20.1 SOLUTIONS 1339 CHAPTER TWENTY - Valenciafd.valenciacollege.edu/file/tsmith143/ch20.pdf · 20.1...

20.1 SOLUTIONS 1339

CHAPTER TWENTYSolutions for Section 20.1

Exercises

1. The first vector field appears to be diverging more at the origin, since both fields are zero at the origin and the vectors nearthe origin are larger in field (I) than they are in field (II).

2. div ~F =∂

∂x(−x) +

∂

∂y(y) = −1 + 1 = 0

3. div ~F =∂

∂x(−y) +

∂

∂y(x) = 0

4. div ~F = ∂∂x

(x2 − y2) + ∂∂y

(2xy) = 2x+ 2x = 4x

5. div ~F =∂

∂x(−x+ y) +

∂

∂y(y + z) +

∂

∂z(−z + x) = −1 + 1− 1 = −1

6. Using the formula for ~a × ~r in Cartesian coordinates, we get

div ~F =∂

∂x(a2z − a3y) +

∂

∂y(a3x− a1z) +

∂

∂z(a1y − a2x) = 0

7. Taking partial derivatives, we get

div ~F =∂

∂x

(−y

(x2 + y2)

)+

∂

∂y

(x

(x2 + y2)

)=

2xy

(x2 + y2)2− 2yx

(x2 + y2)2= 0.

8. In coordinates, we have

~F (x, y, z) =(x− x0)√

(x− x0)2 + (y − y0)2 + (z − z0)2

~i +(y − y0)√

(x− x0)2 + (y − y0)2 + (z − z0)2

~j

+(z − z0)√

(x− x0)2 + (y − y0)2 + (z − z0)2

~k .

So if (x, y, z) 6= (x0, y0, z0), then

div ~F =

(1√

(x− x0)2 + (y − y0)2 + (z − z0)2− (x− x0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

+

(1√

(x− x0)2 + (y − y0)2 + (z − z0)2− (y − y0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

+

(1√

(x− x0)2 + (y − y0)2 + (z − z0)2− (z − z0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

=

((x− x0)2 + (y − y0)2 + (z − z0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2− (x− x0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

+

((x− x0)2 + (y − y0)2 + (z − z0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2− (y − y0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

+

((x− x0)2 + (y − y0)2 + (z − z0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2− (z − z0)2

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

)

=3((x− x0)2 + (y − y0)2 + (z − z0)2)− ((x− x0)2 + (y − y0)2 + (z − z0)2)

((x− x0)2 + (y − y0)2 + (z − z0)2)3/2

=2√

(x− x0)2 + (y − y0)2 + (z − z0)2=

2

‖~r − ~r 0‖.

1340 Chapter Twenty /SOLUTIONS

9. (a) Positive. The inflow from the lower left is less than the outflow from the upper right. The net outflow is positive.(b) Zero. The inflow on the right side is equal to outflow on the left.(c) Negative. The inflow from above is greater than the outflow below. The net outflow is negative.

10. Two vector fields that have positive divergence everywhere are in Figures 20.1 and 20.2.

x

y

Figure 20.1

x

y

Figure 20.2

11. Two vector fields that have negative divergence everywhere are in Figures 20.3 and 20.4.

x

y

Figure 20.3

x

y

Figure 20.4

12. Two vector fields that have zero divergence everywhere are in Figures 20.5 and 20.6.

x

y

Figure 20.5

x

y

Figure 20.6

20.1 SOLUTIONS 1341

Problems

13. Since divF (1, 2, 3) is the flux density out of a small region surrounding the point (1, 2, 3), we have

div ~F (1, 2, 3) ≈ Flux out of small region around (1, 2, 3)

Volume of region.

So

Flux out of region ≈ (div ~F (1, 2, 3)) · Volume of region

= 5 · 4

3π(0.01)3

=0.00002π

3.

14. Since div ~F at a point is approximately equal to the flux density out of a small region around the point, at (1, 5, 7), wehave

div ~F ≈ Flux out of small regionVolume of region

=0.0025

(0.01)3= 2500.

15. (a) By the definition of divergence as flux density, we have

Flux out of small region ≈ Divergence · Volume.

(i) At (2, 1, 1), we have div ~F = 22 + 12 − 12 = 4, so

Flux ≈ 4 · 4

3π(0.1)3 =

0.016

3π = 0.0168.

(ii) At (0, 0, 1), we have div ~F = 02 + 02 − 12 = −1, so

Flux ≈ −1 · 4

3π(0.1)3 = −0.004.

(b) The fact that the flux in part (i) is positive tells us that the vector field is pointing, on average, outward (rather thaninward) on the sphere around (2, 1, 1). The fact that the flux in part (ii) is negative tells us that, on average, the vectorfield is pointing inward (rather than outward) on the sphere around the point (0, 0, 1).

16. (a) We have

div ~F =∂z

∂z= 1.

(b) Above the xy plane, the vector field consists of vectors pointing vertically upward, getting longer as you go up.Below the xy plane, it consists of vectors pointing vertically downward, getting longer as you go down. You canclearly see the divergence on the xy plane, since vectors on either side of it point in opposite directions, but it is notso clear elsewhere. However, the fact that the vectors are getting longer as you go up means that the flux through acube situated above the xy plane will be non-zero, since the flux out of its top face will be greater than the flux intothe bottom face.

17. (a) In Cartesian coordinates,

~F (x, y, z) =x

(x2 + y2 + z2)3/2~i +

y

(x2 + y2 + z2)3/2~j +

z

(x2 + y2 + z2)3/2~k .

So if (x, y, z) 6= (0, 0, 0), then

div ~F (x, y, z) =

(1

(x2 + y2 + z2)3/2− 3x2

(x2 + y2 + z2)5/2

)+

(1

(x2 + y2 + z2)3/2− 3y2

(x2 + y2 + z2)5/2

)

+

(1

(x2 + y2 + z2)3/2− 3z2

(x2 + y2 + z2)5/2

)

=

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3x2

(x2 + y2 + z2)5/2

)


+

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3y2

(x2 + y2 + z2)5/2

)

+

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3z2

(x2 + y2 + z2)5/2

)

=3(x2 + y2 + z2)− 3(x2 + y2 + z2)

(x2 + y2 + z2)5/2

= 0.

(b)

x

z

Figure 20.7: The vector field ~F (~r ) =~r

‖~r ‖3shown in the xz-plane

The vector field is radial (all the arrows point out), so you might think that it is has non-zero divergence. (SeeFigure 20.7.) However the fact that the divergence is 0 at every point shows that flux density out of any small volumearound a point must be 0. This is possible because the arrows also get shorter as you go out.

18. (a) The cube is in Figure 20.8. The vector field is parallel to the x-axis and zero on the yz-plane. Thus the only contri-bution to the flux is from S2. On S2, x = c, the normal is outward. Since ~F is constant on S2, the flux through faceS2 is

∫

S2

~F · d ~A = ~F · ~A S2

= c~i · c2~i= c3.

Thus, total flux through box = c3.(b) Using the geometric definition of convergence

div ~F = limc→0

(Flux through boxVolume of box

)

= limc→0

(c3

c3

)

= 1

(c) Using partial derivatives,

div ~F =∂

∂x(x) +

∂

∂y(0) +

∂

∂z(0) = 1 + 0 + 0 = 1.

20.1 SOLUTIONS 1343

x

y

z

S2

S6

S3

?S1 (back)

IS5 (bottom)

� S4 (back)

Figure 20.8

19. See Figure 20.8.

(a) Since 2~i + 3~k is a constant field, its contribution to the flux is zero (flux in cancels flux out). Therefore∫~F · d ~A =∫

(y~j ) · d ~A =∫S3y~j · d ~A +

∫S4y~j · d ~A since only S3 and S4 are perpendicular to y~j . On S3, y = 0 so∫

S3y~j · d ~A = 0. On S4, y = c and normal is in the positive y-direction, so

∫S4y~j · d ~A = c(Area of S4) =

c · c2 = c3. Thus, total flux = c3.(b) Using the geometric definition of convergence

div ~F = limc→0


)

= limc→0

(c3

c3

)= 1.

(c)∂

∂x(2) +

∂

∂y(y) +

∂

∂z(3) = 0 + 1 + 0 = 1.

20. See Figure 20.8.

(a) This vector field points radially outward from the z-axis. Thus, the vector field is parallel to the surface on S1, S3, S5

and S6, so the only contributions to the flux integral are from S2 and S4.On S2, x = c and normal is in the positive x-direction, so the flux is

∫

S2

~F · d ~A =

∫

S2

(c~i + y~j ) · (dA~i ) =

∫

S2

c dA = c(Area of S2) = c3.

Similarly, the flux through S4 is∫

S4

~F · d ~A =

∫

S4

(x~i + c~j ) · (dA~j ) =

∫

S4

c dA = c(Area of S4) = c3.

Thus, the total flux through the box = 2c3.(b) Using the geometric definition of convergence, we have

div ~F = limc→0

(Flux through surface of boxVolume of box

)

= limc→0

(2c3

c3

)= 2.

(c)∂

∂x(x) +

∂

∂y(y) +

∂

∂z(0) = 1 + 1 + 0 = 2.


21. (a) div ~B =∂

∂x(−y) +

∂

∂y(x) +

∂

∂z(x+ y) = 0, so this could be a magnetic field.

(b) div ~B =∂

∂x(−z) +

∂

∂y(y) +

∂

∂z(x) = 0 + 1 + 0 = 1, so this could not be a magnetic field.

(c) div ~B =∂

∂x(x2− y2−x) +

∂

∂y(y− 2xy) +

∂

∂z(0) = 2x− 1 + 1− 2x+ 0 = 0, so this could be a magnetic field.

22. (a) Since a is a constant,

div grad f =∂

∂x

(∂f

∂x

)+

∂

∂y

(∂f

∂y

)=

∂

∂x(ay + 2axy) +

∂

∂y(ax+ ax2 + 3y2) = 2ay + 6y.

(b) Since div grad f = (2a+ 6)y, we have div grad f = 0 for all x, y if a = −3.

23. Let ~a = a1~i + a2

~j + a3~k with a1, a2, and a3 constant. Then f~a = f(x, y, z)(a1

~i + a2~j + a3

~k ) = f(x, y, z)a1~i +

f(x, y, z)a2~j + f(x, y, z)a3

~k = fa1~i + fa2

~j + fa3~k . So

div(f~a ) =∂(fa1)

∂x+∂(fa2)

∂y+∂(fa3)

∂z

= a1∂f

∂x+ a2

∂f

∂y+ a3

∂f

∂zsince a1, a2, a3 are constants

= (∂f

∂x~i +

∂f

∂y~j +

∂f

∂z~k ) · (a1

~i + a2~j + a3

~k )

= (grad f) · ~a .

24. Let ~F = F1~i + F2

~j + F3~k . Then

div(g ~F ) = div(gF1~i + gF2

~j + gF3~k )

=∂

∂x(gF1) +

∂

∂y(gF2) +

∂

∂z(gF3)

=∂g

∂xF1 + g

∂F1

∂x+∂g

∂yF2 + g

∂F2

∂y+∂g

∂zF3 + g

∂F3

∂z

=∂g

∂xF1 +

∂g

∂yF2 +

∂g

∂zF3 + g

(∂F1

∂x+∂F2

∂y+∂F3

∂z

)

= (grad g) · ~F + g div ~F .

25. Now grad f = fx~i + fy~j + fz~k and grad g is similar. Thus

grad f × grad g =

∣∣∣∣∣∣

~i ~j ~k

fx fy fz

gx gy gz

∣∣∣∣∣∣= (fygz − fzgy)~i − (fxgz − fzgx)~j + (fxgy − fygx)~k .

Thereforediv(grad f × grad g) =

∂

∂x(fygz − fzgy) +

∂

∂y(fzgx − fxgz) +

∂

∂z(fxgy − fygx).

Expanding using the product rule gives

div(grad f × grad g) = fyxgz + fygzx − fzxgy − fzgyx + fzygx + fzgxy

−fxygz − fxgzy + fxzgy + fxgyz − fyzgx − fygxz.

Now consider pairs of terms such as fyxgz − fxygz . Since fyx = fxy provided the second derivatives are continuous,these two terms cancel out. All the other terms cancel in pairs, showing that

div(grad f × grad g) = 0.

20.1 SOLUTIONS 1345

26. (a) Using r = (x2+y2)1/2, we calculate rx = (1/2)(x2+y2)−1/22x.Notice that rx = x/r and, by a similar argument,ry = y/r. We have

div(rA(x~i + y~j )) =∂

∂x(rAx) +

∂

∂y(rAy) = rA +AxrA−1rx + rA +AyrA−1ry

= 2rA +ArA−1 (xrx + yry)

= 2rA +ArA−1

(x2

r+y2

r

)

= 2rA +ArA−1

(x2 + y2

r

)

= 2rA +ArA−1

(r2

r

)

= (2 +A)rA.

(b) The divergence is positive for A = −1, zero for A = −2, and negative for A = −3.Notice that the diverging flow lines in the figures might have suggested positive divergence in all three cases.

However, the shortening of the vector field as you follow the flowlines when A = −2 and A = −3 brings abouta compression, which suggests negative divergence. This shortening is pronounced enough when A = −3 to makedivergence negative.

(c) Think of the divergence at a point P as the limit of the flux density out of a small sphere centered at P as the sphereshrinks to zero. Then the sign of the divergence tells us that the flux out of a small sphere centered at (1, 1, 1) ispositive for A = −1, zero for A = −2, and negative for A = −3. (This assumes that the sphere is small enough thatit does not contain the origin, where the vector fields are not defined.)

Since the vector fields and their divergences are not defined at (0, 0, 0), the divergence calculated in part (b)does not tell us anything about the flux through a sphere centered at (0, 0, 0).

27. Calculate rx = (1/2)(x2 + y2)−1/22x = x/r and ry = y/r. We have

divh(r)

r2(x~i + y~j ) =

∂

∂x

xh(r)

r2+

∂

∂y

yh(r)

r2

=h(r)

r2+ x

r2h′(r)rx − 2h(r)rrxr4

+h(r)

r2+ y

r2h′(r)ry − 2h(r)rryr4

= 2h(r)

r2+

(h′(r)

r2− 2h(r)

r3

)(xrx + yry)

= 2h(r)

r2+

(h′(r)

r2− 2h(r)

r3

)(x2 + y2

r

)

= 2h(r)

r2+

(h′(r)

r2− 2h(r)

r3

)(r2

r

)

=h′(r)

r.

28. We must compute div ~F where ~F =h(r)

r2(x~i + y~j ) and h(r) = r2e−r . Since

h′(r) = (2r − r2)e−r,

we have

div~F =h′(r)

r= (2− r)e−r.

Thus div ~F is positive for 0 ≤ r < 2, equals zero for r = 2 and is negative for r > 2. See Figure 20.9.


x

y

Figure 20.9: ~F = e−r(x~i + y~j )

29. We have now our temperature a function depending on t, x, y, z, hence T = T (t, x, y, z). For a fixed moment, say t0, Tis a function of only x, y, z. For this moment, t = t0, we have:

Rate of heat lossfrom volume V = k

∫

S

(gradT ) · d ~A .

where gradT =(∂T∂x~i + ∂T

∂y~j + ∂T

∂z~k)∣∣∣t=t0

. Now the rate of change, with respect to time, in the average temperature

in the region, at t = t0, is proportional to the average rate at which heat is being lost per unit volume at t = t0, so

∂Tavg∂t

∣∣∣∣t=t0

= −c(Rate heat lost

Volume V

)t=t0

=−ck

∫S

(gradT ) · d ~AVolume V

Taking the limit as V shrinks around the point, the average temperature through the region becomes the temperature atthat point. Thus using the definition of the divergence (with respect to x, y, z), we have

∂T

∂t

∣∣∣∣t=t0

= −ck limV→0

(∫S

(gradT ) · d ~AVolume V

)

= (−ck div gradT )t=t0

As this holds at every moment t0, one has:∂T

∂t= B · div gradT,

whereB = −ck is a function of time only, and the gradient and divergence are taken with respect to the variables x, y, z.

30. The charges that produce this electric field are concentrated along two vertical lines, one near x = −1 and the anotherone near x = 1. This is seen by the change in direction of the field at those lines. Near x = −1 the field is being repulsedby the line (seen by the field going away from the line), and the charge is therefore positive. Near x = 1 the field is beingattracted to the line (seen by the field going toward the line), and the charge is therefore negative.

31. (a) Translating the vector field into rectangular coordinates gives, if (x, y, z) 6= (0, 0, 0)

~E (x, y, z) =kx

(x2 + y2 + z2)3/2~i +

ky

(x2 + y2 + z2)3/2~j +

kz

(x2 + y2 + z2)3/2~k .

We now take the divergence of this to get

div ~E = k

(−3

x2 + y2 + z2

(x2 + y2 + z2)5/2+

3

(x2 + y2 + z2)3/2

)

= 0.

(b) Let S be the surface of a sphere centered at the origin. We have seen that for this field, the flux∫~E · d ~A is the same

for all such spheres, regardless of their radii. So let the constant c stand for∫~E · d ~A . Then

div ~E (0, 0, 0) = limvol→0

∫~E · d ~A

Volume insideS= lim

vol→0

c

Volume.

(c) For a point charge, the charge density is not defined. The charge density is 0 everywhere else.

20.1 SOLUTIONS 1347

32. (a) At any point ~r = x~i + y~j , the direction of the vector field ~v is pointing away from the origin, which means it is ofthe form ~v = f~r for some positive function f , whose value can vary depending on ~r . The magnitude of ~v dependsonly on the distance r, thus f must be a function depending only on r, which is equivalent to depending only on r2

since r ≥ 0. So ~v = f(r2)~r =(f(x2 + y2)

)(x~i + y~j ).

(b) At (x, y) 6= (0, 0) the divergence of ~v is

div~v =∂(K(x2 + y2)−1x)

∂x+∂(K(x2 + y2)−1y)

∂y=Ky2 −Kx2

(x2 + y2)2+Kx2 −Ky2

(x2 + y2)2= 0.

Therefore, ~v is a point source at the origin.(c) The magnitude of ~v is

‖~v ‖ = K(x2 + y2)−1|x~i + y~j | = K(x2 + y2)−1(x2 + y2)1/2 = K(x2 + y2)−1/2 =K

r.

(d) The vector field looks like that in Figure 20.10:

x

y

Figure 20.10

(e) We need to show that gradφ = ~v .

gradφ =∂

∂x(K

2log(x2 + y2))~i +

∂

∂y(K

2log(x2 + y2))~j

=Kx

x2 + y2~i +

Ky

x2 + y2~j

= K(x2 + y2)−1(x~i + y~j )

= ~v

33. (a) At any point ~r = x~i + y~j , the direction of the vector field ~v is pointing toward the origin, which means it is ofthe form ~v = f~r for some negative function f whose value can vary depending on ~r . The magnitude of ~v dependsonly on the distance r, thus f must be a function depending only on r, which is equivalent to depending only on r2

since r ≥ 0. So ~v = f(r2)~r =(f(x2 + y2)

)(x~i + y~j ).

