθ, giving

∫

S

~F · d ~A =∫ 3

1

∫ π

0

((− 2a cos θ

)~i + (2

a sin θ)~j)· ∂~r∂a× ∂~r∂θ

dθ da

=

∫ 3

1

∫ π

0

(4a cos a2 − 4a cos a2) dθ da = 0.

5. Using cylindrical coordinates, we see that the surface S is parameterized by

~r (r, θ) = r cos θ~i + r sin θ~j + r~k .

We have

∂~r

∂r× ∂~r∂θ

=

∣∣∣∣∣∣

~i ~j ~k

cos θ sin θ 1

−r sin θ r cos θ 0

∣∣∣∣∣∣= −r cos θ~i − r sin θ~j + r~k .

Since the vector ∂~r /∂r × ∂~r /∂θ points upward, in the direction opposite to the specified orientation, we use d ~A =− (∂~r /∂r × ∂~r /∂θ) dr dθ. Hence

∫

S

~F · d ~A = −∫ 2π

0

∫ R

0

(r5 cos2 θ sin2 θ~k ) · (−r cos θ~i − r sin θ~j + r~k ) dr dθ

= −∫ 2π

0

∫ R

0

r6 cos2 θ sin2 θ dr dθ

19.3 SOLUTIONS 1323

= −R7

7

∫ 2π

0

sin2 θ cos2 θ dθ

= −R7

7

∫ 2π

0

sin2 θ(1− sin2 θ) dθ

= −R7

7

∫ 2π

0

(sin2 θ − sin4 θ) dθ

= −(R7

7)(π

4) =−π28

R7.

The cone is not differentiable at the point (0, 0). However the flux integral, which is improper, converges.

Problems

6. The plane is parameterized by

~r = x~i + y~j + z~k = x~i + y~j + (2− 2x− y)~k ,

where (x, y) is in the disk R lying inside the circle x2 + y2 = 2x. By completing the square, this circle can be rewrittenas (x− 1)2 + y2 = 1 and so the disk has area π.

We have dA = ‖ ∂~r∂x× ∂~r

∂y‖ dxdy, where

∂~r

∂x× ∂~r∂y

=

∣∣∣∣∣∣

~i ~j ~k

1 0 −20 1 −1

∣∣∣∣∣∣= 2~i +~j + ~k

and so ∥∥∥∥∂~r

∂x× ∂~r∂y

∥∥∥∥ =√

6.

Thus, the surface area of the ellipse S is given by

Surface area =∫

S

1 dA =

∫

R

√6 dxdy

=√

6× (Area of disk x2 + y2 = 2x)=√

6π.

x

y

z

S

Figure 19.5

7. The elliptic cylindrical surface is parameterized by

~r = x~i + y~j + z~k = a cos θ~i + b sin θ~j + z~k where 0 ≤ θ ≤ 2π,−c ≤ z ≤ c.

We have

∂~r

∂θ× ∂~r∂z

=

∣∣∣∣∣∣

~i ~j ~k

−a sin θ b cos θ 00 0 1

∣∣∣∣∣∣= b cos θ~i + a sin θ~j .


This vector points away from the z-axis, so we use d ~A = (b cos θ~i + a sin θ~j ) dθdz, giving∫

S

~F · d ~A =∫ c

−c

∫ 2π

0

(b

a(a cos θ)~i +

a

b(b sin θ~j )) · (b cos θ~i + a sin θ~j ) dθ dz

=

∫ c

−c

∫ 2π

0

(b2 cos2 θ + a2 sin2 θ) dθdz

= 2πc(a2 + b2).

8. The surface S is parameterized by

~r = x~i + f(x) cos θ~j + f(x) sin θ~k , a ≤ x ≤ b, 0 ≤ θ ≤ 2π.

The area element on A is

dA =

∥∥∥∥∂~r

∂x× ∂~r∂θ

∥∥∥∥ dx dθ

= ‖(~i + f ′(x) cos θ~j + f ′(x) sin θ~k )× (−f(x) sin θ~j + f(x) cos θ~k )‖ dx dθ= ‖f(x)f ′(x)~i − f(x) cos θ~j − f(x) sin θ~k ‖ dx dθ= f(x)

√f ′(x)2 + cos2 θ + sin2 θ dx dθ

= f(x)√

1 + f ′(x)2 dx dθ.

So

Surface area =∫

S

dA =

∫ 2π

0

∫ b

a

f(x)√

1 + f ′(x)2 dx dθ = 2π

∫ b

a

f(x)√

1 + f ′(x)2 dx.

9. If S is the part of the graph of z = f(x, y) lying over a region R in the xy-plane, then S is parameterized by

~r (x, y) = x~i + y~j + f(x, y)~k , (x, y) in R.

So∂~r

∂x× ∂~r∂y

= (~i + fx~k )× (~j + fy~k ) = −fx~i − fy~j + ~k .

Since the ~k component is positive, this points upward, so if S is oriented upward

d ~A = (−fx~i − fy~j + ~k ) dx dy

and therefore we have the expression for the flux integral obtained on page 925:∫

S

~F · d ~A =∫

R

~F (x, y, f(x, y)) · (−fx~i − fy~k + ~k ) dx dy.

10. If S is the part of the cylinder of radiusR corresponding to the region T in θz-space, then S is parameterized in cylindricalcoordinates by

~r (θ, z) = R cos θ~i +R sin θ~j + z~k , (θ, z) in T .