(b) At (x, y) 6= (0, 0) the divergence of ~v is

div~v =∂(K(x2 + y2)−1x)

∂x+∂(K(x2 + y2)−1y)

∂y=Ky2 −Kx2

(x2 + y2)2+Kx2 −Ky2

(x2 + y2)2= 0.

Therefore, ~v is a point sink at the origin.(c) The magnitude of ~v is

‖~v ‖ = |K|(x2 + y2)−1|x~i + y~j | = |K|(x2 + y2)−1(x2 + y2)1/2 = |K|(x2 + y2)−1/2 =|K|r.

(remember, K < 0)


(d) The vector field looks like the following:

Figure 20.11

(e) We need to show that gradφ = ~v .

gradφ =∂

∂x(K

2log(x2 + y2))~i +

∂

∂y(K

2log(x2 + y2))~j

=Kx

x2 + y2~i +

Ky

x2 + y2~j

= K(x2 + y2)−1(x~i + y~j )

= ~v

34. The chain rule for partial differentiation of the formulas u = F cosφ and v = F sinφ gives

ux = (cosφ)Fx − F (sinφ)φx

vy = (sinφ)Fy + F (cosφ)φy.

We have

div ~F = ux + vy

= ((cosφ)Fx − F (sinφ)φx) + ((sinφ)Fy + F (cosφ)φy)

= (−vφx + uφy) +1

F(uFx + vFy)

= gradφ · (−v~i + u~j ) +1

FgradF · ~F

The definition of directional derivative tells us that the derivative of φ in the direction of ~N is given by φ ~N = gradφ · ~N .Similarly F~T = gradF · ~T . Since ~T = (1/F ) ~F and ~N = (1/F )(−v~i + u~j ), we have

div ~F = gradφ · (−v~i + u~j ) +1

FgradF · ~F

= F gradφ · 1

F(−v~i + u~j ) + gradF · 1

F~F

= F gradφ · ~N + gradF · ~T= Fφ ~N + F~T .

Solutions for Section 20.2

Exercises

1. No, because the surface S is not a closed surface.

20.2 SOLUTIONS 1349

2. First directly: On the faces x = 0, y = 0, z = 0, the flux is zero. On the face x = 2, a unit normal is~i and d ~A = dA~i .So ∫

Sx=2

~r · d ~A =

∫

Sx=2

(2~i + y~j + z~k ) · (dA~i )

(since on that face, x = 2)

=

∫

Sx=2

2dA = 2 · (Area of face) = 2 · 4 = 8.

In exactly the same way, you get ∫

Sy=2

~r · d ~A =

∫

Sz=2

~r · d ~A = 8,

so ∫

S

~r · −→dA = 3 · 8 = 24.

Now using divergence:

div ~F =∂x

∂x+∂y

∂y+∂z

∂z= 3,

so

Flux =

∫ 2

0

∫ 2

0

∫ 2

0

3 dx dy dz = 3 · (Volume of Cube) = 3 · 8 = 24

3. Finding flux directly:1) On bottom face, z = 0 so ~F = x2~i + 2y2~j is parallel to face so flux is zero.2) On front face, y = 0 so ~F = x2~i + 3z2~k is parallel to face so flux is zero.3) On back face, y = 1 so ~F = x2~i + 2~j + 3z2~k and ~A = ~j so flux is 2.4) On top face, z = 1 so ~F = x2~i + 2y2~j + 3~k and ~A = ~k so flux is 3.5) On side x = 1, ~F =~i + 2y2~j + 3z2~k and ~A = −~i so flux is −1.6) On side x = 2, ~F = 4~i + 2y2~j + 3z2~k and ~A =~i so flux is 4.Total flux is thus 8.

By the Divergence Theorem:div ~F = 2x+ 4y + 6z

So, if W is the interior of the box, we have∫

S

~F · d ~A =

∫

W

(2x+ 4y + 6z)dV = 2

∫ 2

1

∫ 1

0

∫ 1

0

(x+ 2y + 3z)dz dy dx

= 2

∫ 2

1

∫ 1

0

[xz + 2yz +

3z2

2

]1

0

dy dx = 2

∫ 2

1

∫ 1

0

(x+ 2y +

3

2

)dy dx

= 2

∫ 2

1

[xy + y2 +

3y

2

]1

0dx = 2

∫ 2

1

(x+ 1 +

3

2

)dx

=

∫ 2

1

(2x+ 5)dx = (x2 + 5x)

∣∣∣∣2

1

= 8

4. The location of the pyramid has not been completely specified. For instance, where is it centered on the xy plane? Howis base oriented with respect to the axes? Thus, we cannot compute the flux by direct integration with the information wehave. However, we can calculate it using the divergence theorem. First we calculate the divergence of ~F .

div ~F =∂(−z)∂x

+∂0

∂y+∂x

∂z= 0 + 0 + 0 = 0

Thus for any closed surface the flux will be zero, so the flux through our pyramid, regardless of its location or orientation,is zero.


5. First directly, since the vector field is totally in the ~j direction, there is no flux through the ends. On the side of thecylinder, a normal vector at (x, y, z) is x~i + y~j . This is in fact a unit normal, since x2 + y2 = 1 (the cylinder has radius1). Also, using x = cos θ, y = sin θ, in this case, the element of area dA equals 1dθdz. So

Flux =

∫~F · d ~A =

∫ 2

0

∫ 2π

0

(y~j ) · (x~i + y~j ) dθ dz

=

∫ 2

0

∫ 2π

0

y2 dθ dz =

∫ 2

0

∫ 2π

0

sin2 θ dθ dz =

∫ 2

0

πdz = 2π.

Now we calculate the flux using the divergence theorem. The divergence of the field is given by the sum of the

respective partials of the components, so the divergence is simply∂y

∂y= 1. Since the divergence is constant, we can

simply calculate the volume of the cylinder and multiply by the divergence

Flux = 1πr2h = 2π

6. First compute div ~F = 11 at origin, where ~F = x2~i + (y − 2xy)~j + 10z~k . By the Divergence Theorem,∫

sphere~F ·

d ~A =∫R

11dV = 11(volume of R) = 11(4π53/3) = 5500π/3 ≈ 5759.6

7. The divergence of the field is

div ~F =∂(−z)∂x

+∂(0)

∂y+∂x

∂z= 0.

Hence, ∫

Sphere

~F · d ~A =

∫

Balldiv ~F dV =

∫

Ball0 dV = 0,

and so the flux through the sphere is zero. This makes sense, because, as Figure 20.12 shows, the vector field is flowingaround the y-axis and is always tangent to the sphere.

x

y

z

Figure 20.12

8. Since the surface is closed, the flux of a constant vector field out of it is 0.

9. Since div ~G = 1, if W is the interior of the box, the Divergence Theorem gives

Flux =

∫

W

1 dV = 1 · Volume of box = 1 · 2 · 3 · 4 = 24.

10. Since div ~H = y, if W is the interior of the box, the Divergence Theorem gives

Flux =

∫

W

y dV =

∫ 4

0

∫ 3

0

∫ 2

0

y dx dy dz = 4 · 2 · y2

2

∣∣∣∣3

0

= 36.

20.2 SOLUTIONS 1351

11. Since div ~J = 2xy, if W is the interior of the box, the Divergence Theorem gives

Flux =

∫

W

2xy dV =

∫ 4

0

∫ 3

0

∫ 2

0

2xy dx dy dz = x2

∣∣∣∣2

0

y2

2

∣∣∣∣3

0

z

∣∣∣∣4

0

= 72.

12. Since div ~N = 0, the flux through the closed surface of the box is 0.

Problems

13. By the Divergence Theorem,∫S~F · d ~A =

∫W

div ~F dV =∫W

0dV = 0 for a closed surface S, where W is the regionenclosed by S.

14. Sincediv ~F = 0 + 1 + 0 = 1,

we have ∫

S

~F · d ~A = (div ~F ) · Vol =4

3π33 = 36π.

15. Since div ~F = y + z + x, the flux is given by∫

Sphere

~F · d ~A =

∫

Sphere(x+ y + z)dV.

We calculate the first term of the integral

∫

Spherex dV =

∫ 1

−1

∫ √1−z2

−√

1−z2

∫ √1−y2−z2

−√

1−y2−z2

x dx dy dz

=

∫ 1

−1

∫ √1−z2

−√

1−z2

x2

2

∣∣∣√

1−y2−z2

−√

1−y2−z2dy dz

=

∫ 1

−1

∫ √1−z2

−√

1−z2

((√

1− y2 − z2)2

2− (−

√1− y2 − z2)2

2

)dy dz

=

∫ 1

−1

∫ √1−z2

−√

1−z2

0 dydz = 0.

The other terms in the integral are zero by a similar calculation.The same result can be obtained by a symmetry argument, which is much shorter: Since x, y, and z each take equal

positive and negative values on half the sphere, the integral of each term is 0. Thus, the flux is zero:∫

Sphere

~F · d ~A =

∫

Sphere(x+ y + z)dV = 0.

16. Since div ~F = 3x2 + 3y2 + 3z2, the Divergence Theorem gives∫

S

~F · d ~A =

∫

W

div ~F dV =

∫

W

3(x2 + y2 + z2) dV.

Since W is the interior of a cylinder of radius 2 centered on the z-axis, we use cylindrical coordinates, giving∫

S

~F · d ~A =

∫

W

3(x2 + y2 + z2) dV = 3

∫ 2π

0

∫ 2

0

∫ 5

0

(r2 + z2)r dz dr dθ

= 3

∫ 2π

0

∫ 2

0

(r3z +

rz3

3

)∣∣∣∣5

0

dr dθ = 3

∫ 2π

0

∫ 2

0

(5r3 +125

3r) dr dθ

= 3 · 2π(

5r4

4+

125

3· r

2

2

)∣∣∣∣2

0

= 620π.


17. Since div ~F = 3x2 + 3y2 + 3z2, the Divergence Theorem gives

Flux =

∫

S

~F · d ~A =

∫

W

(3x2 + 3y2 + 3z2) dV.

In spherical coordinates, the region W lies between the spheres ρ = 2 and ρ = 3 and inside the cone φ = π/4. Since3x2 + 3y2 + 3z2 = 3ρ2, we have

Flux =

∫

S

~F · d ~A =

∫ 2π

0

∫ π/4

0

∫ 3

2

3ρ2 · ρ2 sinφ dρ dφ dθ

= 2π · 3

5ρ5

∣∣∣∣3

2

(− cosφ)

∣∣∣∣π/4

0

=633(2−

√2)

5π = 232.98.

18. By the Divergence Theorem, if W is the cylinder and S is its surface:∫

S

~F · d ~A =

∫

W

div ~F dV =

∫

W

10 dV = 10 · Volume of cylinder = 10πa3.

19. Apply the Divergence Theorem to the solid cone, whose interior we call W . The surface of W consists of S and D. Thus∫

S

~F · d ~A +

∫

D

~F · d ~A =

∫

W

div ~F dV.

But div ~F = 0 everywhere, since ~F is constant. Thus∫

D

~F · d ~A = −∫

S

~F · d ~A = −3.22.

20. Using the Divergence theorem, we estimate the integral of div ~F through W , the interior of the box:

Flux =

∫

S

~F · d ~A =

∫

W

div ~F dV.

The region W can be divided into four boxes, with sides ∆x = ∆y = 0.5 and ∆z = 1. Since div ~F is monotonic, oneach box the lower bound for div ~F is at the corner closest to the origin, and the upper bound is at the corner farthestfrom the origin. Thus, we use Table 20.1 for the lower bound and Table 20.2 for the upper bound. The volume of eachbox is (0.5)(0.5)1 = 0.25. Thus,

Lower bound for flux = 2(0.25) + 3(0.25) + 4(0.25) + 5(0.25) = 3.5.

Upper bound for flux = 6(0.25) + 8(0.25) + 7(0.25) + 9(0.25) = 7.5.

21. We close the cylinder, S, by adding the circular disk, S1, at the top, z = 3. The surface S+S1 is oriented outward, so S1

is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have∫

S+S1

~F · d ~A =

∫

W

div ~F dV

∫

S

~F · d ~A +

∫

S1

~F · d ~A =

∫div ~F dV.

Sincediv ~F = div(z2~i + x2~j + 5~k ) = 0,

we have ∫

S

~F · d ~A = −∫

S1

~F · d ~A .

Only the ~k -component of ~F contributes to the flux through S1, and d ~A = ~k dx dy on S1, so∫

S

~F · d ~A = −∫

S1

(z2~i + x2~j + 5~k ) · ~k dx dy = −∫

S1

5 dx dy = −5 · Area of S1 = −5π(√

2)2 = −10π.

20.2 SOLUTIONS 1353



S+S1

~F · d ~A =

∫

W

div ~F dV

∫

S

~F · d ~A +

∫

S1

~F · d ~A =

∫div ~F dV.

Sincediv ~F = div(y2~i + z2~j + (x2 + y2)~k ) = 0,

we have ∫

S

~F · d ~A = −∫

S1

~F · d ~A .


S

~F · d ~A = −∫

S1

(y2~i + z2~j + (x2 + y2)~k ) · ~k dx dy = −∫

S1

(x2 + y2) dx dy.

Converting to polar coordinates, since the cylinder has radius√

2, we have

∫

S

~F · d ~A = −∫ 2π

0

∫ √2

0

r2 · r dr dθ = −2πr4

4

∣∣∣∣

√2

0

= −2π.



S+S1

~F · d ~A =

∫

W

div ~F dV

∫

S

~F · d ~A +

∫

S1

~F · d ~A =

∫div ~F dV.

Sincediv ~F = div(z~i + x~j + y~k ) = 0,

we have ∫

S

~F · d ~A = −∫

S1

~F · d ~A .


S

~F · d ~A = −∫

S1

(z~i + x~j + y~k ) · ~k dx dy = −∫

S1

y dx dy.

Since y is an odd function, by symmetry, its integral over S1 is zero. Thus,∫

S

~F · d ~A = −∫

S

y dx dy = 0.



S+S1

~F · d ~A =

∫

W

div ~F dV

∫

S

~F · d ~A +

∫

S1

~F · d ~A =

∫div ~F dV.

Sincediv ~F = div(y2~i + x2~j + 7z~k ) = 7,


we have ∫

S

~F · d ~A =

∫

W

7 dV −∫

S1

~F · d ~A = 7 · Volume of cylinder −∫

S1

~F · d ~A

= 7 · π(√

2)26−∫

S1

~F · d ~A = 84π −∫

S1

~F · d ~A .

Only the ~k -component of ~F contributes to the flux through S1, and d ~A = ~k dx dy and z = 3 on S1, so∫

S

~F · d ~A = 84π −∫

S1

(y2~i + x2~j + 7 · 3~k ) · ~k dx dy = 84π −∫

S1

21 dx dy

= 84π − 21 · Area of S1 = 84π − 21 · π(√

2)2 = 42π.



S+S1

~F · d ~A =

∫

W

div ~F dV

∫

S

~F · d ~A +

∫

S1

~F · d ~A =

∫div ~F dV.

Sincediv ~F = div(x3~i + y3~j + ~k ) = 3x2 + 3y2,

we have ∫

S

~F · d ~A =

∫

W

(3x2 + 3y2) dV −∫

S1

~F · d ~A .

To find the integral over W , we use cylindrical coordinates. For the integral over S1, we use the fact that d ~A = ~k dx dy,so only the k-component of ~F contributes to the flux.

∫

S

~F · d ~A =

∫ 2π

0

∫ 3

−3

∫ √2

0

3r2 · r dr dz dθ −∫

S1

(x3~i + y3~j + ~k ) · ~k dx dy

= θ

∣∣∣∣∣

2π

0

z

∣∣∣∣∣

3

−3

3

4r4

∣∣∣∣∣

√2

0

−∫

S1

dx dy

= 2π · 6 · 3

4(√

2)4 − Area of S1

= 36π − π(√

2)2 = 34π.

26. (a) True. Flux that goes in one face goes out the other, since the vector field is constant and the surface is closed.(b) True. The flux out of S1 along the face shared with S2 cancels with the flux out of S2 over the same face. (The

normals are in opposite directions.) The other five faces of S1 and the other five faces of S2 are each faces of S.

27. (a) Taking partial derivatives, we have for ~r 6= ~0 ,

div ~F = div

(x~i + y~j + z~k

(x2 + y2 + z2)3/2

)

= − 3(x2 + y2 + z2)

(x2 + y2 + z2)5/2+

3

(x2 + y2 + z2)3/2= 0.

(b) We could compute the flux out of the box S by computing the flux out of each side separately. However, sincediv ~F = ~0 everywhere except the origin, we instead consider a region W between the box S and a sphere Sb ofradius b centered at the origin and which fits inside the box. If the sphere is oriented inward, since div ~F = ~0throughout W , the Divergence Theorem says

0 =

∫

W

div ~F dV =

∫

S

~F · d ~A +

∫

Sb

~F · d ~A .

20.2 SOLUTIONS 1355

The flux of ~F through Sb is easier to calculate than the flux through the box. Since Sb is oriented inward,∫

Sb

~F · d ~A = −∫

Sb

‖~F ‖‖d ~A ‖ = −∫

Sb

1

b2‖dA‖

= − 1

b2· Surface area of Sb = − 1

b2· 4πb2 = −4π.

Thus, ∫

S

~F · d ~A = −∫

Sb

~F · d ~A = 4π.

28. We can rewrite ~F in terms of (x, y, z) as

~F =−Gmx√

(x2 + y2 + z2)3

~i +−Gmy√

(x2 + y2 + z2)3

~j +−Gmz√

(x2 + y2 + z2)3

~k

Now we find the divergence of ~F in the usual manner

div ~F =∂

∂x

−Gmx√(x2 + y2 + z2)3

+∂

∂y

−Gmy√(x2 + y2 + z2)3

+∂

∂z

−Gmz√(x2 + y2 + z2)3

= −3Gmx2 + y2 + z2

(x2 + y2 + z2)5/2+

3Gm

(x2 + y2 + z2)3/2

= 0.

Thus, div ~F = 0 for all points except the origin. We consider a region enclosed by two concentric spheres. Since thedivergence of the field is zero at all points except the origin, the volume enclosed contains only points with zero divergence.Consequently, the flux through the surface of the enclosed volume must be zero. Since the field is always inward pointing,this is equivalent to saying that the flux into the outer sphere must equal the flux out of the inner sphere, and so we seethat for any two spheres, the flux must be equal, which shows that the flux is independent of the radius of the spheres.

29. Since both surfaces, S and B, are closed, we use the Divergence Theorem with

div ~F =∂

∂x

(1

3(x3 + y3)

)+

∂

∂y

(1

3(y3 + z3)

)+

∂

∂z

(1

3(z3 + x3)− z

)= x2 + y2 + z2 − 1.

(a) Since div ~F < 0 at every point inside S, which is the sphere x2 + y2 + z2 − 1 = 0, we have∫

S

~F · d ~A =

∫

Interior of S

div ~F dV < 0.

(b) Since div ~F > 0 at every point outside S, and B is entirely outside S, we have∫

B

~F · d ~A =

∫

Interior of B

div ~F dV > 0.

30. (a) First we calculate

~r × ~a =

∣∣∣∣∣∣

~i ~j ~k

x y z

a1 a2 a3

∣∣∣∣∣∣= (ya3 − za2)~i + (za1 − xa3)~j + (xa2 − ya1)~k .

Then we calculate

div(~r × ~a ) = div((ya3 − za2)~i + (za1 − xa2)~j + (xa2 − ya1)~k

)= 0.

(b) The cube is closed and oriented outward. By the Divergence Theorem,∫

Box(~r × ~a ) · d ~A =

∫

Interior of boxdiv(~r × ~a )dV = 0.


31. (a) Since ~a · ~r = a1x+ a2y + a3z, we have

div ((~a · ~r )~r ) = div((a1x+ a2y + a3z)(x~i + y~j + z~k )

),

Now∂

∂x((a1x+ a2y + a3z)x) = a1(1)x+ (a1x+ a2y + a3z)1

= 2a1x+ a2y + a3z,

so, performing similar calculations to find ∂/∂y and ∂/∂z, we have

div ((~a · ~r )~r ) = 2a1x+ a2y + a3z + a1x+ 2a2y + a3z + a1x+ a2y + 2a3z

= 4a1x+ 4a2y + 4a3z = 4~a · ~r .(b) By the Divergence Theorem,

Flux =

∫

Sphere(~a · ~r )~r · d ~A =

∫

Interior of spherediv ((~a · ~r )~r ) dV

=

∫

Interior of sphere4~a · ~r dV = 4

∫

Interior of sphere(a1x+ a2y + a3z) dV.

We calculate the first term in the last integral:∫

Spherea1x dV = a1

∫ 1

−1

∫ √1−z2

−√

1−z2

∫ √1−y2−z2

−√

1−y2−z2

x dx dy dz

= a1

∫ 1

−1

∫ √1−z2

−√

1−z2

x2

2

∣∣∣√

1−y2−z2

−√

1−y2−z2dy dz

= a1

∫ 1

−1

∫ √1−z2

−√

1−z2

((√

1− y2 − z2)2

2− (−

√1− y2 − z2)2

2

)dy dz

= a1

∫ 1

−1

∫ √1−z2

−√

1−z2

0 dydz = 0.

By a similar calculation, the other terms in the integral are 0. Thus the whole integral on the right-hand side is zero.This result can also be obtained by a symmetry argument, which is much shorter: Since x, y, and z each take

equal positive and negative values on half the sphere, the integral of each term is 0. Thus, the flux is 0:∫

Sphere

~F · d ~A = 4

∫

Sphere~a · ~r dV = 0.

32. (a) At the point (1, 2, 1), we have div ~F = 1 · 2 · 12 = 2.(b) Since the box is small, we use the approximation

div ~F = Flux density ≈ Flux out of boxVolume of box

.

ThusFlux out of box ≈ (div ~F ) · (Volume of box) = 2(0.2)3 = 0.016.

(c) To calculate the flux exactly, we use the Divergence Theorem,

Flux out of box =

∫

Box

div ~F dV =

∫

Box

xyz2 dV.

Since the box has side 0.2, it is given by 0.9 < x < 1.1, 1.9 < y < 2.1, 0.9 < z < 1.1, so

Flux =

∫ 1.1

0.9

∫ 2.1

1.9

∫ 1.1

0.9

xyz2 dzdydx =

∫ 1.1

0.9

∫ 2.1

1.9

xyz3

3

∣∣∣∣1.1

0.9

dydx

=(1.1)3 − (0.9)3

3

∫ 1.1

0.9

xy2

2

∣∣∣∣2.1

1.9

dx =(1.1)3 − (0.9)3

3· (2.1)2 − (1.9)2

2· x

2

2

∣∣∣∣1.1

0.9

=(1.1)3 − (0.9)3

3· (2.1)2 − (1.9)2

2· (1.1)2 − (0.9)2

2= 0.016053 . . . .

20.2 SOLUTIONS 1357

Notice that you can calculate the flux without knowing the vector field, ~F .

33. Any closed surface, S, oriented inward, will work. Then,∫

S(inward)

~F · d ~A = −∫

S(outward)

~F · d ~A ,

so, by the Divergence Theorem, with W representing the region inside S,∫

S(inward)

~F · d ~A = −∫

W

div ~F dV = −∫

W

(x2 + y2 + 3)dV.

The integral on the right is positive because the integrand is positive everywhere. Therefore the flux through S orientedinward is negative.

34. Let S3 the sphere of radius 3 and let S5 be the sphere of radius 5. Then we know that∫

S3

~G · d ~A = 8π.

We consider the region W between S3 and S5. For 2 ≤ ‖~r ‖ ≤ 7, we have div ~G = 3. We apply the DivergenceTheorem to the region W between S3 and S5, with S5 oriented outward and S3 oriented inward:

∫

S5

~G · d ~A −∫

S3

~G · d ~A =

∫

W

div ~G dV = 3 · Volume of W

= 3(

4

3π53 − 4

3π33)

= 392π.

Thus, ∫

S5

~G · d ~A =

∫

S3

~G · d ~A +

∫

W

div ~G dV = 8π + 392π = 400π

35. Using the Divergence Theorem, where W is the interior of S,

Flux =

∫

S

~F · d ~A =

∫

W

div ~F dV,

we see that ∫

W

2 dV ≤ Flux ≤∫

W

5 dV

2 · Volume of sphere ≤ Flux ≤ 5 · Volume of sphere

2 · 4π 33

3≤ Flux ≤ 5 · 4π 33

372π ≤ Flux ≤ 180π.

36. (a) We cannot use the Divergence Theorem to calculate the flux through the sphere of radius 2 because ~G is not definedthroughout the interior of the sphere. We calculate the flux directly. Since ~G is parallel to the area vector at thesurface of the sphere, and since || ~G || = 4 · 22 · 2 = 32 on the surface, we have

Flux =

∫

S

~G · d ~A = 32 · Area of surface = 32 · 4π22 = 512π.

(b) The fact that the vectors of ~G get longer as we go away from the origin suggests that div ~G > 0. This is confirmedby calculating the divergence using the formula

~G = 4r2~r = 4(x2 + y2 + z2)(x~i + y~j + z~k ).

Since∂G1

∂x= 4(3x2 + y2 + z2), and so on, div ~G is positive everywhere outside the unit sphere.


The sphere is completely contained within the box. Apply the Divergence Theorem to the region, W , betweenthe sphere and the box. This region has surface area the sphere (oriented inward) and the box (oriented outward). TheDivergence Theorem gives

∫

Box (outward)

~F · d ~A +

∫

Sphere (inward)

~F · d ~A =

∫

W

div ~G dV > 0.

So ∫

Box (outward)

~F · d ~A −∫

Sphere (outward)

~F · d ~A =

∫

W

div ~G dV > 0.

Thus, ∫

Box (outward)

~F · d ~A =

∫

Sphere (outward)

~F · d ~A +

∫

W

div ~G dV.

So the flux through the box is larger than the flux through the sphere.

37. Since the divergence is zero at all points not containing the charge, the flux must be zero through any closed surfacecontaining no charge. We imagine a surface composed of two concentric cylinders and their end-caps, where the axis ofboth cylinders is the z-axis. Then, since no charge is contained in the region enclosed, the flux through the surface mustbe zero. Now, we know that the field points away from the axis, which means it is parallel to the end-caps. Consequently,there must be no flux through the end-caps. This implies that the flux through the inner cylinder must equal the flux outof the outer cylinder. Since the strength of the field only depends upon the distance from the z axis, the flux through eachcylinder is a constant. This implies that the following equation must hold

Flux through each cylinder = Ea2πraL = Eb2πrbL

where Ea and Eb are the strengths of the field at ra and rb respectively, and L is the length of the cylinders. Dividingthrough, we can arrive at the following relationship:

Ea/Eb = rb/ra

If we take Eb to be a constant at a fixed value of rb, then the equation can be simplified to

Ea = k/ra

where k = Ebrb. Thus we see that the strength of the field is proportional to 1/r.

38. (a) Using the expression given for the force, we have

Force in~i direction = ~F ·~i = −∫

S

δgz~i · d ~A

= −δg∫

S

z~i · d ~A .

Now apply the Divergence Theorem to this integral. (Notice that in order to do this, you need to orient S outward,hence the minus sign disappears.)

~F ·~i = δg

∫

V

∂z

∂xdV = 0.

Similarly:

Force in ~j direction = ~F ·~j = −δg∫

S

z~j · d ~A

= δg

∫

V

∂z

∂ydV = 0

(b)

Force in ~k direction = ~F · ~k = −δg∫

S

z~k · d ~A

= δg

∫

V

∂z

∂zdV = δg

∫

V

dV = δgV.

20.2 SOLUTIONS 1359

39. (a) The rate at which heat is generated at any point in the earth is div ~F at that point. So div ~F = 30 watts/km3.(b) Differentiating gives div(α(x~i + y~j + z~k )) = α(1 + 1 + 1) = 3α so α = 30/3 = 10 watts/km3. Thus, ~F = α~r

has constant divergence. Note that ~F = α~r has flow lines going radially outward, and symmetric about the origin.(c) The vector gradT gives the direction of greatest increase in temperature. Thus, −gradT gives the direction of

greatest decrease in temperature. The equation ~F = −k gradT says that heat will flow in the direction of greatestdecrease in temperature (i.e. from hot regions to cold), and at a rate proportional to the temperature gradient.

(d) We assume that ~F is given by the answer to part (b). Then, using part (c), we have

~F = 10(x~i + y~j + z~k ) = −30,000 gradT,

sogradT = − 10

30,000(x~i + y~j + z~k ).

Integrating we get

T =−10

2(30,000)(x2 + y2 + z2) + C.

At the surface of the earth, x2 + y2 + z2 = 64002, and T = 20◦C, so

T =−1

6000(64002) + C = 20.

Thus,

C = 20 +64002

6000= 6847.

At the center of the earth, x2 + y2 + z2 = 0, so

T = 6847◦C.

40. (a) Taking partial derivatives of ~E gives

∂E1

∂x=

∂

∂x[qx(x2 + y2 + z2)−3/2] = q[(x2 + y2 + z2)−3/2 + x(−3/2)(2x)(x2 + y2 + z2)−5/2]

= q(y2 + z2 − 2x2)(x2 + y2 + z2)−5/2.

Similarly,

∂E2

∂x= q(x2 + z2 − 2y2)(x2 + y2 + z2)−5/2

∂E3

∂x= q(x2 + y2 − 2z2)(x2 + y2 + z2)−5/2.

Summing, we obtain div ~E = 0.(b) Since on the surface of the sphere, the vector field ~E and the area vector ∆ ~A are parallel,

~E ·∆ ~A = ‖ ~E ‖‖∆ ~A ‖.

Now, on the surface of a sphere of radius a,

‖ ~E ‖ =q‖~r ‖‖~r ‖3 =

q

a2.

Thus, ∫

Sa

~E · d ~A =

∫q

a2‖d ~A ‖ =

q

a2· Surface area of sphere =

q

a2· 4πa2 = 4πq.

(c) It is not possible to apply the Divergence Theorem in part (b) since ~E is not defined at the origin (which lies insidethe region of space bounded by Sa), and the Divergence Theorem requires that the vector field be defined everywhereinside S.


(d) Let R be the solid region lying between a small sphere Sa, centered at the origin, and the surface S. Applying theDivergence Theorem and the result of part (a), we get:

0 =

∫

R

div ~E dV =

∫

Sa

~E · d ~A +

∫

S

~E · d ~A ,

where S is oriented with the outward normal vector, and Sa with the inward normal vector (since this is “outward”with respect to the region R). Since

∫

Sa, inward

~E · d ~A = −∫

Sa, outward

~E · d ~A ,

the result of part (b) yields ∫

S

~E · d ~A = 4πq.

[Note: It is legitimate to apply the Divergence Theorem to the regionR since the vector field ~E is defined everywherein R.]

41. Check that div ~E = 0 by taking partial derivatives. For instance,

∂E1

∂x=

∂

∂x[q(x− x0)[(x− x0)2 + (y − y0)2 + (z − z0)2]−3/2]

= q[(y − y0)2 + (z − z0)2 − 2(x− x0)2][(x− x0)2 + (y − y0)2 + (z − z0)2]−5/2

and similarly,

∂E2

∂y= q[(x− x0)2 + (z − z0)2 − 2(y − y0)2][(x− x0)2 + (y − y0)2 + (z − z0)2]−5/2

∂E3

∂z= q[(x− x0)2 + (y − y0)2 − 2(z − z0)2][(x− x0)2 + (y − y0)2 + (z − z0)2]−5/2.

Therefore,∂E1

∂x+∂E2

∂y+∂E3

∂z= 0.

The vector field ~E is defined everywhere but at the point with position vector ~r 0. If this point lies outside the surface S,the Divergence Theorem can be applied to the region R enclosed by S, yielding:

∫

S

~E · d ~A =

∫

R

div ~E dV = 0.

If the charge q is located inside S, consider a small sphere Sa centered at q and contained in R. The Divergence Theoremfor the region R′ between the two spheres yields:

∫

S

~E · d ~A +

∫

Sa

~E · d ~A =

∫

R′div ~E dV = 0.

In this formula, the Divergence Theorem requires S to be given the outward orientation, and Sa the inward orientation. Tocompute

∫Sa

~E ·d ~A , we use the fact that on the surface of the sphere, ~E and ∆ ~A are parallel and in opposite directions,so

~E ·∆ ~A = −‖ ~E ‖‖∆ ~A ‖since on the surface of a sphere of radius a,

‖ ~E ‖ = q‖~r − ~r 0‖‖~r − ~r 0‖3

=q

a2.

Then, ∫

Sa

~E · d ~A =

∫− q

a2‖d ~A ‖ =

−qa2· Surface area of sphere = − q

a2· 4πa2 = −4πq.

∫

Sa

~E · d ~A = −4πq.

∫

S

~E · d ~A −∫

Sa

~E · d ~A = 4πq.

20.3 SOLUTIONS 1361

42. (a) If ~p = p1~i + p2

~j + p3~k we have

~E (~r ) = 3(xp1 + yp2 + zp3)~r

(x2 + y2 + z2)5/2− ~p

(x2 + y2 + z2)3/2

Thus, taking partial derivatives,

∂E1

∂x=

∂

∂x

3x(xp1 + yp2 + zp3)− p1(x2 + y2 + z2)

(x2 + y2 + z2)3/2

=∂

∂x

2p1x2 + 3(p2y + p3z)x− p1(y2 + z2)

(x2 + y2 + z2)3/2

=(4p1x+ 3p2y + 3p3z)(x

2 + y2 + z2)3/2

(x2 + y2 + z2)3

− [2p1x2 + 3(p2y + p3z)x− p1(y2 + z2)](3/2)(2x)(x2 + y2 + z2)1/2

(x2 + y2 + z2)3

=p1x(−2x2 + y2 + z2) + p2y(−6x2 + 3y2 + 3z2) + p3z(−6x2 + 3y2 + 3z2)

(x2 + y2 + z2)5/2

The expressions for ∂E2/∂y and ∂E3/∂z are obtained similarly; Summing, we obtain

div ~E =∂E1

∂x+∂E2

∂y+∂E3

∂z= 0.

(b) The Divergence Theorem cannot be applied directly since ~E is not defined at the origin (which lies inside S) andthe theorem requires that the vector field be defined everywhere inside S. However, it can be applied to the region Rbounded by S and a small sphere Sa with center at the origin and contained inside S. We conclude that

∫

R

div ~E dV =

∫

S

~E · d ~A −∫

Sa

~E · d ~A ,

where S and Sa are outward oriented. Since div ~E = 0 inside R, it follows that∫

S

~E · d ~A =

∫

Sa

~E · d ~A .

The integral on the right-hand side was computed in Problem 38 on page 936 of the text, and is equal to zero nomatter what the radius a is. Therefore, ∫

S

~E · d ~A = 0.


Exercises

1. This vector field shows no rotation, and the circulation around any closed curve appears to be zero, so we suspect a zerocurl here.

2. This vector field is definitely swirling, so we suspect a nonzero curl here.

3. The circulation around the boundary of a square in quadrant one is positive, because the vectors on the top of the squareare larger than those at the bottom. Therefore there is a nonzero curl.

4. This vector field shows no rotation, and the circulation around any closed curve appears to be zero, so the vector field haszero curl.

5. This vector field shows no rotation, so we expect a zero curl.

6. The circulation around the boundary of a square with sides parallel to the axes and enclosing the point is nonzero becausethe vectors at the bottom are larger than those at the top, so we expect a non-zero curl at this point.


7. Using the definition in Cartesian coordinates, we have

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x2 − y2 2xy 0

∣∣∣∣∣∣

=

(∂

∂y(0)− ∂

∂z(2xy)

)~i +

(− ∂

∂x(0) +

∂

∂z(x2 − y2)

)~j +

(∂

∂x(2xy)− ∂

∂y(x2 − y2)

)~k

= 4y~k .

8. Using the definition of Cartesian coordinates,

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x2 y3 z4

∣∣∣∣∣∣

=

(∂

∂y(z4)− ∂

∂z(y3)

)~i +

(− ∂

∂x(z4) +

∂

∂z(x2)

)~j +

(∂

∂x(y3)− ∂

∂y(x2)

)~k

= ~0 .


curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

ex cos y ez2

∣∣∣∣∣∣

=

(∂

∂y(ez

2

)− ∂

∂z(cos y)

)~i +

(− ∂

∂x(ez

2

) +∂

∂z(ex)

)~j +

(∂

∂x(cos y)− ∂

∂y(ex)

)~k

= ~0 .


curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

2yz 3xz 7xy

∣∣∣∣∣∣

= (7x− 3x)~i − (7y − 2y)~j + (3z − 2z)~k

= 4x~i − 5y~j + z~k .


curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

(−x+ y) (y + z) (−z + x)

∣∣∣∣∣∣

=

(∂

∂y(−z + x)− ∂

∂z(y + z)

)~i +

(− ∂

∂x(−z + x) +

∂

∂z(−x+ y)

)~j

+

(∂

∂x(y + z)− ∂

∂y(−x+ y)

)~k

= −~i −~j − ~k .

20.3 SOLUTIONS 1363

12. Using the definition in Cartesian coordinates

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x+ yz y2 + xzy zx3y2 + x7y6

∣∣∣∣∣∣

= (2x3yz + 6x7y5 − xy)~i + (−3x2y2z − 7x6y6 + y)~j + (yz − z)~k

13. This vector field points radically outward and has unit length everywhere (except the origin). Thus, we would expect itscurl to be ~0 . Computing the curl directly we get

curl

(~r

‖~r ‖

)=

∣∣∣∣∣∣∣

~i ~j ~k∂

∂x∂∂y

∂∂z

x

(x2+y2+z2)1/2y

(x2+y2+z2)1/2z

(x2+y2+z2)1/2

∣∣∣∣∣∣∣

The~i -component is given by =

(−1

2· 2yz

(x2 + y2 + z2)3/2−(−1

2· 2yz

(x2 + y2 + z2)1/2

))~i

= ~0

Similarly, the ~j and ~k components are also both ~0 .