So∂~r

∂θ× ∂~r∂z

= (−R sin θ~i +R cos θ~j )× ~k = R cos θ~i +R sin θ~j .This points outward, so

d ~A = (R cos θ~i +R sin θ~j ) dθ dz = (cos θ~i + sin θ~j )Rdθ dz

and therefore we obtain the expression for the flux integral in cylindrical coordinates on page 927:∫

S

~F · d ~A =∫

T

~F (R cos θ,R sin θ, z) · (cos θ~i + sin θ~j )Rdθ dz.

19.3 SOLUTIONS 1325

11. If S is the part of the sphere of radius R corresponding to the region T in θφ-space, then S is parameterized in sphericalcoordinates by

~r (θ, φ) = R sinφ cos θ~i +R sinφ sin θ~j +R cosφ~k , (θ, φ) in T .

So

∂~r

∂θ× ∂~r∂φ

=

∣∣∣∣∣∣

~i ~j ~k

−R sinφ sin θ R sinφ cos θ 0R cosφ cos θ R cosφ sin θ −R sinφ

∣∣∣∣∣∣

= −R2 sin2 φ cos θ~i −R2 sin2 φ sin θ~j −R2 sinφ cosφ~k= −R2 sinφ(sinφ cos θ~i + sinφ sin θ~j + cosφ~k ).

This points inward, so the outward area element is

d ~A = (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )R2 sinφ dθ dφ,

and therefore we obtain the expression for the flux integral in spherical coordinates on page 928:∫

S

~F · d ~A

=

∫

T

~F (R sinφ cos θ,R sinφ sin θ,R cosφ) · (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )R2 sinφ dθ dφ.

12. In terms of the st-parameterization,

d ~A =∂~r

∂s× ∂~r∂t

ds dt.

By the chain rule, we have

∂~r

∂s=∂~r

∂u

∂u

∂s+∂~r

∂v

∂v

∂s∂~r

∂t=∂~r

∂u

∂u

∂t+∂~r

∂v

∂v

∂t.

So taking the cross product, we get

∂~r

∂s× ∂~r∂t

=

(∂~r

∂u

∂u

∂s+∂~r

∂v

∂v

∂s

)×(∂~r

∂u

∂u

∂t+∂~r

∂v

∂v

∂t

)

=(∂u

∂s

∂v

∂t− ∂u∂t

∂v

∂s

)∂~r

∂u× ∂~r∂v

.

Now suppose we are going to change variables in a double integral from uv-coordinates to st-coordinates. TheJacobian is

∂(u, v)

∂(s, t)=

∣∣∣∣∂u∂s

∂v∂s

∂u∂t

∂v∂t

∣∣∣∣ =∂u

∂s

∂v

∂t− ∂u∂t

∂v

∂s.

Since the Jacobian is assumed to be positive, converting from a uv-integral to an st-integral gives:∫

T

~F · ∂~r∂u× ∂~r∂v

dudv =

∫

R

~F · ∂~r∂u× ∂~r∂v

∂(u, v)

∂(s, t)dsdt

=

∫

R

~F · ∂~r∂u× ∂~r∂v

(∂u

∂s

∂v

∂t− ∂u∂t

∂v

∂s

)dsdt.

However, we know that this gives us∫

T

~F · ∂~r∂u× ∂~r∂v

du dv =

∫

R

~F · ∂~r∂u× ∂~r∂v

(∂u

∂s

∂v

∂t− ∂u∂t

∂v

∂s

)dsdt =

∫

R

~F · ∂~r∂s× ∂~r∂t

dsdt.

Thus, the flux integral in uv-coordinates equals the flux integral in st-coordinates.


Solutions for Chapter 19 Review

Exercises

1. Since the surface is closed, the flux of a constant vector field out of it is 0.2. The only contribution to the flux is from the face x = 1, since the vector field is zero or parallel to the other faces. On this

face, ~G =~i . This face has area 6, so its area vector ~A = 6~i . Thus

Flux =~i · ~A = 6.

3. The only contribution to the flux is from the face z = 3, since the vector field is zero or parallel to the other faces. On thisface, ~H = 3x~k . The vector field is everywhere perpendicular to the face z = 3 but varies in magnitude from point topoint. On this surface, d ~A = ~k dx dy. Thus

Flux =∫ 2

0

∫ 1

0

3x~k · ~k dx dy =∫ 2

0

∫ 1

0

3x dx dy =3x2

2

∣∣∣∣1

0

· y∣∣∣∣2

0

= 3.

4. Since the surface is closed and the vector field is constant, the flux is zero.5. Since the surface is in the plane x = 2, only the ~i -component contributes to the flux. The area vector of the surface isπ12~i = π~i . Thus, ∫

S

(2~i + 3~j − 5~k ) · d ~A = 2~i · π~i = 2π.

6. Since the surface is in the plane x+ y + x = 1, whose normal vector is~i +~j + ~k , a unit normal in the direction of theorientation is (~i +~j + ~k )/

√3. Thus, the area vector of the surface is π12(~i +~j + ~k )/

√3 = π(~i +~j + ~k )/

√3. The

flux is given by∫

S

(2~i + 3~j + 5~k ) · d ~A = (2~i + 3~j + 5~k ) · π~i +~j + ~k√

3=π(2 + 3 + 5)√

3=

10π√3.

7. Since the vector field is everywhere perpendicular to the surface of the sphere, and || ~F || = π on the surface, we have∫

S

~F · d ~A = ||~F || · Area of sphere = π · 4π(π)2 = 4π4.

8. Since ~F = ~r = x~i + y~j + z~k , only the ~k component contributes to the flux. (The~i and ~j components have a zerodot product with the area vector of the square.) On the square, z = 2. The square has side 3, area 9, and area vector 9~k .Thus, ∫

Square

~F · d ~A = (x~i + y~j + 2~k ) · Area vector of square = 2~k · 9~k = 18.

9. Since ~F = ~r = x~i + y~j + z~k , only the~i component contributes to the flux. (The ~j and ~k components have a zero dotproduct with the area vector of the disk.) On the disk, x = 5. The disk has radius 3, area 9π, and area vector 9π~i . Thus,

∫

S

~F · d ~A = (5~i + y~j + z~k ) · Area vector of disk = 5~i · 9π~i = 45π.

10. Since ~G is constant, the net flux through the sphere is 0, so∫

S

~G · d ~A = 0.

11. The area vector of the surface is −49~j , so∫

S

~G · d ~A = (2~j + 3~k ) · (−49~j ) = −98.

SOLUTIONS to Review Problems for Chapter Nineteen 1327

12. The area vector of the surface is π(52)~k = 25π~k , so∫

S

~G · d ~A = (2~j + 3~k ) · 25π~k = 75π.

13. A normal vector to S is −~j + ~k , so a unit normal is ~n = (−~j + ~k )/√

2.∫

S

~G · d ~A = (2~j + 3~k ) · (−~j + ~k )√

2· (Area of surface) = (−2 + 3)√

2·√

2 = 1.

14. Only the~i component of ~F contributes to the flux. On the disk, d ~A =~i dA, so we have

Flux =∫

~F · d ~A = 2~i · (~i Area of disk) = 2π(52) = 50π.

15. Only the~k component contributes to the flux. In the plane z = 4, we have ~F = 2~i +3~j +4~k . On the square d ~A = ~k dA,so we have

Flux =∫

~F · d ~A = 4~k · (~k Area of square) = 4(52) = 100.

16. Only the~i and ~j components of ~F contribute to the flux. On the disk, d ~A = ~n dA where ~n is the unit normal to theplane in the direction of orientation, ~n = (~i +~j )/

√2. Thus d ~A = (~i +~j )dA/

√2, so

Flux =∫

~F · d ~A = (2~i + 3~j ) ·(

(~i +~j )√2

Area of disk

)=

5√2π(52) =

125√2π.

17. Only the~i -component contributes to the flux, so∫

S

~F · d ~A = 7 · Area of disk = 7 · π22 = 28π.

18. In the plane y = 3, we have ~F = x~i + 6~j + 3z~k . Only the ~j -component contributes to the flux, so∫

S

~F · d ~A = 6 · Area of square = 6 · 22 = 24.

19. Since the vector field is constant, the flux is zero.20. On the sphere of radius 2, the vector field has || ~F || = 10 and points inward everywhere (opposite to the orientation of

the surface). So

Flux =∫

S

~F · d ~A = −|| ~F || · Area of sphere = −10 · 4π22 = −160π.

21. We have d ~A = ~k dA, and z = 4, so,∫

S

~F · d ~A =∫

S

(x~i + y~j + (42 + 3)~k ) · ~k dA =∫

S

19 dA

= 19(Area of rectangle) = 19(6) = 114.

22. We have d ~A = ~k dA, so∫

S

~F · d ~A =∫

S

(z~i + y~j + 2x~k ) · ~k dA =∫

S

2x dA

=

∫ 3

0

∫ 2

0

2x dxdy = 12.


23. We have d ~A =~i dA, so∫

S

~F · d ~A =∫

S

((2 + cos z)~i + y~j + 2x~k ) ·~i dA =∫

S

(2 + cos z) dA

=

∫ 4

0

∫ 3

0

(2 + cos z) dydz = 3(8 + sin 4)

24. On the surface S, y is constant, y = −1, and d ~A = −~j dA, so,∫

S

~F · d ~A =∫

S

(x2~i + (x+ e−1)~j − ~k ) · (−~j ) dA = −∫

S

(x+ e−1) dA

= −∫ 4

0

∫ 2

0

(x+ e−1) dx dz = −4(2 + 2e−1) = −8(1 + e−1).

25. Observe that the ~j and ~k components of ~F are parallel to the surface S, so they contribute nothing to the flux integral.On the surface S, the~i component of ~F equals 5~i , because x = 0 on S. Since 5~i is normal to S and in the direction ofthe orientation of S,

∫S~F · d ~A =

∫S

5~i · d ~A = ‖5~i ‖(Area of S) = 20.26. There is no flux through the base or top of the cylinder because the vector field is parallel to these faces. For the curved

surface, consider a small patch with area ∆ ~A . The vector field is pointing radially outward from the z-axis and so isparallel to ∆ ~A . Since ‖ ~F ‖ =

√x2 + y2 = 2 on the curved surface of the cylinder, we have ~F ·∆ ~A = ‖ ~F ‖‖∆ ~A ‖ =

2∆A. Replacing ∆A with dA, we get∫

S

~F · d ~A =∫

Curvedsurface

2 dA = 2(Area of curved surface) = 2(2π · 2 · 3) = 24π.