14. (a) Using the definition of curl in Cartesian coordinates,

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

xy + z2 x2 xz − z

∣∣∣∣∣∣

= 0~i + (2z − z)~j + (2x− x)~k

= z~j + x~k

At the point (0,−1, 0), we have curl ~F = ~0 .(b) The expression shows curl ~F = ~0 if and only if x = 0 and z = 0, so curl ~F is not ~0 everywhere. Thus, the vector

field is not irrotational.

Problems

15. The conjecture is that when the first component of ~F depends only on x, the second component depends only on y, andthe third component depends only on z, that is, if

~F = F1(x)~i + F2(y)~j + F3(z)~k

thencurl ~F = ~0

The reason for this is that if ~F = F1(x)~i + F2(y)~j + F3(z)~k , then

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

F1(x) F2(y) F3(z)

∣∣∣∣∣∣

=

(∂

∂yF3(z)− ∂

∂zF2(y)

)~i +

(− ∂

∂xF3(z) +

∂

∂zF1(x)

)~j +

(∂

∂xF2(y)− ∂

∂yF1(x)

)~k

= ~0 .


16. The part of this vector field in the xy-plane looks like Figure 20.1 on page 940, and shows no rotational tendency. Thuswe expect the curl to be ~0 . In fact it is, because the circulation around every closed curve C is zero, since

~F = x~i + y~j + z~k = grad(x2/2 + y2/2 + z2/2),

so ~F is a gradient field. Thus the circulation density is zero in any direction, and hence curl ~F (P ) = ~0 for every pointP . Using the formula, we see that

curl ~F = ∇× ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x y z

∣∣∣∣∣∣

=

(∂z

∂y− ∂y

∂z

)~i +

(∂x

∂z− ∂z

∂x

)~j +

(∂y

∂x− ∂x

∂y

)~k = 0~i + 0~j + 0~k = ~0 .

17. (a) No. See Figure 20.13. The line integral along C1 is zero, while the line integral along C2 is positive. Therefore ~F isnot path independent and not a gradient field.

C1

C2

PQx

y

Figure 20.13

(b) Positive. Since ~F (x, y) has no z-component, the direction of curl ~F is ~k , so curl ~F · ~k > 0.(c) A possible formula for ~F is ~F = x~j .(d) Using the formula in part (c), we have curl ~F = ~k .

18. (a) The vector field always points perpendicularly to both ~k and ~r , in the direction determined by the right hand rule,and its magnitude is twice the magnitude of ~r . Thus

~F (~r ) = 2~k × ~r = −2y~i + 2x~j .

(b) Using the definition in Cartesian coordinates

curl~v =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−2y 2x 0

∣∣∣∣∣∣

= (∂

∂y(0)− ∂

∂z(2x))~i + (

∂

∂z(−2y)− ∂

∂x(0))~j + (

∂

∂x(2x)− ∂

∂y(−2y))~k = 4~k .

This makes sense, because we computed the circulation density of this vector field in the z-direction and found it was4, and we would expect the z-direction to give the maximum circulation density from the symmetry of the vectorfield.

19. The curl is defined in such a way that if ~n is a unit vector and C is a small circle in the plane perpendicular to ~n and withorientation induced by ~n , then

(curl ~G ) · ~n = Circulation density

≈∫C~G · d~r

Area inside Cso

Circulation =

∫

C

~G · d~r ≈((curl ~G ) · ~n

)· Area inside C.

20.3 SOLUTIONS 1365

(a) Let C be the circle in the xy-plane, and let ~n = ~k . Then

Circulation ≈ (2~i − 3~j + 5~k ) · ~k · π(0.01)2

= 0.0005π.

(b) By a similar argument to part (a), with ~n =~i , we find the circulation around the circle in the yz-plane:

Circulation ≈ (2~i − 3~j + 5~k ) ·~i · π(0.01)2

= 0.0002π.

(c) Similarly for circulations around the circle in the xz-plane,

Circulation ≈ −0.0003π.

20. The vector curl ~F has its component in the x-direction given by

(curl ~F )x ≈ Circulation around small circle around x-axisArea inside circle

=Circulation around C2

Area inside C2=

0.5π

π(0.1)2= 50.

Similar reasoning leads to

(curl ~F )y ≈ Circulation around C3

Area inside C3=

3π

π(0.1)2= 300,

(curl ~F )z ≈ Circulation around C1

Area inside C1=

0.02π

π(0.1)2= 2.

Thus,curl ~F ≈ 50~i + 300~j + 2~k .

21. (a) A thin twig at the origin along the x-axis would only feel the velocity along that axis, and thus go counterclockwise.(b) Clockwise.(c) Using the Cartesian coordinate definition, we get

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y x 0

∣∣∣∣∣∣= 0~i + 0~j + (1− 1)~k = ~0 .

This is as expected, since a paddle-wheel (instead of a twig) placed in the field would not rotate at all.

22. We have~F (t, x, y, z) = (cos t~j + sin t~k )× ~r = (z cos t− y sin t)~i + x sin t~j − x cos t~k ,

so(curl ~F )(t, x, y, z) = 2 cos t~j + 2 sin t~k .

(a) At t = 0 the vector (curl ~F )(0, x, y, z) = 2~j is horizontal, pointing in the y direction.(b) At t = π/2 the vector (curl ~F )(π/2, x, y, z) = 2~k is vertical, pointing in the z direction.(c) At 0 < t < π/2 the vector (curl ~F )(t, x, y, z) = 2 cos t~j + 2 sin t~k is parallel to the yz-plane, making an angle of

t radians with the horizontal plane. Thus as t goes from 0 to π/2, the curl goes steadily from horizontal to vertical.

23. Investigate the velocity vector field of the atmosphere near the fire. If the curl of this vector field is non-zero, there iscirculatory motion. Consequently, if the magnitude of the curl of this vector field is large near the fire, a fire storm hasprobably developed.


24. Let ~C = a~i + b~j + c~k . Then

curl(~F + ~C ) =

(∂

∂y(F3 + c)− ∂

∂z(F2 + b)

)~i +

(∂

∂z(F1 + a)− ∂

∂x(F3 + c)

)~j

+

(∂

∂x(F2 + b)− ∂

∂y(F1 + a)

)~k

=

(∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k

= curl ~F .

25.

curl ~F = (∂F3

∂y− ∂F2

∂z)~i + (

∂F1

∂z− ∂F3

∂x)~j + (

∂F2

∂x− ∂F1

∂y~k )

div curl ~F =∂

∂x(∂F3

∂y− ∂F2

∂z) +

∂

∂y(∂F1

∂z− ∂F3

∂x) +

∂

∂z(∂F2

∂x− ∂F1

∂y)

=∂2F3

∂x∂y− ∂2F2

∂x∂z+∂2F1

∂y∂z− ∂2F3

∂y∂x+∂2F2

∂z∂x− ∂2F1

∂z∂y

Since, if ~F has continuous second partial derivatives,

∂2F3

∂x∂y=∂2F3

∂y∂x,

∂2F2

∂x∂z=∂2F2

∂z∂x, and

∂2F1

∂y∂z=∂2F1

∂z∂y

everything cancels out and we get div curl ~F = ~0 .

26. The Fundamental Theorem of Calculus for Line Integrals states that if C is a path from P to Q, then∫

C

grad f · d~r = f(Q)− f(P ).

Since C is a closed path we have ∫

c

grad f · d~r = f(P )− f(P ) = 0

(a) For any unit vector ~n

circ~n grad f = limArea→0

(∫grad f · d~rArea of C

)= lim

Area→0

(0

Area

)= 0

where the limit is taken over curves C in a plane perpendicular to ~n , and oriented by the right hand rule. Thus thecirculation density of grad f is zero in every direction, and hence curl grad f = ~0 .

(b) Using the Cartesian coordinate definition

curl grad f = curl(∂f

∂x~i +

∂f

∂y~j +

∂f

∂z~k )

=

(∂2f

∂y∂z− ∂2f

∂z∂y

)~i +

(∂2f

∂z∂x− ∂2f

∂x∂z

)~j +

(∂2f

∂x∂y− ∂2f

∂y∂x

)~k

= 0~i + 0~j + 0~k = ~0 .

20.3 SOLUTIONS 1367

27. Let ~c = c1~i + c2~j + c3~k and ~F = F1~i + F2

~j + F3~k . We then show the desired result as follows:

div(~F × ~c ) = div((F1~i + F2

~j + F3~k )× (c1~i + c2~j + c3~k ))

= div((F2c3 − F3c2)~i + (F3c1 − F1c3)~j + (F1c2 − F2c1)~k )

=∂

∂x(F2c3 − F3c2) +

∂

∂y(F3c1 − F1c3) +

∂

∂z(F1c2 − F2c1)

= c3∂F2

∂x− c2 ∂F3

∂x+ c1

∂F3

∂y− c3 ∂F1

∂y+ c2

∂F1

∂z− c1 ∂F2

∂z

= c1(∂F3

∂y− ∂F2

∂z) + c2(

∂F1

∂z− ∂F3

∂x) + c3(

∂F2

∂x− ∂F1

∂y)

= (c1~i + c2~j + c3~k ) · ((∂F3

∂y− ∂F2

∂z)~i + (

∂F1

∂z− ∂F3

∂x)~j + (

∂F2

∂x− ∂F1

∂y)~k )

= ~c · curl ~F .

28.

curl(φ~F )

=

(∂

∂y(φF3)− ∂

∂z(φF2)

)~i +

(∂

∂z(φF1)− ∂

∂x(φF3)

)~j +

(∂

∂x(φF2)− ∂

∂y(φF1)

)~k

=

(φ∂F3

∂y+∂φ

∂yF3 − φ∂F2

∂z− ∂φ

∂zF2

)~i +

(φ∂F1

∂z+∂φ

∂zF1 − φ∂F3

∂x− ∂φ

∂xF3

)~j

+

(φ∂F2

∂x+∂φ

∂xF2 − φ∂F1

∂y− ∂φ

∂yF1

)~k

= φ

((∂F3

∂y− ∂F2

∂z

)~i +

(∂F1

∂z− ∂F3

∂x

)~j +

(∂F2

∂x− ∂F1

∂y

)~k

)

+

((∂φ

∂yF3 − ∂φ

∂zF2

)~i +

(∂φ

∂zF1 − ∂φ

∂xF3

)~j +

(∂φ

∂xF2 − ∂φ

∂yF1

)~k

)

= φ curl ~F +

(∂φ

∂x~i +

∂φ

∂y~j +

∂φ

∂z~k

)× (F1

~i + F2~j + F3

~k )

= φ curl ~F + (gradφ)× ~F .

29. By Problem 28, curl ~F = grad f × grad g + f curl grad g = grad f × grad g, since curl grad g = 0. Since the crossproduct of two vectors is perpendicular to both vectors, curl ~F is perpendicular to grad g. But ~F is a scalar times grad g,so curl ~F is perpendicular to ~F .

30. If ~F = F1~i + F2

~j + F3~k , ~u = u1

~i + u2~j + u3

~k , ~v = v1~i + v2

~j + v3~k , then

grad( ~F · ~v ) · ~u − grad( ~F · ~u ) · ~v= grad(F1v1 + F2v2 + F3v3) · (u1

~i + u2~j + u3

~k )− grad(F1u1 + F2u2 + F3u3) · (v1~i + v2

~j + v3~k )

=∂F1

∂xv1u1 +

∂F2

∂xv2u1 +

∂F3

∂xv3u1 +

∂F1

∂yv1u2 +

∂F2

∂yv2u2 +

∂F3

∂yv3u2 +

∂F1

∂zv1u3 +

∂F2

∂zv2u3 +

∂F3

∂zv3u3 −

(∂F1

∂xu1v1 +

∂F2

∂xu2v1 +

∂F3

∂xu3v1 +

∂F1

∂yu1v2 +

∂F2

∂yu2v2 +

∂F3

∂yu3v2+

∂F1

∂zu1v3 +

∂F2

∂zu2v3 +

∂F3

∂zu3v3

)

=

(∂F3

∂y− ∂F2

∂z

)(u2v3 − u3v2) +

(∂F1

∂z− ∂F3

∂x

)(u3v1 − u1v3) +

(∂F2

∂x− ∂F1

∂y

)(u1v2 − u2v1)

= (curl ~F ) · ~u × ~v .


31. (a) Figure 20.14 shows a cross-section of the vector field in xy-plane with ω = 1, so ~v = −y~i + x~j .

y

x

Figure 20.14: ~v = −y~i + x~j

Figure 20.15 shows a cross-section of vector field in xy-plane with ω = −1, so ~v = y~i − x~j .y

x

Figure 20.15: ~v = y~i − x~j

(b) The distance from the center of the vortex is given by r =√x2 + y2. The velocity of the vortex at any point

is −ωy~i + ωx~j , and the speed of the vortex at any point is the magnitude of the velocity, or s = ‖~v ‖ =√(−ωy)2 + (ωx)2 = |ω|

√x2 + y2 = |ω| r.

(c) The divergence of the velocity field is given by:

div~v =∂(−ωy)

∂x+∂(ωx)

∂y= 0

The curl of the field is:

curl~v = curl(−ωy~i + ωx~j ) =

(∂

∂x(ωx)− ∂

∂y(−ωy)

)~k = 2ω~k

(d) We know that ~v has constant magnitude |ω|R everywhere on the circle and is everywhere tangential to the circle.In addition, if ω > 0, the vector field rotates counterclockwise; if ω < 0, the vector field rotates clockwise. Thus ifω > 0, ~v and ∆~r are parallel and in the same direction, so

∫

C

~v · ~dr = |~v | · (Length of C) = ωR · 2πR = 2πωR2

If ω < 0, then |ω| = −ω and ~v and ∆~r are in opposite directions, so∫

C

~v · ~dr = − |~v | · (Length of C) = − |ω|R · (2πR) = 2πωR2.

20.3 SOLUTIONS 1369

32. (a) Using r = (x2+y2)1/2, we calculate rx = (1/2)(x2+y2)−1/22x. Notice that rx = x/r and, by a similar argument,ry = y/r. We have

curl(rA · (−y~i + x~j )

)=

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−rAy rAx 0

∣∣∣∣∣∣

=

(∂(0)

∂y− ∂(rAx)

∂z

)~i +

(∂(−rAy)

∂z− ∂(0)

∂x

)~j +

(∂(rAx)

∂x− ∂(−rAy)

∂y

)~k

= 0~i + 0~j + c~k = c~k

where

c =∂

∂x(xrA)− ∂

∂y(−yrA) = rA +AxrA−1rx + rA +AyrA−1ry

= 2rA +ArA−1(xrx + yry)

= 2rA +ArA−1

(x2

r+y2

r

)

= 2rA +ArA−1

(x2 + y2

r

)

= 2rA +ArA−1

(r2

r

)

= (2 +A)rA.

(b) The curl is in the direction of ~k for A = −1, is the zero vector for A = −2, and is in the direction of −~k forA = −3.

The counterclockwise flow in the figure suggests the curl is in the +~k direction in all three cases. However, itis not easy to see all the effects in the picture. The counterclockwise flow indicates that the circulation on any patharound ~k is positive, but the curl is the circulation density, not the circulation, in the ~k -direction. To understand thecurl, we must see how the circulation around the ~k -direction changes when the circles get smaller and smaller. Weneed to take into account the changing length of the vector field as well as the changing direction. This is too subtlean effect to be measured accurately by eye. The three vector fields differ in the rate at which the magnitude of thevector field changes as you move perpendicular to the flow (the shear) and this accounts for the differing directionsof their curls. When A = −1 the magnitude of the vector field is constant, when A = −2 or −3 the magnitudedecreases as you move farther from the origin, and the decrease is most rapid for A = −3.

(c) The curl has a component only in the ~k direction. Think of the this component as the limit of the circulation densityaround a small circle in the xy-plane, as the circle shrinks to zero. Thus the sign of the component of the curl tells usthat this circulation around the circle centered at (1, 1, 1) is positive for A = −1, zero for A = −2, and negative forA = −3. (This assumes that the circle is small enough that it does not go around the origin, where the vector fieldsare not defined.)

Since the vector fields and their curls are not defined at (0, 0, 0), the curl calculated in part (b) does not tell usanything about the circulation around a circle centered at (0, 0, 0).

33. (a) Since ~F is in the xy-plane, curl ~F is parallel to ~k (because F3 = 0 and F1, F2 have no z-dependence). Imaginecomputing the circulation of ~F counterclockwise around a small rectangle R at the point P with sides of length hparallel to ~F and sides of length t perpendicular to ~F as shown in Figure 20.16. Since ~F is perpendicular to C2 andC4, the line integral over these two sides is zero. Assuming that ~F is approximately constant on C1 and C3, its valueon these sides is F (Q)~T and −F (P )~T , respectively. Thus, since ~F is parallel to C1 and C3, the line integral overC1 is approximately F (Q)h and the line integral over C3 is approximately −F (P )h. Finally

curl ~F (P ) ≈ Circulation around RArea of R

≈ F (Q)h− F (P )h

ht=F (Q)− F (P )

t

≈ Directional derivative of F in the direction of −−→PQ.


Q

P

C1

h

ht

tC4

C2C3

3~F

Figure 20.16: Path R used to find curl ~F at P

(b) Since ~F = F (x, y)~T = F (x, y)a~i + F (x, y)b~j , with a, b constant, we have

curl ~F = (bFx − aFy)~k .

Also ~T × ~k = (a~i + b~j )× ~k = b~i − a~j , so

bFx − aFy = (gradF ) · (b~i − a~j ) = gradF · ((a~i + b~j )× ~k ) = F~T ×~k ,

where F~T ×~k is the directional derivative of F in the direction of the unit vector ~T × ~k , which is perpendicular to~F . The right-hand rule applied to ~T × ~k shows that ~T × ~k is obtained by a clockwise rotation of ~T through 90◦.


Exercises

1. No, because the curve C over which the integral is taken is not a closed curve, and so it is not the boundary of a surface.

2. By Stokes’ theorem, the circulation of ~F around C is the flux of curl ~F through the disk S in the yz-plane enclosed byC. By the right hand rule, a positive normal vector to the disk points in the direction of the negative x-axis, −~i . Thuscurl ~F · d ~A = curlF · (−~i )dA = −dA, so the flux through S is negative. So the circulation is negative.

3. First C is parameterized by~r (θ) = 2 cos θ~i + 2 sin θ~j + ~k .

Note that C bounds the disk S given by x2 + y2 ≤ 4, z = 1. Then

~r ′(θ) = −2 sin θ~i + 2 cos θ~j .

Now,

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

z − 2y 3x− 4y z + 3y

∣∣∣∣∣∣= 3~i +~j + 5~k ,

and d ~A = ~k dA. Using Stokes’ Theorem we get∫

C

~F · d~r =

∫

S

curl ~F · d ~A

=

∫

S

(3~i +~j + 5~k ) · ~k dA =

∫

S

5dA

= 5(Area of circle) = 5(4π) = 20π.

4. First note that curl = 2~k .

(a) By Stokes’ Theorem,∫C~F · d~r =

∫S

2~k · d ~A where S is the disk of radius 10 in the xy-plane centered at the

origin, oriented downward. Since this orientation is opposite to 2~k ,∫S

2~k · d ~A = −‖2~k ‖(area of S) = −200π.