27. The vector field ~F = −y~i + x~j + z~k is tangent to the curved surface of the cylinder. (The area vector is parallel to thevector pointing radially outward from the z-axis, namely x~i + y~j and (−y~i + x~j + z~k ) · (x~i + y~j ) = 0.) Thus theonly contributions to the flux integral are from the top and the bottom. On the top, z = 1 and d ~A = dA~k , so

~F · d ~A = (−y~i + x~j + ~k ) · dA~k = dA.

Thus ∫

Top

~F · d ~A =∫

TopdA = Area of top = π(1)2 = π.

Similarly, on the base, z = −1 and d ~A = (−dA~k ), so

~F · d ~A = (−y~i + x~j − ~k ) · (−dA~k ) = dA.∫

Base

~F · d ~A =∫

BasedA = Area of base = π.

Therefore,Total flux through cylinder = Flux through top + Flux through base = 2π.

28. First we havezx =

x√x2 + y2

zy =y√

x2 + y2.

Although z is not a smooth function of x and y at (0, 0), the improper integral that we get converges:∫

S

~F · d ~A =∫

S

(x2~i + y2~j +√x2 + y2~k ) · (− x√

x2 + y2~i − y√

x2 + y2~j + ~k ) dA

=

∫

S

(− x

3 + y3√x2 + y2

+√x2 + y2

)dA


Changing to polar coordinates we have∫

S

~F · d ~A =∫ π/2

0

∫ 1

0

(−r2 cos3 θ − r2 sin3 θ + r)r drdθ

=

∫ π/2

0

(−r

4

4(cos3 θ + sin3 θ) +

1

3r3∣∣∣∣r=1

r=0

)dθ

=

∫ π/2

0

(−1

4(cos3 θ + sin3 θ) +

1

3

)dθ

=

∫ π/2

0

(−1

4(cos θ − cos θ sin2 θ + sin θ − sin θ cos2 θ) + 1

3

)dθ

= −14

(sin θ − 13

sin3 θ − cos θ + 13

cos3 θ) +θ

3

∣∣∣∣π/2

0

=π

6− 1

3.

Problems

29.

x

y

z

S1

(0, 0, 2)

�

�

1

Figure 19.6

z

−x

−y

S2

(0, 0, 3)� �1

Figure 19.7

x

yz

S3

�

��

�

√2

√2

Figure 19.8

x

y

z

(1, 0, 1)

�

�

√2

�

�

√2

Figure 19.9

Flux through S1 = ~F · ~A = (−~i −~j + ~k ) · (~k ) = 1Flux through S2 = ~F · ~A = (−~i −~j + ~k ) · (~k ) = 1Flux through S3 = ~F · ~A = (−~i −~j + ~k ) · (−2~j ) = 2


For S4, a normal is −~i + ~k and the area is 2, so ~A = −√

2~i +√

2~k

Flux through S4 = ~F · ~A = (−~i −~j + ~k ) · (−√

2~i +√

2~k ) = 2√

2.

So,Flux through S1 = Flux through S2 < Flux through S3 < Flux through S4.

30. (a) The vector field points radially outward from the origin and the vectors increase in length as the distance from theorigin increases.

(b) The vector field on the sphere points outward everywhere, so the flux is positive.(c) On the surface of the sphere, the vector field and the normal vector are parallel to each other. Everywhere on the

surface || ~F || = c||~r || = 3c, so

Flux =∫

Sphere

~F · d ~A = 3c · Area of sphere = 3c · 4π(3)2 = 108cπ.

31. The vector field ~D has constant magnitude on S, equal to Q/4πR2, and points radially outward, so∫

S

~D · d ~A = Q4πR2

· 4πR2 = Q.

32. The vector field ~D has constant magnitude on S, equal to ρ0R/3, and points radially outward, so∫

S

~D · d ~A = ρ0R3· 4πR2 = 4

3πR3ρ0.

33. The vector x~i + y~j points perpendicularly out of the side of the cylinder and has constant magnitude R on the cylinder.So ∫

S

~E · d ~A = ρ2�0

R · Area of S = ρ2�0

R(2πRh) =ρ

�0πR2h.

The flux of ~E across the top or bottom of the cylinder is zero, so including the ends would not change the answer.

34. Notice that the speed is 3 cm/sec at the center of the pipe and 0 cm/sec at the sides. Suppose ~i is the unit vector parallelto the direction of flow. Then, at a distance r from the center of the pipe, the velocity is given by

~v =(

3− 34r2)~i cm/sec.

Divide the circular cross-section into concentric rings of width ∆r, so that the velocity is approximately constant oneach one. The area of a typical ring is ∆A ≈ 2πr∆r. Then since ~v and ∆ ~A are parallel (see Figure 19.10), we have

Flux through ring ≈ ~v ∆ ~A = ‖~v ‖‖∆ ~A ‖ ≈(

3− 34r2) cm

sec· (2πr∆r) cm2.

6

?

4 cm6?6?r

� ∆r

∆ ~A

~v

Figure 19.10: Flux through pipe whenvelocity varies with distance from the center

Thus, the flux through the circular cross-section of the pipe is given by

Flux = lim‖∆ ~A ‖→0

∑~v ·∆ ~A

= lim∆r→0

∑(3− 3

4r2)

2πr∆r

=

∫ r=2

r=0

(3− 3

4r2)

2πr dr = 6π

∫ 2

0

(r − r

3

4

)dr = 6π cm3/sec.