(b) By Stokes’ Theorem,∫C~F · d~r =

∫S

2~k · d ~A where S is the disk of radius 10 in the yz-plane centered at the

origin, oriented in the negative x direction. Since the vector field 2~k is parallel to the surface S, its flux through thesurface is zero.

20.4 SOLUTIONS 1371

5. The graph of ~F = ~r /‖~r ‖3 consists of vectors pointing radially outward. There is no swirl, so curl ~F = ~0 . From Stokes’Theorem, ∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

~0 · d ~A = 0

6. The circulation is the line integral∫C~F · d~r which can be evaluated directly by parameterizing the circle, C. Or, since

C is the boundary of a flat disk S, we can use Stokes’ Theorem:∫

C

~F · d~r =

∫

S

curl ~F · d ~A

where S is the disk x2 +y2 ≤ 1, z = 2 and is oriented upward (using the right hand rule). Then curl ~F = −y~i −x~j +~kand the unit normal to S is ~k . So

∫

S

curl ~F · d ~A =

∫

S

(−y~i − x~j + ~k ) · ~k dxdy

=

∫

S

1 dxdy

= Area of S = π

7. To calculate∫C~F · d~r directly, we compute the integral along each of the sides C1, C2, C3 in Figure 20.17. Now C1 is

parameterized byx(t) = t, y(t) = 0, z(t) = 0 for 0 ≤ t ≤ 5, so r′(t) =~i .

Similarly, C2 is parameterized by

x(t) = 5, y(t) = t, z(t) = 0 for 0 ≤ t ≤ 5, so ~r ′(t) = ~j .

Also, C3 is parameterized by

x(t) = 5− t, y(t) = 5− t, z(t) = 0 for 0 ≤ t ≤ 5, so ~r ′(t) = −~i −~j .

Thus∫

C

~F · d~r =

∫

C1

~F · d~r +

∫

C2

~F · d~r +

∫

C3

~F · d~r

=

∫ 5

0

t(~i +~j ) ·~i dt+

∫ 5

0

(5− t)(~i +~j ) ·~j dt+

∫ 5

0

((5− t)− (5− t))(~i +~j ) · (−~i −~j ) dt

=

∫ 5

0

t dt+

∫ 5

0

5− t dt+

∫ 5

0

0 dt =

∫ 5

0

5 dt = 25.

To calculate∫C~F · d~r using Stokes’ Theorem, we find

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x− y + z x− y + z 0

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

For Stokes’ Theorem, the triangular region S in Figure 20.17 is oriented upward, so d ~A = ~k dx dy. Thus∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(−~i +~j + 2~k ) · ~k dx dy

=

∫

S

2 dx dy = 2 · Area of triangle = 2 · 1

2· 5 · 5 = 25.


(5, 5, 0)

(5, 0, 0)(0, 0, 0)

SC3

C2

C1

x

y

Figure 20.17

8. To calculate∫C~F · d~r directly, we compute the integral along the paths C1 and C2 in Figure 20.18. Now C1 is parame-

terized byx(t) = 3− t, y(t) = 0 z(t) = 0 for 0 ≤ t ≤ 6, so r′(t) = −~i ,

and C2 is parameterized by

x(t) = −3 cos t, y(t) = 0, z(t) = 3 sin t for 0 ≤ t ≤ π, so ~r ′(t) = 3 sin t~i + 3 cos t~k .

Thus∫

C

~F · d~r =

∫

C1

~F · d~r +

∫

C2

~F · d~r

=

∫ 6

0

(3− t)(~i +~j ) · (−~i ) dt+

∫ π

0

((−3 cos t+ 3 sin t)~i − 3 cos t~j ) · (3 sin t~i + 3 cos t~k ) dt

= −∫ 6

0

(3− t) dt+ 9

∫ π

0

(− sin t cos t+ sin2 t) dt

= −3t+t2

2

∣∣∣∣∣

6

0

+ 9

(− sin2 t

2− 1

2sin t cos t+

t

2

)∣∣∣∣π

0

=9π

2.

Formula IV-17 was used to calculate the last integral.To calculate

∫C~F · d~r using Stokes’ Theorem, we find

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x+ z x y

∣∣∣∣∣∣=~i +~j + ~k .

For Stokes’ Theorem, the semicircular region S in Figure 20.18 is oriented into the page, so d ~A = ~j dx dz. Thus,∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(~i +~j + ~k ) ·~j dx dz =

∫

S

dx dz

= Area of semicircle =1

2π32 =

9

2π.

(−3, 0, 0) (3, 0, 0)

S

C2

C1

x

z

Figure 20.18

20.4 SOLUTIONS 1373

9. Since C is the curve x2 + y2 = 4, oriented counterclockwise, we calculate∫C~F · d~r directly using the parameterization

x(t) = 2 cos t, y(t) = 2 sin t, z(t) = 0, 0 ≤ t ≤ 2π, so ~r ′(t) = −2 sin t~i + 2 cos t~j .

Thus,∫

C

~F · d~r =

∫ 2π

0

(2 sin t~i − 2 cos t~j ) · (−2 sin t~i + 2 cos t~j ) dt

= −4

∫ 2π

0

(sin2 t+ cos2 t) dt = −4 · 2π = −8π.

Since S is given by z = 4− x2 − y2, we have

∂z

∂x= −2x and

∂z

∂y= −2y, so d ~A = (2x~i + 2y~j + ~k ) dx dy.

Also

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y −x 0

∣∣∣∣∣∣= −2~k ,

so, writing R for the disk below S in the xy-plane, Stokes’ Theorem gives∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

R

−2~k · (2x~i + 2y~j + ~k ) dx dy

= −2

∫dx dy = −2 · Area of disk = −2 · π22 = −8π.

10. A sketch of the surface S and curve C which is the union of four curves C1, C2, C3, and C4, and the region R is shownin Figure 20.19.

xy

z

2

1C3

C1

C2C4

x

y

R

Figure 20.19

To compute the flux integral, we find d ~A , oriented upward.

d ~A = (2x~i + ~k )dxdy and curl ~F = −y~i − z~j − x~k .

Thus,∫

S

curl ~F · d ~A =

∫

S

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

=

∫ 1

0

∫ 2

−2

(−2xy − x)dydx = −2.


The line integral∫

S

~F · d~r is the sum of four integrals along C1, C2, C3, and C4.

On C1: x = 1, z = 0, dx = 0, dz = 0, so∫

S

~F · d~r =

∫ 2

−2

0 dy = 0.

On C2: y = 2, z = 1− x2, dy = 0, dz = −2xdx, so∫

C2

~F · d~r =

∫

C2

2xdx+ 2(1− x2)0 + x(1− x2)(−2x)dx =

∫ 0

1

(2x− 2x2 + 2x4)dx = −11

15.

On C3: x = 0, z = 1, dx = 0, dz = 0, so∫

C3

~F · d~r =

∫

C3

0 + ydy + 0 =

∫ −2

2

ydy = 0.

On C4: y = −2, z = 1− x2, dy = 0, dz = −2xdx, so∫

C4

~F · d~r =

∫

C4

−2xdx− 2(1− x2)0 + x(1− x2)(−2x)dx =

∫ 1

0

(−2x− 2x2 + 2x4)dx = −19

15.

Hence ∫

C

~F · d~r = 0− 11

15+ 0− 19

15= −2.

Thus, ∫

C

~F · d~r =

∫

S

curl ~F · d ~A .

11. The boundary of S isC, the circle x2 +y2 = 1, z = 0, oriented counterclockwise and parameterized in polar coordinatesby

~r (θ) = cos θ~i + sin θ~j , 0 ≤ θ ≤ 2π,

so,~r ′(θ) = − sin θ~i + cos θ~j .

Hence∫

C

~F · d~r =

∫ 2π

0

(sin θ~i + 0~j + cos θ~k ) · (− sin θ~i + cos θ~j + 0~k )dθ

=

∫ 2π

0

− sin2 θdθ = −π.

Now consider the integral∫S

curl ~F · d ~A . Here curl ~F = −~i − ~j − ~k and the area vector d ~A , oriented upward, isgiven by

d ~A = 2x~i + 2y~j + ~k dxdy.

If R is the disk x2 + y2 ≤ 1, then we have∫

S

curl ~F · d ~A =

∫

R

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

Converting to polar coordinates gives:∫

S

curl ~F · d ~A =

∫ 2π

0

∫ 1

0

(−~i −~j − ~k ) · (2r cos θ~i + 2r sin θ~j + ~k )rdrdθ

=

∫ 2π

0

∫ 1

0

(−2r cos θ − 2r sin θ − 1)rdrdθ

=

∫ 2π

0

(2

3(− cos θ − sin θ)− 1

2

)dθ

= −π.

20.4 SOLUTIONS 1375

Thus, we confirm that ∫

C

~F · d~r =

∫

S

curl ~F · d ~A .

Problems

12. (a) It appears that div ~F < 0, and div ~G < 0; div ~G is larger in magnitude (more negative) if the scales are the same.(b) curl ~F and curl ~G both appear to be zero at the origin (and elsewhere).(c) Yes, the cylinder with axis along the z-axis will have negative flux through it (ends parallel to xy-plane).(d) Same as part(c).(e) No, you cannot draw a closed curve around the origin such that ~F has a non-zero circulation around it because curl

is zero. By Stokes’ theorem, circulation equals the integral of the curl over the surface bounded by the curve.(f) Same as part(e)

13. (a) We have

curl(y~i + z~j + x~k ) =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y z x

∣∣∣∣∣∣= −~i −~j − ~k .

(b) Let S be the triangular interior of the curve C, oriented upward. Then, by Stokes’ Theorem,∫

C

(y~i + z~j + x~k ) · d~r =

∫

S

curl(y~i + z~j + x~k ) · d ~A =

∫

S

−(~i +~j + ~k ) · d ~A .

On the triangle d ~A = ~k dA, so∫

S

−(~i +~j + ~k ) · d ~A =

∫

S

−(~i +~j + ~k ) · ~k dA = − Area of triangle = −1

2· 4 · 3 = −6.

14. (a) We have

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y z x

∣∣∣∣∣∣=~i (−1)−~j (1) + ~k (−1) = −~i −~j − ~k .

(b) (i) Using Stokes’ Theorem, with S representing the disk inside the circle, oriented upward, we have∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(−~i −~j − ~k ) · ~k dA = − Area of disk = −4π.

(ii) This is a right triangle in the plane x = 2; it has height 5 and base length 3. Using Stokes’ Theorem, with Srepresenting the triangle, oriented toward the origin (in the direction −~i ), we have∫

C

~F ·d~r =

∫

S

curl ~F ·d ~A =

∫

S

(−~i −~j −~k ) · (−~i dA) =

∫

S

dA = Area of triangle =1

2· 3 · 5 =

15

2.

15. (a) We have

curl(z~i + x~j + y~k ) =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

z x y

∣∣∣∣∣∣=~i +~j + ~k .

(b) Let S be the square interior of the curve C, oriented toward the origin. Then, by Stokes’ Theorem,∫

C

(z~i + x~j + y~k ) · d~r =

∫

S

curl(z~i + x~j + y~k ) · d ~A =

∫

S

(~i +~j + ~k ) · d ~A .

Since the plane has normal vector ~n = ~i + ~j + ~k , a unit normal in this direction is (~i + ~j + ~k )/√

3. But S isoriented toward the origin, so d ~A = −(~i +~j + ~k )/

√3 dA. Thus

∫

S

(~i +~j + ~k ) · d ~A =

∫

S

(~i +~j + ~k ) · −(~i +~j + ~k )√3

dA = − 3√3· Area of square = − 3√

3· 22 = −4

√3.


16. Since ~F is constant, curl ~F = ~0 , so if S is the disk in the plane enclosed by the circle, Stokes’ Theorem gives∫

C

~F · d~r =

∫

S

curl ~F · d ~A = 0.

17. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−y x z

∣∣∣∣∣∣=

(∂

∂x(x)− ∂

∂y(−y)

)~k = 2~k ,

writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

2~k · d ~A .

Now d ~A = ~n dA, where ~n is the unit vector perpendicular to the plane, so

~n =1√3

(~i +~j + ~k ).

Thus∫

C

~F d~r =

∫

S

2~k · 1√3

(~i +~j + ~k ) dA =2√3

∫

S

dA =2√3· Area of disk =

2√3· π22 =

8π√3.

18. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y −x y − x

∣∣∣∣∣∣=

∂

∂y(y − x)~i − ∂

∂x(y − x)~j +

(∂

∂x(−x)− ∂

∂y(y)

)~k =~i +~j − 2~k ,


C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(~i +~j − 2~k ) · d ~A .


~n =1√3

(~i +~j + ~k ).

Thus ∫

C

~F d~r =

∫

S

(~i +~j − 2~k ) ·~i +~j + ~k√

3dA =

∫

S

0√3dA = 0.

19. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

2y + ex (sin y)− x 2y − x+ cos z2

∣∣∣∣∣∣= 2~i − (−1)~j + (−1− 2)~k = 2~i +~j − 3~k ,


C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(2~i +~j − 3~k ) · d ~A .


~n =1√3

(~i +~j + ~k ).

Thus ∫

C

~F d~r =

∫

S

(2~i +~j − 3~k ) ·~i +~j + ~k√

3dA =

∫

S

0 dA = 0.

20.4 SOLUTIONS 1377

20. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

0 −z y

∣∣∣∣∣∣=

(∂

∂y(y)− ∂

∂z(−z)

)~i − 0~j + 0~k = 2~i ,


C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

2~i · d ~A .


~n =1√3

(~i +~j + ~k ).

Thus ∫

C

~F d~r =

∫

S

2~i ·~i +~j + ~k√

3dA =

∫

S

2√3dA =

2√3· Area of disk =

2√3π22 =

8π√3.

21. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

(z − y) (x− z) (y − x)

∣∣∣∣∣∣= 2~i + 2~j + 2~k ,


C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(2~i + 2~j + 2~k ) · d ~A .


~n =1√3

(~i +~j + ~k ).

Thus∫

C

~F d~r =

∫

S

(2~i + 2~j + 2~k ) ·~i +~j + ~k√

3dA =

∫

S

6√3dA = 2

√3 · Area of disk = 2

√3π22 = 8

√3π.

22. The curl ~F (x, y, z) of this vector field is equal to −2~j . Notice, as a check, that this field rotates in a direction oppositeto the direction of C. Therefore we expect a negative line integral. The surface S is the disk parallel to the xz plane withradius 2. The curl points in the opposite direction to the normal vector to the surface with this orientation, so by Stokes’Theorem,

Circulation around S =

∫

S

curl ~F · ~dA =

∫

S

(−2~j ) · d ~A =

∫

S

−2 dA

= −2(Area of circle) = −2π22 = −8π.

23. We calculate

curl ~G =

∣∣∣∣∣∣∣

~i ~j ~k∂

∂x

∂

∂y

∂

∂zxy z 3y

∣∣∣∣∣∣∣= (3− 1)~i − (0− 0)~j + (0− x)~k = 2~i − x~k .

Let S be the surface of the square, oriented in the positive x-direction. Then, by Stokes’ Theorem∫

C

~G · d~r =

∫

S

(curl ~G ) · d ~A =

∫

S

(2~i − x~k ) · d ~A .

On the square we have, d ~A =~i dydz, so∫

C

~G · d~r =

∫

S

(2~i − x~k ) ·~i dydz = 2 · Area of square = 2 · 36 = 72.


24. Use Stokes’ theorem, applied to the surface R, oriented upward. Since curl ~F = ~k for ~F = 12(−y~i + x~j ), we have

12

∫C

(−y~i + x~j ) · d~r =∫R~k · d ~A = ‖~k ‖(area of R) = area of R.

25. (a) We calculate

curl ~F =

∣∣∣∣∣∣∣

~i ~j ~k∂

∂x

∂

∂y

∂

∂zy − z x+ z xy

∣∣∣∣∣∣∣= (x− 1)~i − (y + 1)~j + (1− 1)~k = (x− 1)~i − (y + 1)~j .

Since the circle is in the xy-plane and curl ~F has no ~k component, the line integral is zero.(b) Even though the line integral around this closed curve is 0, we do not know that the line integral around every closed

curve is zero, so we cannot conclude that ~F is path-independent (conservative). Since curl ~F 6= ~0 , we know that ~Fis not path-independent (conservative).

26. (a) ~b × ~r =

∣∣∣∣∣∣

~i ~j ~k

b1 b2 b3

x y z

∣∣∣∣∣∣= (b2z − b3y)~i − (b1z − b3x)~j + (b1y − b2x)~k .

curl(~b × ~r ) =

∣∣∣∣∣∣

~i ~j ~k

∂/∂x ∂/∂y ∂/∂z

b2z − b3y b3x− b1z b1y − b2x

∣∣∣∣∣∣

= (b1 − (−b1))~i − (−b2 − b2)~j + (b3 − (−b3))~k

= 2b1~i + 2b2~j + 2b3~k

= 2~b .

(b) If D is the disk inside the circle, oriented upward, then, by Stokes’ Theorem∫

C

(~b × ~r ) · d~r =

∫

D

curl (~b × ~r ) · d ~A =

∫

D

2~b · d ~A .

Since the disk is flat, with normal vector pointing upward in the ~k direction, we have∫

D

2~b · d ~A =

∫

D

2~b · ~k dA = 2b3

∫

D

dA = 2b3 · Area of disk

= 2b3 · π(5)2 = 50πb3.

27. Using Stokes’ Theorem, the flux integral∫S

curl ~F · d ~A has the same value as the line integral∫C~F · d~r , where C is

the boundary curve of S with the appropriate orientation. Here C is the unit circle x2 + y2 = 1, z = 0 oriented clockwisewhen viewed from above. We can parameterize C by ~r (t) = cos t~i − sin t~j with 0 ≤ t ≤ 2π.

Then ~r ′(t) = − sin t~i − cos t~j , so∫

C

~F · d~r =

∫ 2π

0

(sin t~i + cos t~j ) · (− sin t~i − cos t~j )dt

=

∫ 2π

0

(− sin2 t− cos2 t) dt

=

∫ 2π

0

−1dt

= −2π

28. (a) At every point of C the vector field ~F is tangent to the curve C and of magnitude a. Since ~F is in the directionopposite to the orientation of C, we have

∫C~F · d~r = −‖ ~F ‖(length of C) = −a(2πa) = −2πa2.

(b) Since curl ~F = 2~i , which is normal to S and in the direction in which S is oriented,∫S

curl~F ·d ~A = ‖2~i ‖(area of S) =

2πa2.(c) The orientations of C and S are not related by the right-hand rule, so Stokes’ Theorem does not apply.

20.4 SOLUTIONS 1379

29. (a) Computing curl ~G , we get

curl ~G =

(∂(xz)

∂y− ∂(3x)

∂z

)~i +

(∂(4yz2)

∂z− ∂(xz)

∂x

)~j +

(∂(3x)

∂x− ∂(4yz2)

∂y

)~k

= 0~i + (8yz − z)~j + (3− 4z2)~k

= ~F

(b) Because ~F is a curl field, the flux of ~F through S is the same as its flux through any surface with the same boundaryas S. Let us replace S by the circular disk S1 of radius 5 in the xy-plane. On S1, z = 0, and so the vector fieldreduces to ~F = 3~k . Since the vector 3~k is normal to S1 and in the direction of the orientation of S1, we have

∫

S

~F · d ~A =

∫

S1

~F · d ~A =

∫

S1

3~k · d ~A = ‖3~k ‖(area of S1) = 75π.

30. Suppose we have a surface S which completely bounds a solid region W . Pick a fixed point P on the surface. Imaginea closed curve C which lies on S and encloses P . Consider the surface S ′ which is S with the patch around P cut out.Then C can be thought of as the boundary of S ′, so from Stokes’ Theorem,

∫

S′curl ~F · d ~A =

∫

C

~F · d~r

Now let the length of C shrink to zero. The surface S ′ becomes S, and C becomes the point P , the line integral of ~Faround C becomes zero, and so ∫

S

curl ~F · d ~A = 0

Thus, if Q is a point in the domain of curl ~F and S denotes closed surfaces around Q,

div curl ~F (Q) = limVol→0

∫S

curl ~F · d ~AVol enclosed by S

= limVol→0

0

Vol= 0

31. Let ~F = F1(x, y)~i + F2(x, y)~j be a continuously differentiable vector function in a domain in the xy-plane whichcontains a closed curve C surrounding a region R. Assume that the orientation of C is chosen so that R is on the left aswe move around C. Then, R is oriented upward, so the area element d ~A is given by

d ~A = ~k dxdy.

Thus,

curl ~F · d ~A = curl ~F · ~k dxdy =

(∂F2

∂x− ∂F1

∂y

)dxdy.

Thus the formula in Stokes’ Theorem takes the form∫

C

~F · d~r =

∫

C

(∂F2

∂x− ∂F1

∂y

)dxdy.

This shows that Green’s Theorem in the plane is a special case of Stokes’ Theorem.

32. (a) You can’t say anything, because any surface bounded by the circle must intersect the z-axis. Since curl ~F is notdefined on the z-axis, the surface integral in Stokes’ Theorem is not defined.

(b) In this case curl ~F is defined and equal to ~0 on a surface S bounded by the the circle, so Stokes’ Theorem says that∫

C

~F · d~r =

∫

S

curl ~F · d ~A = 0.


33. Notice that ~F , div ~F , and curl ~F are defined and continuous everywhere.

(a) By the Divergence Theorem, the fact that the flux through any closed surface is 0 tells us that everywhere

div ~F = 0.

Since div ~F = a, we know a = 0. We do not have any information about b, c,m.(b) Since the circulation around any closed curve is 0, by Stokes’ Theorem, we have

curl ~F = ~0 .

Now

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

ax+ by + 5z x+ cz 3y +mx

∣∣∣∣∣∣= (3− c)~i − (m− 5)~j + (1− b)~k ,

so curl ~F = 0 means c = 3,m = 5 and b = 1. We do not have any information about a.

34. (a) Notice that the denominators (z2 + 1)2 and (z2 + 1) are always positive and so affect the magnitude (but not thedirection) of the motion of each of the terms.

The (−y~i + x~j ) term represents rotation around the z-axis (counterclockwise when viewed from above). The−z(x~i + y~j ) term represents radial motion (toward the z-axis when z > 0 and away when z < 0). The ~k termis downward motion. So ~F is a flow rotating inward and downward around the z-axis (for z > 0), like an actualbathtub drain.

(b) Let D be the disk representing the drain, oriented downward. Then the rate at which the water is leaving the bathtubis the flux of water flowing out of the drain:

∫

D

~F · d ~A =

∫

D

~F · (−dA~k ) =

∫

D

1

z2 + 1dA.

Because D is in the xy-plane, z = 0, so

Flux out of D =

∫

D

dA = π cm3/sec.

(c) We have, in units/sec,

div ~F = − z

(z2 + 1)2− z

(z2 + 1)2+

2z

(z2 + 1)2= 0.

(d) Let W be the closed region bounded by the hemisphere S of radius 1 lying below the xy-plane and the disk, D, inthe xy-plane representing the drain. Both S and D are oriented downward, so by the Divergence Theorem, we have:

0 =

∫

W

div ~F dV = Flux out of W − Flux into W

=

∫

S

~F · d ~A −∫

D

~F · d ~A

=

∫

S

~F · d ~A − π.

Thus, ∫

S

~F · d ~A = π cm3/sec.

(e) Since C is oriented clockwise, we parameterize the circle by ~r (t) = (sin t)~i + (cos t)~j . In addition, on the drain,z = 0. Thus,

∫

C

~G · d~r =

∫

C

1

2(y~i − x~j − (x2 + y2)~k ) · d~r

=1

2

∫ 2π

0

(cos t~i − sin t~j − ~k ) · (cos t~i − sin t~j ) dt

=1

2

∫ 2π

0

(sin2 t+ cos2 t)dt = π.

20.4 SOLUTIONS 1381

So, ∫

C

~G · d~r = π cm3/sec.

(f) Computing the partial derivatives, we find that

curl ~G = − y + xz

(z2 + 1)2~i − yz − x

(z2 + 1)2~j − 1

z2 + 1~k = ~F .

(g) By Stokes’ Theorem, we have:∫

C

~G · d~r =

∫

D

curl ~G · d ~A =

∫

D

~F · d ~A .

Thus, since curl ~G = ~F , Stoke’s Theorem tells us that the answers to parts (d) and (e) should be equal.

35. (a) We have

curl ~F =

∣∣∣∣∣∣∣∣

~i ~j ~k∂

∂x

∂

∂y

∂

∂z−y

x2 + y2

x

x2 + y20

∣∣∣∣∣∣∣∣= 0~i + 0~j +

(∂

∂x

(x

x2 + y2

)+

∂

∂y

(y

x2 + y2

))~k .

Since∂

∂x

(x

x2 + y2

)=

1

x2 + y2− x(2x)

(x2 + y2)2=x2 + y2 − 2x2

(x2 + y2)2=

y2 − x2

(x2 + y2)2

and similarly∂

∂y

(y

x2 + y2

)=

x2 − y2

(x2 + y2)2, we have, provided x2 + y2 6= 0,

curl ~F = ~0 .

The domain of curl ~F is all points in 3-space except the z-axis.(b) On C1, the unit circle x2 + y2 = 1 in the xy-plane, the vector field ~F is tangent to the circle and || ~F || = 1. Thus

Circulation =

∫

C1

~F · d~r = || ~F || · Perimeter of circle = 2π.

Note that Stokes’ Theorem cannot be used to calculate this circulation since the z-axis pierces any surface which hasthis circle as boundary.

(c) Consider the disk (x− 3)2 + y2 ≤ 1 in the plane z = 4. This disk has C2 as boundary and curl ~F = ~0 everywhereon this disk. Thus, by Stokes’ Theorem

∫C2

~F · d~r = 0.(d) The square S has an interior region which is pierced by the z-axis, so we cannot use Stokes’ Theorem. We consider

the region, D, between the circle C1 and the square S. See Figure 20.20. Stokes’ Theorem applies to the region D,provided C1 is oriented clockwise. Then we have

∫

C1(clockwise)

~F · d~r +

∫

S

~F · d~r =

∫

D

curl ~F · d ~A = 0.

Thus, ∫

S

~F · d~r = −∫

C1(clockwise)

~F · d~r =

∫

C1(counterclockwise)

~F · d~r = 2π.

(e) If a simple closed curve goes around the z-axis, then it contains a circle C of the form x2 + y2 = a2. The circulationaround C is 2π or −2π, depending on its orientation. A calculation similar to that in part (d) then shows that thecirculation around the curve is 2π or −2π, again depending on its orientation. If the closed curve does not go aroundthe z-axis, then curl ~F = ~0 everywhere on its interior and the circulation is zero.


(2, 2, 0)(−2, 2, 0)

(−2,−2, 0) (2,−2, 0)

C1

S

D

x

y

Figure 20.20

36. (a) The radius vector at the point (x, y) is ~r = x~i + y~j . A vector perpendicular to ~r and pointing counterclockwisearound the origin is −y~i + x~j . This is not a unit vector, so we divide by its length, giving

~T =−y~i + x~j√x2 + y2

.

(b) Since ~T is tangent to the circle and || ~F || = ||~r ||n = an on the circle C, the line integral is given by∫

C

~F · d~r = || ~F || · Circumference of circle = an · 2πa = 2πan+1.

(c) We have

~F = (x2 + y2)n/2(−y~i + x~j )√

x2 + y2=(−y(x2 + y2)(n−1)/2

)~i +

(x(x2 + y2)(n−1)/2

)~j ,

so

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−y(x2 + y2)(n−1)/2 x(x2 + y2)(n−1)/2 0

∣∣∣∣∣∣

=

(∂

∂x

(x(x2 + y2)(n−1)/2

)− ∂

∂y

(−y(x2 + y2)(n−1)/2

))~k

=

((x2 + y2)(n−1)/2 +

2(n− 1)

2x2(x2 + y2)(n−3)/2+

(x2 + y2)(n−1)/2 +2(n− 1)

2y2(x2 + y2)(n−3)/2

)~k

= (x2 + y2)(n−3)/2(x2 + y2 + (n− 1)x2 + x2 + y2 + (n− 1)y2)~k

= (n+ 1)(x2 + y2)(n−1)/2~k .

(d) Using Stokes’ Theorem, if D is the disk in the xy-plane inside the circle, we have∫

C

~F · d~r =

∫

D

curl ~F · d ~A =

∫

D

(n+ 1)(x2 + y2)(n−1)/2~k · d ~A .

Since d ~A = ~k dx dy,∫

C

~F · d~r =

∫

D

(n+ 1)(x2 + y2)(n−1)/2~k · ~k dx dy =

∫

D

(n+ 1)(x2 + y2)(n−1)/2dx dy.

Converting to polar coordinates,∫

C

~F · d~r =

∫ 2π

0

∫ a

0

(n+ 1)(r2)(n−1)/2r dr dθ =

∫ 2π

0

∫ a

0

(n+ 1)rn−1r dr dθ

=

∫ 2π

0

∫ a

0

(n+ 1)rn dr dθ = θ

∣∣∣∣∣

2π

0

rn+1

∣∣∣∣∣

a

0

= 2πan+1.

20.5 SOLUTIONS 1383


Exercises

1. Since curl ~F = ~0 and ~F is defined everywhere, we know by the curl test that ~F is a gradient field. In fact, ~F = gradf ,where f(x, y, z) = xyz + yz2, so f is a potential function for ~F .

2. Since curl ~G = 2~k 6= ~0 , the vector field ~G is not a gradient field.

3. (a) We calculate the curl of each of these vector fields.

curl ~A = curl(−by~i ) =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−by 0 0

∣∣∣∣∣∣= − ∂

∂y(−by)~k = b~k .

(b)

curl ~A = curl(bx~j ) =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

0 bx 0

∣∣∣∣∣∣=

∂

∂x(bx)~k = b~k .

(c)

−1

2~r × ~B = −1

2yb~i +

1

2xb~j

curl ~A = curl(

1

2~B × ~r

)

=

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

−(1/2)by (1/2)bx 0

∣∣∣∣∣∣

=

(∂

∂x

((1

2

)bx)− ∂

∂y

(−(

1

2

)by))

~k = b~k .

Problems

4. Note thatdiv(2y~i + 4x~j ) =

∂

∂x(2y) +

∂

∂y(4x) = 0

and

curl(3x~i + 9y~j ) =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

3x 9y 0

∣∣∣∣∣∣= ~0 ,

so (3x~i + 9y~j ) + (2y~i + 4x~j ) is the required decomposition.

5. Let ~v = a~i + b~j + c~k and try

~F = ~v × ~r = (a~i + b~j + c~k )× (x~i + y~j + z~k ) = (bz − cy)~i + (cx− az)~j + (ay − bx)~k .

Then

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

bz − cy cx− az ay − bx

∣∣∣∣∣∣= 2a~i + 2b~j + 2c~k .

Taking a = 1, b = − 32, c = 2 gives curl ~F = 2~i − 3~j + 4~k , so the desired vector field is ~F = (− 3

2z− 2y)~i + (2x−

z)~j + (y + 32x)~k .

6. In Example 3 on page 957 we showed that curl(~b × ~r ) = 2~b . Thus (1/2)~b × ~r is a vector potential for ~B .


7. We must show curl ~A = ~B .

curl ~A =∂

∂y

(−Ic

ln(x2 + y2))~i − ∂

∂x

(−Ic

ln(x2 + y2))~j

=−Ic

(2y

x2 + y2

)~i +

I

c

(2x

(x2 + y2)

)~j

=2I

c

(−y~i + x~j

x2 + y2

)

= ~B .

8. Since div(y~i + x~j ) = 0, there is such a vector field ~G . In fact, ~G = (1/2)(y2 − x2)~k is one possibility, though thereare others.

9. Since div ~F = 2 + 3− 5 = 0, a vector potential does exist. One such is the vector field ~H = (−xy + 5yz)~i + (2xy +

xz2)~k , but there are many others.

10. Since div ~G = 2x+ 2y + 2z 6= 0, there is not a vector potential for ~G .

11. (a) Yes. To show this, we use a version of the product rule for curl (Problem 28 on page 1367):

curl(φ~F ) = φ curl ~F + (gradφ)× ~F ,

where φ is a scalar function and ~F is a vector field. So

curl

(q

~r

‖~r ‖3)

= curl

(q

‖~r ‖3 ~r)

=q

‖~r ‖3 curl~r + grad

(q

‖~r ‖3)× ~r

= ~0 + q grad

(1

‖~r ‖3)× ~r

Since the level surfaces of 1/‖~r ‖3 are spheres centered at the origin, grad(1/‖~r ‖3) is parallel to ~r , so grad(1/‖~r ‖3)×~r = ~0 . Thus, curl ~E = ~0 .

(b) Yes. The domain of ~E is 3-space minus (0, 0, 0). Any closed curve in this region is the boundary of a surfacecontained entirely in the region. (If the first surface you pick happens to contain (0, 0, 0), change its shape slightly toavoid it.)

(c) Yes. Since ~E satisfies both conditions of the curl test, it must be a gradient field. In fact,

~E = grad

(−q 1

‖~r ‖

).

12. (a) Yes. This is the case p = 2 of Example 5 on page 958.(b) No. The domain of ~B is 3-space minus the z-axis. A closed curveC which surrounds the z-axis cannot be contracted

to a point without hitting the z-axis, so it cannot remain at all times within the domain.(c) No. In Example 2 on page 964 we found that if C is a circle around the origin,

∫

C

~B · d~r =4πI

c.

Thus ~B has non-zero circulation around C, and hence cannot be a gradient field.

13. (a) Using the product rule from Problem 28 on page 1367, we find

curl ~E = curl

(~r

‖~r ‖p)

=1

‖~r ‖p curl~r + grad

(1

‖~r ‖p)× ~r .

Now curl~r = ~0 and grad(

1‖~r ‖p

)is parallel to ~r , so both terms are zero. Thus curl ~E = ~0 .

(b) The domain of ~E is 3-space minus the origin if p > 0, and it is all of 3-space if p ≤ 0.

SOLUTIONS to Review Problems for Chapter Twenty 1385

(c) Both domains have the property that any closed curve can be contracted to a point without hitting the origin, so ~Esatisfies the curl test for all p. Since ~E has constant magnitude r1−p on the sphere of radius r centered at the origin,and is parallel to the outward normal at every point of the sphere, the sphere must be a level surface of the potentialfunction φ, that is, φ is a function of r alone. Further, since ‖ ~E ‖ = r1−p, a good guess is

φ(r) =

∫r1−p dr,

that is,

φ(r) =

{r2−p2−p if p 6= 2

ln r if p = 2.

You can check that this is indeed a potential function for ~E by checking that gradφ = ~E .

14. (a) Although div ~B = 0, ~B does not satisfy the divergence test because its domain is 3-space minus the origin, whichdoes not have the required property that every closed surface is the boundary of a solid region which is entirelycontained within the domain. For example, the solid region inside a sphere centered at the origin contains the origin,hence is not in the domain of ~B .

(b) Using the product rule from Problem 28 on page 1367, we find

curl ~A =

(1

‖~r ‖3)

curl(µ× ~r ) + grad

(1

‖~r ‖3)× (~µ × ~r ).

By Example 3 on page 957,curl(~µ × ~r ) = 2~µ

and by Problem 40 on page 718

grad

(1

‖~r ‖3)

= −31

‖~r ‖5 ~r .

So

curl ~A = 2~µ

‖~r ‖3 − 3

(1

‖~r ‖5)~r × (~µ × ~r ).

From Problem 32 on page 679, we have

~r × (~µ × ~r ) = ‖~r ‖2~µ − (~µ · ~r )~r .

So

curl ~A = 2~µ

‖~r ‖3 − 3

(1

‖~r ‖5)

(‖~r ‖2~µ − (~µ · ~r )~r ) = − ~µ

‖~r ‖3 +3(~µ · ~r )~r

‖~r ‖5 .

(c) No. The divergence test says a vector field must be a curl field if it satisfies the conditions of the test; it does not saythe vector field cannot be a curl field if the vector field fails to satisfy the test.

15. (a) Since curl gradψ = 0 for any function ψ, curl( ~A + gradψ) = curl ~A + curl gradψ = curl ~A = ~B .(b) We have

div( ~A + gradψ) = div ~A + div gradψ = div ~A +∇2ψ.

Thus ψ should be chosen to satisfy the partial differential equation

∇2ψ = − div ~A .

Solutions for Chapter 20 Review

Exercises

1. Fields with zero curl: (c), (d), (f) because these don’t appear to be swirling.

2. Fields (a), (c), (e) because they do not appear to be exploding or collapsing.


3. C2, C3, C4, C6, since line integrals around C1 and C5 are clearly nonzero. You can see directly that∫C2

~F · d~r and∫C6

~F · d~r are zero, because C2 and C6 are perpendicular to their fields at every point.

4. We havediv ~F =

∂

∂x

(x2)

+∂

∂y

(y3)

+∂

∂z

(z4)

= 2x+ 3y2 + 4z3,

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x2 y3 z4

∣∣∣∣∣∣= ~0 .

So ~F is not solenoidal, but ~F is irrotational.

5. We have

div ~F =∂

∂x(xy) +

∂

∂y(yz) +

∂

∂z(zx) = y + z + x,

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

xy yz zx

∣∣∣∣∣∣= −y~i − z~j − x~k .

So ~F is not solenoidal and not irrotational.

6. We havediv ~F =

∂

∂x(cosx) +

∂

∂y(ey) +

∂

∂z(x+ y + z) = − sinx+ ey + 1

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

cosx ey x+ y + z

∣∣∣∣∣∣=~i −~j .

So ~F is not solenoidal and not irrotational.

7. We havediv ~F =

∂

∂x(ey+z) +

∂

∂y(sin(x+ z)) +

∂

∂z(x2 + y2) = 0

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

ey+z sin(x+ z) x2 + y2

∣∣∣∣∣∣= (2y − cos(x+ z))~i −

(2x− ey+z

)~j +

(cos(x+ z)− ey+z

)~k .

So ~F is solenoidal, but ~F is not irrotational.