35. (a) Consider two opposite faces of the cube, S1 and S2. The corresponding area vectors are ~A 1 = 4~ı and ~A 2 = −4~ı(since the side of the cube has length 2). Since ~E is constant, we find the flux by taking the dot product, giving

Flux through S1 = ~E · ~A 1 = (a~i + b~j + c~k ) · 4~i = 4a.

Flux through S2 = ~E · ~A 2 = (a~i + b~j + c~k ) · (−4~i ) = −4a.Thus the fluxes through S1 and S2 cancel. Arguing similarly, we conclude that, for any pair of opposite faces, thesum of the fluxes of ~E through these faces is zero. Hence, by addition,

∫S~E · d ~A = 0.

(b) The basic idea is the same as in part (a), except that we now need to use Riemann sums. First divide S into twohemispheresH1 andH2 by the equatorC located in a plane perpendicular to ~E . For a tiny patch S1 in the hemisphereH1, consider the patch S2 in the opposite hemisphere which is symmetric to S1 with respect to the center O of thesphere. The area vectors ∆ ~A 1 and ∆ ~A 2 satisfy ∆ ~A 2 = −∆ ~A 1, so if we consider S1 and S2 to be approximatelyflat, then ~E · ∆ ~A 1 = − ~E · ∆ ~A 2. By decomposing H1 and H2 into small patches as above and using Riemannsums, we get ∫

H1

~E · d ~A = −∫

H2

~E · d ~A , so∫

S

~E · d ~A = 0.

(c) The reasoning in part (b) can be used to prove that the flux of ~E through any surface with a center of symmetry iszero. For instance, in the case of the cylinder, cut it in half with a plane z = 1 and denote the two halves by H1 andH2. Just as before, take patches in H1 and H2 with ∆A1 = −∆A2, so that ~E ·∆A1 = − ~E ·∆ ~A 2. Thus, we get

∫

H1

~E · d ~A = −∫

H2

~E · d ~A ,

which shows that ∫

S

~E · d ~A = 0.

36. There are two possible methods:(a) The flux of ~E through Sa is given by

∫

Sa

~E · d ~A =∫

Sa

q~r

‖~r ‖3 ·~r

‖~r ‖ dA,

since d ~A = ~n dA, where ~n is the outward pointing unit normal vector field on S and ~n = ~r /‖~r ‖, and dA is thescalar area element. On the sphere Sa, we have ‖~r ‖ = a and therefore

∫

Sa

~E · d ~A =∫

Sa

q

‖~r ‖2 dA

=q

a2

∫

Sa

dA

=q

a2· 4πa2

= 4πq.

(b) Alternatively, we may compute the flux of ~E through Sa by using the definition of the flux integral as a limit ofRiemann sums. Divide the sphere into approximately flat patches Pi. The vector area ∆ ~A i of the patch Pi has thedirection of the outward normal vector to the sphere at the point with position vector ~r i and the magnitude area(Pi).The outward normal vector at any point of the sphere is proportional to the position vector of the point, hence wemust have

∆ ~A i = Area(Pi)~r i‖~r i‖

.

The Riemann sum corresponding to the above division of the sphere is therefore∑

i

~E (~r i) ·∆ ~A i =∑

i

q~r i‖~r i‖3

Area(Pi)~r i‖~r i‖

=∑

i

qArea(Pi)

a2

since ‖~r i‖ = a for any i. But∑

iArea(Pi) = Area(Sa) = 4πa2, so we get

∑

i

~E (~r i) ·∆ ~A i = 4πq.


The right-hand side is independent of the way the sphere is divided, therefore in the limit, as ‖∆ ~A i‖ → 0, we get:∫

Sa

~E · d ~A = lim‖∆ ~A i‖→0

∑

i

~E (~r i) ·∆ ~A i = 4πq.

37. (a) The vector field ~B is sketched in Figure 19.11 for I > 0.

−2−4 2 4

−2

−4

2

4

x

y

Figure 19.11

(b) The disk S can be parameterized as z = h (viewed as a constant function of x and y), for x, y in the region{x2 + y2 ≤ a2}. Hence

∫

S

~B · d ~A =∫

{x2+y2≤a2}

I

2π· −y~ı + x~x2 + y2

· ~k dx dy = 0,

since~ı · ~k = 0 and ~ · ~k = 0. The answer is as we would expect, since the vector field ~B is tangent to the surfaceS, hence there is no flux through S.

(c) The flux of ~B through S2 is given by∫S2~B · d ~A . On S2 we have

~B (x, y, z) =I

2π· −y~ıy2

= − I2πy

~ı ,

andd ~A = ~n dA = −~ı dy dz

Hence,∫

S2

~B · d ~A =∫ h

0

∫ b

a

I

2π· (−~ı )

y· (−~ı ) dy dz

=I

2π

∫ h

0

∫ b

a

1

ydy dz

=I

2π

∫ h

0

[ln |y|]ba dz

=I

2π

∫ h

0

(ln |b| − ln |a|) dz

=I

2πh(

ln

∣∣∣ ba

∣∣∣).

This time we get a non-zero flux since the direction of ~B is everywhere parallel to the orientation of S2. For 0 <a < b the flux is positive since |b/a| > 1 and increases as the area S2 increases. This is as Figure 19.11 would leadus to expect. For a < b < 0, the flux is negative since |b/a| < 1. If a < 0 < b, the flux can be either positive ornegative.


38. (a) If ~r = x~ı + y~ + z~k is the position vector of a point on the sphere, then

~D (~r ) =3zp

a5~r − p

a3~k .

The second term is a constant vector field. Hence, by symmetry,∫

S

(− pa3~k)· d ~A = 0.

(See the solution to Problem 35 on page 1331). Let us also apply a symmetry argument to∫

S

(3zp

a5~r ) · d ~A .

We will show that the flux of ~D through the upper hemisphere H1 equals minus the flux of ~D through the lowerhemisphere H2. The flux of ~D through H1 and H2 will be computed as limits of Riemann sums.

Consider a small patch P1 in H1 and call its reflection about the xy-plane P2. The contribution of P1 to the fluxof ~D through S is

3z1p

a5~r · d ~A 1 = 3z1p

a5~r · Area(P1) ~r‖~r ‖ =

3z1p

a4· Area(P1),

whereas the contribution from P2 is3z2p

a5~r · d ~A 1 = 3z2p

a4Area(P2).

But Area(P1) = Area(P2) and z2 = −z1, so the contributions from P1 and P2 cancel each other. Dividing H1 andH2 into symmetric patches as above, and taking the limit as the areas of the patches become smaller and smaller, onegets ∫

H1

(3zp

a5~r)· d ~A = −

∫

H2

(3zp

a5~r)· d ~A .

i.e. ∫

S

(3zp

a5~r)· d ~A = 0.

Since we also know that ∫

S

(− pa3~k)· d ~A = 0,

we can conclude that ∫

S

~D · d ~A = 0.

(b) By Gauss’s law, ∫

S

~E · d ~A = 4π(q − q) = 0,

which is the same as the flux of ~D through S.

CAS Challenge Problems

39. (a) When x > 0, the vector x~i points in the positive x-direction, and when x < 0 it points in the negative x-direction.Thus it always points from the inside of the ellipsoid to the outside, so we expect the flux integral to be positive. Theupper half of the ellipsoid is the graph of z = f(x, y) = 1√

2(1− x2 − y2), so the flux integral is

∫

S

~F · d ~A =∫ 1/2

−1/2

∫ 1/2

−1/2x~i · (−fx~i − fy~j + ~k ) dxdy

=

∫ 1/2

−1/2

∫ 1/2

−1/2(−xfx) dxdy =

∫ 1/2

−1/2

∫ 1/2

−1/2

x2√1− x2 − y2

dxdy

=−√

2 + 11 arcsin( 1√3) + 10 arctan( 1√

2)− 8 arctan( 5√

2)

12= 0.0958.

Different CASs may give the answer in different forms. Note that we could have predicted the integral was positivewithout evaluating it, since the integrand is positive everywhere in the region of integration.


(b) For x > −1, the quantity x+ 1 is positive, so the vector field (x+ 1)~i always points in the direction of the positivex-axis. It is pointing into the ellipsoid when x < 0 and out of it when x > 0. However, its magnitude is smaller when−1/2 < x < 0 than it is when 0 < x < 1/2, so the net flux out of the ellipsoid should be positive. The flux integralis

∫

S

~F · d ~A =∫ 1/2

−1/2

∫ 1/2

−1/2(x+ 1)~i · (−fx~i − fy~j + ~k ) dxdy

=

∫ 1/2

−1/2

∫ 1/2

−1/2−(x+ 1)fx dxdy =

∫ 1/2

−1/2

∫ 1/2

−1/2

x(1 + x)√1− x2 − y2

dxdy

=

√2− 11 arcsin( 1√

3)− 10 arctan( 1√

2) + 8 arctan( 5√

2)

12= 0.0958

The answer is the same as in part (a). This makes sense because the difference between the integrals in parts (a) and(b) is the integral of

∫ 1/2−1/2

∫ 1/2−1/2(x/

√1− x2 − y2) dxdy, which is zero because the integrand is odd with respect

to x.(c) This integral should be positive for the same reason as in part (a). The vector field y~j points in the positive y-direction

when y > 0 and in the negative y-direction when y < 0, thus it always points out of the ellipsoid. Evaluating theintegral we get

∫

S

~F · d ~A =∫ 1/2

−1/2

∫ 1/2

−1/2y~j · (−fx~i − fy~j + ~k ) dxdy

=

∫ 1/2

−1/2

∫ 1/2

−1/2(−yfy) dxdy =

∫ 1/2

−1/2

∫ 1/2

−1/2

y2√1− x2 − y2

dxdy

=

√2− 2 arcsin( 1√

3)− 19 arctan( 1√

2) + 8 arctan( 5√

2)

12= 0.0958.

The symbolic answer appears different but has the same numerical value as in parts (a) and (b). In fact the answer isthe same because the integral here is the same as in part (a) except that the roles of x and y have been exchanged.Different CASs may give different symbolic forms.