8. (a) We will compute separately the flux of the vector field ~F = x3~i + 2y~j + 3~k through each of the six faces of thecube.

The face SI where x = 1, which has normal vector~i . Only the~i component x3~i = ~i of ~F has flux throughSI .

∫

SI

~F · d ~A =

∫

SI

~i · d ~A = ‖~i ‖(area of SI) = 4.

The face SII where x = −1, which has normal vector −~i . Only the~i component x3~i = −~i of ~F has fluxthrough SII .

∫

SII

~F · d ~A =

∫

SII

−~i · d ~A = ‖ −~i ‖(area of SII) = 4.

The face SIII where y = 1, which has normal vector ~j . Only the ~j component 2y~j = 2~j of ~F has fluxthrough SIII .

∫

SIII

~F · d ~A =

∫

SIII

2~j · d ~A = ‖2~j ‖(area of SIII) = 8.

The face SIV where y = −1, which has normal vector −~j . Only the ~j component 2y~j = −2~j of ~F has fluxthrough SIV .


∫

SIV

~F · d ~A =

∫

SIV

−2~j · d ~A = ‖ − 2~j ‖(area of SIV ) = 8.

The face SV where z = 1, which has normal vector ~k . Only the ~k component 3~k of ~F has flux through SV .∫

SV

~F · d ~A =

∫

SV

3~k · d ~A = ‖3~k ‖(area of SV ) = 12.

The face SV I where z = −1, which has normal vector −~k . Only the ~k component 3~k of ~F has flux throughSV I .

∫

SV I

~F · d ~A =

∫

SV I

3~k · d ~A = −‖3~k ‖(area of SV I) = −12.

(Total flux through S) = 4 + 4 + 8 + 8 + 12− 12 = 24.

(b) Since S is a closed surface the Divergence Theorem applies. Since div ~F = 3x2 + 2,∫

S

~F · d ~A =

∫ 1

x=−1

∫ 1

y=−1

∫ 1

z=−1

(3x2 + 2)dzdydx = 24.

9. (a) By direct calculation, the flux of ~F through the plane x = 0 is 0 because the ~i -component of the vector field is 0there. The flux of ~F through S, the square 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 in the plane x = 1 is

Flux =

∫

S

~F · d ~A =

∫ 1

0

∫ 1

0

(1 · y~i + yz~j + z · 1~k ) ·~i dy dz =

∫ 1

0

∫ 1

0

y dy dz =

∫ 1

0

y2

2

∣∣∣∣1

0

dz =1

2.

Similarly, the faces y = 0 and y = 1 contribute a flux of 0 and 1/2, respectively, as do the faces z = 0 and z = 1.Therefore

Total flux = 0 +1

2+ 0 +

1

2+ 0 +

1

2=

3

2.

(b) Since div ~F = y + z + x, the flux is given by∫

Box

~F · d ~A =

∫

Box(x+ y + z)dV.

We calculate the integral∫

Box(x+ y + z) dV =

∫ 1

0

∫ 1

0

∫ 1

0

(x+ y + z) dx dy dz =

∫ 1

0

∫ 1

0

(x2

2+ yx+ zx

)∣∣∣1

0dy dz

=

∫ 1

0

∫ 1

0

(1

2+ y + z

)dy dz =

∫ 1

0

(y

2+y2

2+ zy

)∣∣∣1

0dz

=

∫ 1

0

(1

2+

1

2+ z)dz =

∫ 1

0

(1 + z) dz =

(z +

z2

2

)∣∣∣1

0=

3

2.

10. (a) Let us parameterize the curve C by ~r (t) = 3 cos t~i + 3 sin t~j , 0 ≤ t ≤ 2π.Then d~r = (−3 sin t~i + 3 cos t~j )dt and so

∫

C

((yz2 − y)~i + (xz2 + x)~j + 2xyz~k ) · d~r =

∫

C

(−3 sin t~i + 3 cos t~j ) · d~r

=

∫ 2π

0

9dt = 18π.

(b) Since C is a closed curve, Stokes’ Theorem applies. We choose the surface S to be the disk in the xy-plane boundedby C, and it must be oriented upward. Since curl ~F = 2~k ,

∫

C

~F · d~r =

∫

S

2~k · d ~A = ‖2~k ‖(area of S) = 2(π32) = 18π.


11. Since div ~F = 2 + 3 + 4 = 9, the Divergence Theorem gives∫

S

~F · d ~A =

∫

Interiorof sphere

div ~F dV = 9 · Volume of sphere = 9 · 4

3π53 = 1500π.

12. Since div ~F = 1 + 2− 1 = 2, the Divergence Theorem gives∫

S

~F · d ~A =

∫

Interiorof sphere


3π13 =

8π

3.

13. Since div ~F = 2x+ 2y, the Divergence Theorem gives∫

S

~F · d ~A =

∫

Interiorof cube

(2x+ 2y) dV

= 2

∫ 3

0

∫ 3

0

∫ 3

0

(x+ y) dx dy dz

= 2

∫ 3

0

∫ 3

0

(x2

2+ xy

)∣∣∣∣3

0

dy dz

= 2

∫ 3

0

∫ 3

0

(9

2+ 3y

)dy dz

= 2

∫ 3

0

(9

2y +

3

2y2) ∣∣∣∣

3

0

dz = 2

∫ 3

0

27 dz = 162.

14. Since div ~F = 3x2 + 3y2, using cylindrical coordinates to calculate the triple integral gives

∫

S

~F · d ~A =

∫

Interiorof cylinder

(3x2 + 3y2) dV = 3

∫ 2π

0

∫ 5

0

∫ 2

0

r2 · r dr dz dθ = 3 · 2π · 5 r4

4

∣∣∣∣2

0

= 120π.

15. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x2 y2 z2

∣∣∣∣∣∣= ~0

the line integral∫C~F · d~r = 0 around any closed curve, including this unit circle.

16. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y − x z − y x− z

∣∣∣∣∣∣= −~i −~j − ~k ,

and the area vector of the disk x2 + y2 ≤ 5 is ~A = π(√

5)2~k = 5π~k , Stokes’ Theorem gives∫

C

~F · d~r =

∫

Disk

(−~i −~j − ~k ) · d ~A = (−~i −~j − ~k ) · Area vector = (−~i −~j − ~k ) · 5π~k = −5π.

17. If C is the circle x2 + y2 = 9 oriented counterclockwise when viewed from above, Stokes’ Theorem gives∫

S

curl ~F · d ~A =

∫

C

~F · d~r .

Only the~i and ~j components of ~F , namely −y~i + x~j , contribute to the line integral. Since∥∥−y~i + x~j

∥∥ = 3 on thecircle, and −y~i + x~j is tangent to the circle and points in the direction of the orientation of C we have

∫

S

curl ~F · d ~A =

∫

C

~F · d~r =∥∥−y~i + x~j

∥∥ · Length of curve = 3 · 2π3 = 18π.

Alternatively, the line integral can be evaluated by parameterizing the curve using x = 3 cos t, y = 3 sin t for 0 ≤ t ≤2π.


18. If C is the rectangular path around the rectangle, traversed counterclockwise when viewed from above, Stokes’ Theoremgives ∫

S

curl ~F · d ~A =

∫

C

~F · d~r .

The ~k component of ~F does not contribute to the line integral, and the ~j component contributes to the line integral onlyalong the segments of the curve parallel to the y-axis. Thus, if we break the line integral into four parts

∫

S

curl ~F · d ~A =

∫ (3,0)

(0,0)

~F · d~r +

∫ (3,2)

(3,0)

~F · d~r +

∫ (0,2)

(3,2)

~F · d~r +

∫ (0,0)

(0,2)

~F · d~r ,

we see that the first and third integrals are zero, and we can replace ~F by its ~j component in the other two

∫

S

curl ~F · d ~A =

∫ (3,2)

(3,0)

(x+ 7)~j · d~r +

∫ (0,0)

(0,2)

(x+ 7)~j · d~r .

Now x = 3 in the first integral and x = 0 in the second integral and the variable of integration is y in both, so∫

S

curl ~F · d ~A =

∫ 2

0

10 dy +

∫ 0

2

7 dy = 20− 14 = 6.

19. (a) We have

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

cosx ey x− y − z

∣∣∣∣∣∣= −~i −~j .

(b) If S is the disk on the plane within the circle C, Stokes’ Theorem gives∫

C

~F · d~r =

∫

S

curl ~F · d ~A .

For Stokes’ Theorem, the disk is oriented upward. Since the unit normal to the plane is (~i + ~j + ~k )/√

3 andthe disk has radius 3, the area vector of the disk is

~A =~i +~j + ~k√

3π(32) = 3

√3π(~i +~j + ~k ).

Thus, using curl ~F = −~i −~j , we have∫

C

~F · d~r =(−~i −~j

)· 3√

3π(~i +~j + ~k ) = −6√

3π.

20. (a) A counterclockwise parameterization of the circle is

x(t) = cos t, y(t) = sin t 0 ≤ t ≤ 2π.

Since x′(t) = − sin t, and y′(t) = cos t, we have∫

C

~F · d~r =

∫ 2π

0

(sin t~i − cos t~j ) · (− sin t~i + cos t~j )dt

=

∫ 2π

0

(− sin2 t− cos2 t)dt = −2π.

(b) curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

y −x 0

∣∣∣∣∣∣= 0~i + 0~j − 2~k = −2~k .

(c) Using Stokes’ Theorem, where R is the region inside the circle oriented upward∫

C

~F · d~r =

∫

R

curl ~F · d ~A =

∫

R

−2~k · d ~A = −2 · Area of circle = −2π.


(d) Stokes’ Theorem, which says that if C is a closed curve which is the boundary of a surface R, and ~F is a smoothvector field, then ∫

C

~F · d~r =

∫

R

curl ~F · d ~A .

Here, the orientations of C and of R are related by the right-hand rule.

Problems

21. Calculating the curl, we obtain∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

2z + ax2y 5z + x3 bx+ cy + z

∣∣∣∣∣∣= (c− 5)~i − (b− 2)~j + (3x2 − ax2)~k .

Thus, the curl of this vector field is ~0 for all x, y, z if a = 3, b = 2, c = 5.

22. To show that the force field is irrotational we must show that its curl is zero. Let us do this in Cartesian coordinates:

~F = f(r)~r = f(√x2 + y2 + z2)(x~i + y~j + z~k )

The third component of curl ~F is

∂

∂x(f(√x2 + y2 + z2)y)− ∂

∂y(f(√x2 + y2 + z2)x)

= f ′(√x2 + y2 + z2) · 2xy

2√x2 + y2 + z2

− f ′(√x2 + y2 + z2) · 2xy

2√x2 + y2 + z2

= 0.

A similar computation shows that the other components of curl ~F are 0 too.

23. (a) Since div ~F at a point is approximately equal to the flux density out of a small region around the point, at (4, 5, 2),we have

div ~F ≈ Flux out of small regionVolume of region

=0.0125

(4/3)π(0.1)3= 2.984.

(b) The flux through the sphere is approximated by

Flux through sphere =

∫

S

~F · d ~A ≈ div ~F · Volume of sphere = 2.984 · 4

3π(0.2)2 = 0.100.

We could also estimate the flux by noticing that it is eight times the original flux, that is, 8(0.0125) = 0.100.

24. Figure 20.21 shows a two dimensional cross-section of the vector field ~v = −2~r . The vector field points radially inward,so if we take S to be a sphere of radius R centered at the origin, oriented outward, we have

~v ·∆ ~A = −2R ‖∆ ~A ‖,

for a small area vector ∆ ~A on the sphere. Therefore,∫

S

~v · d ~A =

∫

S

−2R ‖d ~A ‖ = −2R(Surface area of sphere) = −2R(4πR2) = −8πR3.

Thus, we find that

div~v (0, 0, 0) = limvol→0

( ∫S~v · d ~A

Volume of sphere

)= limR→0

(−8πR3

43πR3

)= −6.

Notice that the divergence is negative. This is what you would expect, since the vector field represents an inward flow atthe origin.

Since ~v = −2~r = −2x~i − 2y~j − 2z~k , the coordinate definition give

div~v =∂

∂x(−2x) +

∂

∂y(−2y) +

∂

∂z(−2z) = −2− 2− 2 = −6.


−R R

x

y

Figure 20.21: The vector field ~v = −2~r

25. (a) On S1, x = a and normal is in negative x-direction, so

~F ·∆ ~A = (2a~i − 3y~j + 5z~k ) · (−∆A~i ) = −2a∆A.

Thus ∫

S1

~F · d ~A =

∫

S1

−2adA = −2a(Area of S1) = −2aw2.

On S2, x = a+ w and normal is in positive x-direction, so

~F ·∆ ~A = (2(a+ w)~i − 3y~j + 5z~k ) · (∆A~i ) = 2(a+ w)∆A.

Thus ∫

S2

~F · d ~A =

∫

S2

2(a+ w)dA = 2(a+ w)(Area of S2) = 2(a+ w)w2

Calculating the flux through the other sides similarly, we get

Total flux

=

∫

S1

~F · d ~A +

∫

S2

~F · d ~A +

∫

S3

~F · d ~A +

∫

S4

~F · d ~A +

∫

S5

~F · d ~A +

∫

S6

~F · d ~A

= −2aw2 + 2(a+ w)w2 + 3bw2 − 3(b+ w)w2 − 5cw2 + 5(c+ w)w2

= (2w − 3w + 5w)w2 = 4w3.

(b) To find div ~F at the point (a, b, c), let the box shrink to the point by letting w → 0. Then

div ~F = limw→0


)

= limw→0

(4w3

w3

)= 4.

(c)

div ~F =∂

∂x(2x) +

∂

∂y(−3y) +

∂

∂z(5z) = 2− 3 + 5 = 4.

26. Using flux: On S1, x = a and normal is in negative x-direction, so

~F ·∆ ~A = ((3a+ 2)~i + 4a~j + (5a+ 1)~k ) · (−∆A~i ) = −(3a+ 2)∆A

Thus ∫

S1

~F · d ~A =

∫

S1

−(3a+ 2)dA = −(3a+ 2)(Area of S1) = −(3a+ 2)w2.

On S2, x = a+ w and normal is in the positive x-direction, so

~F ·∆ ~A = [(3(a+ w) + 2)~i + 4(a+ w)~j + (5(a+ w) + 1)~k ] · (∆A~i ) = (3a+ 3w + 2)∆A

Thus ∫

S2

~F · d ~A =

∫

S2

(3a+ 3w + 2)dA = (3a+ 3w + 2)(Area of S2) = (3a+ 3w + 2)w2.


Next, we have∫S3

~F · d ~A =∫S3−4xdA and

∫S4

~F · d ~A =∫S4

4xdA. Since these two are integrated over the same

region in the xz-plane, the two integrals cancel. Similarly,∫S5

~F · d ~A =∫S5−(5x+ 1)d ~A cancels out

∫S6

~F · d ~A =∫S6

(5x+ 1)d ~A . Therefore,

Total flux

=

∫

S1

~F · d ~A +

∫

S2

~F · d ~A +

∫

S3

~F · d ~A +

∫

S4

~F · d ~A +

∫

S5

~F · d ~A +

∫

S6

~F · d ~A

= −(3a+ 2)w2 + (3a+ 3w + 2)w2 +

∫

S3

−4xdA+

∫

S4

4xdA

+

∫

S5

−(5x+ 1)dA+

∫

S6

(5x+ 1)dA = 3w3.

To find div ~F at the point (a, b, c), let the box shrink to the point by letting w → 0. Then

div ~F = limw→0


)

= limw→0

(3w3

w3

)= 3.

Using partial derivatives:

div ~F =∂

∂x(3x+ 2) +

∂

∂y(4x) +

∂

∂z(5x+ 1) = 3

27. (a) The vector field appears to be conservative. Consider a closed rectangular path with sides parallel to the axes. (SeeFigure 20.22.) Since all the vectors of ~F are parallel to the y-axis, the line integral along the sides AB and DCof the rectangle are zero. Also, the vector field is independent of x, so it is the same along the sides AD and BC.Thus, the integrals along AD and BC cancel out, since they are traversed in opposite directions as you go aroundthe rectangle. So the integral around any closed rectangular path is zero. A similar argument suggests that the lineintegral of this vector field around any closed curve is zero.

x

y

A

C

B

D

Figure 20.22

(b) Since the vector field is conservative, its curl is zero, so curl ~F · ~k = 0.(c) We are told ~F has no z-component and is independent of z. From the figure, we see that the vectors all point parallel

to the y-axis, so ~F is a multiple of ~j . Also, they are independent of x, so ~F is a function of y times ~j . A possibleformula is ~F = −y~j , since the vectors of this field are zero along the x-axis, point downward above it, and getlarger as y gets larger.

(d) We have

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

0 −y 0

∣∣∣∣∣∣= ~0 .


28. Sincediv(3x~i + 4y~j + xy~k ) = 3 + 4 + 0 = 7,

we calculate the flux using the Divergence Theorem:

Flux =

∫

S

(3x~i + 4y~j + xy~k ) · d ~A =

∫

W

7 dV = 7 · Volume of box = 7 · 3 · 5 · 2 = 210.

29. We use the Divergence Theorem.Since div ~F = 3− 1 + 2 = 4 and a closed surface is oriented outward, the Divergence Theorem gives

∫

S

~F · d ~A =

∫

Interiorof sphere


3π13 =

16π

3.

30. We use Stokes’ Theorem. If C is the circle y2 + z2 = 3, oriented counterclockwise when viewed from the positivex-direction, then Stokes’ Theorem gives

∫

S

curl ~F · d ~A =

∫

C

~F · d~r =

∫

C

((z + y)~i − (z + x)~j + (y + x)~k

)· d~r .

Since C is in the yz-plane, the~i component does not contribute to the line integral. On the yz-plane, x = 0 so∫

S

curl ~F · d ~A =

∫

C

~F · d~r =

∫

C

(−z~j + y~k ) · d~r .

On the circle y2 + z2 = 3, we have∥∥−z~j + y~k

∥∥ =√

3 and −z~j + y~k is tangent to the curve pointing in thecounterclockwise direction, so

∫

S

curl ~F · d ~A =

∫

C

(−z~j + y~k ) · d~r =√

3 · Length of curve =√

3 · 2π√

3 = 6π.

31. We use the Divergence Theorem. Since div ~F = 3x2 +3y2 +3z2 and a closed surface is oriented outward, the DivergenceTheorem gives

∫

S

~F · d ~A =

∫

Interiorof sphere

div ~F dV

=

∫

Interiorof sphere

(3x2 + 3y2 + 3z2

)dV

= 3

∫ 2π

0

∫ π

0

∫ 1

0

ρ2 · ρ2 sinφ dρ dφ dθ

= 3 · 2π(− cosφ)

∣∣∣∣π

0

ρ5

5

∣∣∣∣1

0

=12π

5.

32. We use Stokes’ Theorem. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x+ y y + 2z z + 3x

∣∣∣∣∣∣= −2~i − 3~j − ~k ,

if S is the interior of the square, then∫

C

~F · d~r =

∫

S

curl ~F · d ~A =

∫

S

(−2~i − 3~j − ~k ) · d ~A

Since the area vector of S is 49~j , we have∫

C

~F · d~r =

∫

S

(−2~i − 3~j − ~k ) · d ~A = −3~j · 49~j = −147.