40. (a) The surface has a shape of a flower or trumpet opening in the direction of the positive y-axis. See Figure 19.12.The outer rim is a circle of radius 4, so the surface lies above z = −4. Thus z + 4 > 0 on the surface, so thevector field (z + 4)~k points in the positive z direction everywhere on the surface. Thus it crosses the surface in theopposite direction as the orientation when it is below the xy-plane, and in the same direction when it is above thexy-plane. Also, it has smaller magnitude when −2 ≤ z ≤ 2 than it does when 0 ≤ z ≤ 2, so we expect the negativecontribution to the flux integral to be smaller than the positive contribution, so the flux integral should be positive.

−xy

z

Figure 19.12

(b) The area vector element is

∂~r

∂s× ∂~r∂t

= s2 cos t~i − (2s3cos(t)2 + 2s3sin2 t)~j + s2 sin(t)~k .

CHECK YOUR UNDERSTANDING 1335

This points in the direction of the negative y-axis, as required for computing the flux integral. The flux integral is∫

S

~F · d ~A =∫ 2π

0

∫ 2

0

~F (~r (s, t)) ·(∂~r

∂s× ∂~r∂t

)dsdt

=

∫ 2π

0

∫ 2

0

(s2 sin t+ 2)s2 sin t dsdt =32π

5

CHECK YOUR UNDERSTANDING

1. True. By definition, the flux integral is the limit of a sum of dot products, hence is a scalar.2. False. ~A is perpendicular to the flat surface.3. False. The flux integral measures the net flow through the surface. There could be as much flow into the sphere as out,

which would give a flux integral of zero. As an example, the constant field ~F = ~i has zero flux integral over the entiresphere, yet is not the zero vector field.

4. True. The flow of this field is in the same direction as the orientation of the surface everywhere on the surface, so the fluxis positive.

5. True. The flow of this field is in the same direction as the orientation of the surface everywhere on the surface, so the fluxis positive.

6. True. Since the vector field is constant, the negative flux into the bottom of the cube is equal in magnitude to the positiveflux out of the top, so these cancel in the sum defining the flux integral. The other four faces of the cube each have zeroflux from the field, since the field is parallel to each of them.

7. True. Reversing the orientation on S replaces all of the area vectors ∆ ~A in the sum defining the flux integral with theirnegatives, so that the flux integral over −S is the negative of the flux integral over S.

8. False. There is no reason to expect a relationship between the flux integrals over S1 and S2 simply based on their relativeareas. The value of the flux integral over a surface depends both on the shape of the surface and the behavior of the vectorfield at points on the surface. For example, let S1 be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0, oriented upward, and letS2 be the rectangle 0 ≤ y ≤ 1, 0 ≤ z ≤ 2, x = 0 with positive orientation in the~i direction. The area of S1 = 1 and thearea of S2 = 2. Then if ~F =~i we have

∫S1~F · d ~A = 0 (since ~F is parallel to S1) and

∫S2~F · d ~A = 2. These values

do not satisfy 2∫S1~F · d ~A =

∫S2~F · d ~A .

9. True. In the sum defining the flux integral for ~F , we have terms like ~F ·∆ ~A = (2 ~G ) ·∆ ~A = 2( ~G ·∆ ~A ). So eachterm in the sum approximating the flux of ~F is twice the corresponding term in the sum approximating the flux of ~G ,making the sum for ~F twice that of the sum for ~G . Thus the flux of ~F is twice the flux of ~G .

10. False. The flux integral measures the net flow through the surface S. The vector field ~G could be large in magnitude onS (larger than || ~F ||), but be parallel to the surface S, and so contribute nothing to the flux. Put another way, a “small”vector field, flowing directly across S, can have greater flux than a much “larger” field flowing parallel to S.

For example, take S to be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0, oriented upward. Then if ~F = ~k and ~G = 5~i ,the flux integrals have values

∫S~F ·d ~A = 1 and

∫S~G ·d ~A = 0 (since ~G is parallel to S). Thus

∫S~F ·d ~A >

∫S~G ·d ~A ,

but ||~F || = 1 < 5 = || ~G ||.11. False. The fact that ~r · ~F = 0 does not necessarily imply that ~F · d ~A = 0.12. True. The area vector for the graph of f(x, y), parametrized in the usual way, is given by ~A = fx~i + fy~j + ~k . The

surface area is then the double integral of the magnitude of ~A , namely√f2x + f2y + 1 over the given rectangle.

13. False. Both surfaces are oriented upward, so ~A (x, y) and ~B (x, y) both point upward. But they could point in differentdirections, since the graph of z = −f(x, y) is the graph of z = f(x, y) turned upside down.

14. False. The total flux can be 0 without the vector field always being perpendicular to the surface. For example, ifF (x, y, z) =~k , then the flux is zero over the sphere, but F is not perpendicular to the sphere except at the north and south poles.

15. True. The vector 2~i − 4~j + 6~k is the gradient vector for f(x, y, z) = x2 − y2 + z2 at (1, 2, 3) so it is perpendicular tothe surface. Thus it is parallel to the area vector.

16. False. It is true that both d ~A and 3~i +4~j +5~k are perpendicular to the plane at every point, so they are multiples of eachother. However, the ratio between them might not be a constant. For example, x = s3, y = t3, z = (1/5)(7− 3s3− 4t3)is a parameterization of the plane, but

d ~A = (3s2~i − (9/5)s2~k )× (3t2~j − (12/5)t2~k )dsdt


= ((27/5)s2t2~i + (36/5)s2t2~j + 9s2t2~k )dsdt

= (9/5)s2t2(3~i + 4~j + 5~k )dsdt.

PROJECTS FOR CHAPTER NINETEEN

1. (a) (i) Since the direction of the electric field is perpendicular to the surface of any cylinder with the wireas an axis, it is parallel to the surfaces of the two washers. Consequently, there is no flux through thewashers.

(ii) Gauss’ Law tells us that the total flux through the surface must be zero, since no charge is containedwithin it. (Note that the region within the surface S lies between the cylinders.) Since the flux throughthe washers is zero, the flux into the inner surface must equal the flux out of the outer surface in orderfor the net flux through the surface to be zero.

(iii) Since the surface area of a cylinder is given by A = 2πRL where R is the radius of the cylinder andL is its length, and we know that EaAa = EbAb (since the fluxes are equal), we have

Eb(2πbL) = Ea(2πaL)

EbEa

=2πaL

2πbLEbEa

=a

b.

(iv) The equations in part (iii) imply thataEa = bEb.

Since a, b are arbitrary radii we can say:

rEr = Constant

Er = Constant(

1

r

),

for any radius r. This statement tells us that the strength of the electric field at r is proportional to1/r.

(b) Since the electric field points perpendicular to the sheet, it is parallel to all sides of the box, except forthe two sides parallel to the sheet. Additionally, since there is no charge contained in the box, Gauss Lawtells us the net flux through the surface of the box must be zero. This implies that the flux into the nearface must equal the flux out of the far face. Since the faces have the same area, the field must have equalstrengths at the two faces in order for their fluxes to be equal. Since we did not use the values of a or b, wesee that for all points in space on the same side of the sheet, the field has the same magnitude.

2. (a) (i) In cylindrical coordinates, the position vector of a point (R, θ, z) on the cylinder is given by

~r = R cos θ~i +R sin θ~j + z~k .

So ‖~r ‖ =√R2 + z2. For an area element on the cylinder we have

d ~A = (cos θ~i + sin θ~j )Rdz dθ,

so the flux integral is:∫

S

~E · d ~A =∫ 2π

0

∫ H

−Hq~r

‖~r ‖3· (cos θ~i + sin θ~j )Rdz dθ

= q

∫ 2π

0

∫ H

−H

R cos2 θ +R sin2 θ

(R2 + z2)3/2Rdz dθ = 2πq

∫ H

−H

R2 dz

(R2 + z2)3/2.

PROJECTS FOR CHAPTER NINETEEN 1337

To compute this one variable integral, we write:∫ H

−H

R2dz

(R2 + z2)3/2=

∫ H

−H

(R2 + z2)dz

(R2 + z2)3/2−∫ H

−H

z2dz

(R2 + z2)3/2

We calculate the integral∫ H

−H

z2dz

(R2 + z2)3/2=

∫ H

−Hz

zdz

(R2 + z2)3/2using integration by parts:

∫ H

−H

z2dz

(R2 + z2)3/2=

∫ H

−H

dz

(R2 + z2)1/2−(∫ H

−H

dz

(R2 + z2)1/2− z

(R2 + z2)1/2

∣∣∣∣H

−H

)

=2H√

R2 +H2.

Note that the integral∫ H

−H

dz

(R2 + z2)1/2has canceled. Therefore we have

∫

S

~E · d ~A = 4πq H√R2 +H2

.

(ii) • Let R be fixed. We have

limH→0

∫

S

~E · d ~A = limH→0

4πqH√

H2 +R2= 0.

limH→∞

∫

S

~E · d ~A = limH→∞

4πqH√

H2 +R2= 4πq.

• Now let H be fixed. We have

limR→0

∫

S

~E · d ~A = limR→0

4πqH√

H2 +R2= 4πq.

limR→∞

∫

S

~E · d ~A = limR→∞

4πqH√

H2 +R2= 0.

Each of these results is as we would expect from Gauss’ Law.(b) Let S denote the side of the cylinder. We have

∫

T

~E · d ~A =∫

S

~E · d ~A +∫

Top

~E · d ~A +∫

Bottom

~E · d ~A .

The result of part (a) shows that ∫

S

~E · d ~A = 4πq H√R2 +H2

.

Let’s compute∫

Top~E · d ~A . The normal at any point on the top is ~n = ~k and so d ~A = ~k r drdθ. On the

top, z = H so~r = r cos θ~i + r sin θ~j +H~k .

Therefore ‖~r ‖ =√r2 +H2 and we have

∫

Top

~E · d ~A =∫ 2π

0

∫ R

0

q~r

‖~r ‖3 ·~k r dr dθ

= q

∫ 2π

0

∫ R

0

(r cos θ~i + r sin θ~j +H~k ) · ~k(r2 +H2)3/2

r dr dθ

= qH

∫ 2π

0

∫ R

0

r dr dθ

(r2 +H2)3/2= −2πqH 1√

r2 +H2

∣∣∣∣R

0

= −2πq H√R2 +H2

+ 2πq.


Similarly, or using a symmetry argument, we find that the flux through the bottom is given by∫

Bottom

~E · d ~A = −2πq H√R2 +H2

+ 2πq.

Thus, the total flux is given by∫

S

~E · d ~A = 4πq H√R2 +H2

− 2πq H√R2 +H2

+ 2πq − 2πq H√R2 +H2

+ 2πq

= 4πq.

19.1 SOLUTIONS 1307 CHAPTER NINETEEN - Valencia...

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