33. We use Stokes’ Theorem. Since

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x− y3 + z x3 + y + z x+ y + z3

∣∣∣∣∣∣= (3x2 + 3y2)~k ,

if S is the disk x2 + y2 ≤ 10 oriented upward,

∫

C

~F · d~r =

∫

S

(3x2 + 3y2)~k · d ~A = 3

∫ 2π

0

∫ √10

0

r2 · r dr dθ = 3 · 2π · r4

4

∣∣∣∣

√10

0

= 150π.

34. We use the Divergence Theorem. Since div ~F = 3y2 + 3z2, we have∫

S

~F · d ~A =

∫

Interiorof cylinder

(3y2 + 3z2) dV.

Converting to cylindrical coordinates with y = r cos θ, z = r sin θ, we have y2 + z2 = r2, so∫

S

~F · d ~A = 3

∫ 1

−1

∫ 2π

0

∫ 4

0

r2 · r dr dθ dx = 3 · 2 · 2π r4

4

∣∣∣∣4

0

= 768π.

35. We find div ~F = 1 + 1 + 1 = 3. To use the Divergence Theorem, we need to have a closed surface, so we add a circulardisk, D, oriented downward. With S representing the hemisphere and W the interior of the closed region, the DivergenceTheorem gives

∫

Closed surface

~F · d ~A =

∫

S

~F · d ~A +

∫

D

~F · d ~A =

∫

W

div ~F dV = 3 · Volume of region

= 3(

1

2· 4

3π53)

= 250π.

Since D is in the plane z = 0 and is oriented downward, d ~A = −~k dA on D, giving

Flux through disk =

∫

D

~F · d ~A =

∫

D

((x+ cos y)~i + (y + sinx)~j + 3~k

)· (−~k dA)

= −∫

D

3 dA = −3π(52) = −75π.

Thus, ∫

S

~F · d ~A = 250π −∫

D

~F · d ~A = 250π − (−75π) = 325π.

36. We see that div ~F = 0, since the~i component contains no xs, the ~j component contains no ys, and the ~k componentcontains no zs. Thus, if we close the surface by adding a disk, D, oriented downward, to the hemisphere, S, we can usethe Divergence Theorem: ∫

S

~F · d ~A +

∫

D

~F · d ~A =

∫

Interior0 dV = 0.

Since D is horizontal, d ~A = −~k dA on D, and we have

Flux =

∫

S

~F · d ~A = −∫

D

~F · d ~A = −∫

D

((tan z)~i + exz~j + (x2 + y2)~k

)· (−~k dA)

=

∫

D

(x2 + y2) dA.

To evaluate the integral, use polar coordinates. Since x2 + y2 = r2, we have

Flux =

∫

S

~F · d ~A =

∫

D

(x2 + y2) dA =

∫ 2π

0

∫ 5

0

r2 · r drd θ = 2πr4

4

∣∣∣5

0=

625π

2.


37. We use Stokes’ Theorem, withC as the circle bounding the surface, x2 +y2 = 5, oriented counterclockwise when viewedfrom above. If S is the hemisphere,

Flux =

∫

S

curl(−y~i + x~j ) · d ~A =

∫

C

(−y~i + x~j ) · d~r .

Since −y~i + x~j is tangent to the circle, pointing counterclockwise, and has magnitude ‖(−y)2 + x2‖ = 5 on C, wehave

Flux =

∫

C

(−y~i + x~j ) · d~r = || − y~i + x~j || · Length of curve = 5 · 2π(5) = 50π.

38. (a) curl ~F =

∣∣∣∣∣∣∣

~i ~j ~k∂

∂x

∂

∂y

∂

∂zy2z 2xyz xy

∣∣∣∣∣∣∣= (x− 2xy)~i − (y − y2)~j + (2yz − 2yz)~k 6= ~0 . Since curl ~F 6= 0, we know ~F

is not conservative.(b) A line integral round the closed path shown in Figure 20.23 is not zero, so ~F is not conservative.

Note: To show that a vector field is not conservative, we only need to find one path whose line integral is nonzero.To show that a vector field is conservative, we must show the line integral is zero on all paths.

x

y

Figure 20.23

39. The flux of ~E through a small sphere of radius R around the point marked P is negative, because all the arrows arepointing into the sphere. The divergence at P is

div ~E (P ) = limvol→0

( ∫S~E · d ~A

Volume of sphere

)= limR→0

(Negative number

43πR3

)≤ 0.

By a similar argument, the divergence at Q must be positive or zero.

40. (a) Let D be the face of the box which was removed from the yz-plane. We use Stokes’ Theorem with D as the surface.

curl ~F =

∣∣∣∣∣∣

~i ~j ~k∂∂x

∂∂y

∂∂z

x+ y −z y

∣∣∣∣∣∣= 2~i − 2~k ,

so ∫

C

~F · d~r =

∫

D

curl ~F · d ~A =

∫

D

(2~i − 2~k ) · d ~A = 2(Area of D) = 2 · 32 = 18.

(b) To use the Divergence Theorem, we calculate that div ~F = 1. Since the Divergence Theorem requires a closedsurface, we close S by adding D, oriented in the direction of the negative x-axis. Then

∫

S+D

~F · d ~A =

∫

S

~F · d ~A +

∫

D

~F · d ~A =

∫

Interior

1 dv = Volume of box = 27.


Now on D, where x = 0, we have ~F = y~i − z~j + y~k . In addition, on D, we have d ~A = −~i dy dz, so∫

D

~F · d ~A =

∫(y~i − z~j + y~k ) · (−i dy dz) = −

∫ 3

0

∫ 3

0

y dy dz

= −∫ 3

0

y2

2

∣∣∣∣∣

3

0

dz =−9

2

∫ 3

0

dz = −27

2.

Thus, ∫

S

~F · d ~A = 27−∫

D

~F · d ~A = 27−(−27

2

)=

81

2.

41. (a) We calculate

~r × ~c =

∣∣∣∣∣∣

~i ~j ~k

x y z

c1 c2 c3

∣∣∣∣∣∣= (c3y − c2z)~i − (c3x− c1z)~j + (c2x− c1y)~k .

Thusdiv(~r × ~c ) = 0.

(b) By the Divergence Theorem ∫

S

(~r × ~c ) · d ~A = 0.

42. (a) Let W1 be the ball inside S1. By the Divergence Theorem,∫

S1

~F · d ~A =

∫

W1

(x2 + y2 + z2 + 3) dV.

Using spherical coordinates, we have∫

S1

~F · d ~A =

∫

W1

(x2 + y2 + z2 + 3) dV =

∫ 2π

0

∫ π

0

∫ 1

0

(ρ2 + 3)ρ2 sinφ dρ dφ dθ

= 2π ·(ρ5

5+ ρ3

)∣∣∣∣1

0

(− cosφ)

∣∣∣∣π

0

= 2π · 6

5· 2 =

24

5π.

(b) If we were to calculate all three integrals using the Divergence Theorem, we would be integrating div ~F throughthe interior of each of these regions. Since S2 lies entirely inside S3, and S3 lies entirely inside S4, and sincediv ~F = x2 + y2 + z2 + 3 is positive everywhere,

∫

S2

~F · d ~A <

∫

S3

~F · d ~A <

∫

S4

~F · d ~A .

43. (a) Since ~F is radial, it is everywhere parallel to the area vector, ∆ ~A . Also, || ~F || = 1 on the surface of the spherex2 + y2 + z2 = 1, so

Flux through the sphere =

∫

S

~F · d ~A = lim||∆ ~A ||→0

∑~F ·∆ ~A

= lim||∆ ~A ||→0

∑||~F || ||∆ ~A || = lim

||∆ ~A ||→0

∑||∆ ~A ||

= Surface area of sphere = 4π · 12 = 4π.

(b) In Cartesian coordinates,

~F (x, y, z) =x

(x2 + y2 + z2)3/2~i +

y

(x2 + y2 + z2)3/2~j +

z

(x2 + y2 + z2)3/2~k .


So,

div ~F (x, y, z) =

(1

(x2 + y2 + z2)3/2− 3x2

(x2 + y2 + z2)5/2

)

+

(1

(x2 + y2 + z2)3/2− 3y2

(x2 + y2 + z2)5/2

)

+

(1

(x2 + y2 + z2)3/2− 3z2

(x2 + y2 + z2)5/2

)

=

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3x2

(x2 + y2 + z2)5/2

)

+

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3y2

(x2 + y2 + z2)5/2

)

+

(x2 + y2 + z2

(x2 + y2 + z2)5/2− 3z2

(x2 + y2 + z2)5/2

)

=3(x2 + y2 + z2)− 3(x2 + y2 + z2)

(x2 + y2 + z2)5/2

= 0.

(c) We cannot apply the Divergence Theorem to the whole region within the box, because the vector field ~F is notdefined at the origin. However, we can apply the Divergence Theorem to the region, W , between the sphere and thebox. Since div ~F = 0 there, the theorem tells us that

∫

Box(outward)

~F · d ~A +

∫

Sphere(inward)

~F · d ~A =

∫

W

div ~F dV = 0.

Therefore, the flux through the box and the sphere are equal if both are oriented outward:∫

Box(outward)

~F · d ~A = −∫

Sphere(inward)

~F · d ~A =

∫

Sphere(outward)

~F · d ~A = 4π.

44. We have:~F (x, y, z) =

x~i + y~j + z~k

(x2 + y2 + z2)3/2

Calculating the flux of ~F through the ellipsoid directly would be difficult. However, since div ~F = 0, we can replace theellipsoid by a sphere. Except at the origin, we have div ~F = 0. Let T be the surface of a sphere centered at the origininside the ellipsoid S, and letW be the region between S and T . Suppose both S and T are oriented away from the origin.By the Divergence Theorem, we have

Flux out of W =

∫

W

div ~F · dV = 0,

and therefore

Flux out of W = (Flux out − Flux in) =

∫

S

~F · d ~A −∫

T

~F · d ~A = 0.

Thus, we have∫

S

~F · d ~A =

∫

T

~F · d ~A = (Magnitude of ~F on sphere) · (Surface area)

= (1

radius2) · (4π · radius) = 4π.


45. Use the Divergence Theorem. Since

~F (x, y, z) =1

(x2 + y2 + z2)3/2(x~i + y~j + z~k ),

we have div ~F = 0, except at the origin.Let T be the surface of a sphere inside the cylinder S, and let W be the region between S and T . By the Divergence

Theorem,

Flux out of W =

∫

S+T

~F · d ~A =

∫

W

div ~F dV = 0.

Since S is oriented outward and T is oriented inward,

Net flux out of W = Flux out− Flux in =

∫

S

~F · d ~A −∫

T

~F · d ~A = 0.

so ∫

S

~F · d ~A =

∫

T

~F · d ~A = Magnitude of ~F on sphere · Surface area

=(

1

Radius2

)· (4π Radius2) = 4π.

46. (a) Since

~F = F1~i + F2

~j + F3~k =

x~i + y~j + z~k

(x2 + y2 + z2)a/2,

we have

∂F1

∂x=

1

(x2 + y2 + z2)a/2− a

2· x(2x)

(x2 + y2 + z2)(a/2)+1

=x2 + y2 + z2 − ax2

(x2 + y2 + z2)(a/2)+1.

Similarly, calculating ∂F2/∂y and ∂F3/∂z and adding gives

div ~F =∂F1

∂x+∂F2

∂y+∂F3

∂z=

3(x2 + y2 + z2)− ax2 − ay2 − az2

(x2 + y2 + z2)(a/2)+1

=3− a

(x2 + y2 + z2)a/2.

(b) div ~F = 0 if a = 3.(c) The vector field ~F is radial, and shows no “swirl”, so we expect curl ~F = ~0 . See Figure 20.24.

x y

z

Figure 20.24


47. Since div ~r = div(x~i + y~j + z~k ) = 3, applying the Divergence Theorem to the vector field ~F = ~r gives∫

S

~r · d ~A =

∫

V

3dV = 3

∫

V

dV = 3V.

Thus 13

∫S~r · d ~A = V.

48. We consider a sphere of radius R centered at the origin and compute the flux of ~r through its surface. Since ~r and d ~Aboth point radially outward, ~r · d ~A = ‖~r ‖‖d ~A ‖ = RdA on the surface of the sphere, so

∫

S

~F · d ~A =

∫

S

RdA = R

∫

S

dA = R(Surface area of a sphere) = R(4πR2) = 4πR3.

Therefore, volume of sphere = 13

∫S~F · d ~A = 4

3πR3.

49. Since ~r is parallel to the slanted edges of the cone, the flux of ~r through the surface is all through the base (See Fig-ure 20.25). On the base, z = h, and the normal is upward, so

∫

S

~r · d ~A =

∫

base(x~i + y~j + z~k ) · (dA~k )

=

∫

baseh dA

= h(Area of base)

= h(πb2)

Thus

V =1

3

∫

S

~r · d ~A =π

3b2h.

z

x

y

-� b6

?

h

Figure 20.25

50. (a) Can be computed. If W is the interior of the sphere, by the Divergence Theorem, we have∫

S

~F · d ~A =

∫

W


3π · 23 =

128π

3.

(b) Cannot be computed.(c) Can be computed. Use the fact that div(curl ~F ) = 0. If W is the inside of the sphere, then by the Divergence

Theorem, ∫

S

curl ~F · d ~A =

∫

W

div(curl ~F )dV =

∫

W

0 dV = 0


51. (a) The velocity vector for the traffic flow would look like:

(b) When 0 ≤ x < 2000, the velocity is decreasing linearly from 55 to 15, so its formula is (55 − x/50)~i mph. Then,when 2000 ≤ x < 7000, the speed is constant, so ~v (x) = 15~i mph. Next, when 7000 ≤ x < 8000, the velocityis increasing linearly from 15 to 55, so ~v (x) = (15 + (x− 7000)/25)~i mph. Finally, when x ≥ 8000, the speed isconstant, so ~v (x) = 55~i mph.

(c) div~v = dv(x)/dx.At x = 1000, v(x) = 55− x/50, so div~v = −1/50.At x = 5000, v(x) = 15, so div~v = 0.At x = 7500, v(x) = 15 + (x− 7000)/25, so div~v = 1/25.At x = 10, 000, v(x) = 55, so div~v = 0.In each case the units of div~v are miles/hour

feet .

52. (a) Usually, the distance between cars is more at higher speeds and less at lower speeds. The cars are traveling the fastestat x = 0, so at that point, the traffic should be the least dense. Thus,

ρ(0) < ρ(1000) < ρ(5000)

(b) Since ρ is in cars/mile, ~v is in miles/hour~v is in km/hour ρ~v is in cars/hour. The vector quantity ρ~v gives the numberof cars passing through a fixed point in a time interval.

(c) Pick any two points on the highway, x = a and x = b (a < b). We expect ρ~v to be the same at both places. Thisis because if more cars pass through a than b, that would mean cars are disappearing (or at least stopping, which weknow is not the case since the velocity field is not 0) between a and b. On the other hand, if more cars pass through bthan a, that would mean cars are being created between a and b. So we expect ρ~v to be the same at a and b. Since aand b were chosen arbitrarily, we can say that ρ~v is constant at all x. This means div(ρ~v ) = 0.

(d) At x = 0, ~v (0) = 55~i and ρ(0) = 75. We have ρ~v (0) = 4125~i = constant. So ρ(x) = ‖ρ~v ‖/‖~v (x)‖ = 4125/v.

ρ(x) =4125

55− x50

if 0 ≤ x < 2000

ρ(x) =4125

15= 275 if 2000 ≤ x < 7000

ρ(x) =4125

15 + x−700025

if 7000 ≤ x < 8000

ρ(x) =4125

55= 75 if x ≥ 8000

(e) We have ρ(0) = 75, ρ(1000) = 118, ρ(5000) = 275, where ρ is given in cars/mile. At x = 0, there are 75 cars ina 1-mile stretch of highway. Since there are two lanes, there are about 38 cars in a mile in one lane. A mile is 5280feet, so that says on average, one car occupies 139 feet. So at x = 0, the distance between two cars is 139 feet.

Similarly, we find that at x = 1000, the distance is 89 feet, and at x = 5000, the distance is 38 feet.

53. (a) Since ~v = gradφ we have

~v =

(1 +

y2 − x2

(x2 + y2)2

)~i +

−2xy

(x2 + y2)2~j

(b) Differentiating the components of ~v , we have

div~v =∂

∂x(1 +

y2 − x2

(x2 + y2)2) +

∂

∂y(−2xy

(x2 + y2)2) =

2x(x2 − 3y2)

(x2 + y2)3+

2x(3y2 − x2)

(x2 + y2)3= 0


(c) The vector ~v is tangent to the circle x2+y2 = 1, if and only if the dot product of the field on the circle with any radiusvector of that circle is zero. Let (x, y) be a point on the circle. We want to show: ~v · ~r = ~v (x, y) · (x~i + y~j ) = 0.We have:

~v (x, y) · (x~i + y~j ) = ((1 +y2 − x2

(x2 + y2)2)~i +

−2xy

(x2 + y2)2~j ) · (x~i + y~j )

= x+ xy2 − x2

(x2 + y2)2− 2xy2

(x2 + y2)2

=x(x2 + y2 − 1)

x2 + y2,

but we know that for any point on the circle, x2 + y2 = 1, thus we have ~v · ~r = 0. Therefore, the velocity field istangent to the circle. Consequently, there is no flow through the circle and any water on the outside of the circle mustflow around it.

(d)

Figure 20.26

54. Set up the wire as the z-axis with the current flowing upward, and let ~B be the magnetic field due to the current in thewire. Then by the right-hand rule, in the xy-plane ~B points counterclockwise around a circle centered around the z-axis.(See Figure 20.27.) Let us take the circulation on the boundary of a ring of outer radius R and inner radius r around thez-axis. We choose the upward orientation for S. (This is arbitrary.) The boundary of this region, C, is a curve in twopieces, CP and CQ. Given the orientation of S, we must have CP oriented clockwise and CQ oriented counterclockwisewhen viewed from above. (See Figure 20.27.) The boundary of S is CP + CQ. From Stokes’ Theorem,

∫

S

curl ~B · d ~A =

∫

C

~B · d~r

x

y

Q

P

S

CP

CQ

r R

*

*

I ~B

Figure 20.27: Current along positive z-axis


Thus, since curl ~B = ~0 , we have

0 =

∫

S

curl ~B · d ~A =

∫

CQ

~B · d~r +

∫

CP

~B · d~r

so ∫

CQ

~B · d~r = −∫

CP

~B · d~r

Since the vector field has constant magnitude on each circle and is parallel to the circle∫

CQ

~B · d~r = ‖B(Q)‖ · Length of CQ = 2πR‖B(Q)‖

and because CP is oriented in the opposite direction∫

CP

~B · d~r = −‖B(P )‖ · Length of CP = −2πr‖B(P )‖

Thus,2πR‖B(Q)‖ = −(−2πr‖B(P )‖)

which simplifies to:‖ ~B (Q)‖‖ ~B (P )‖

=R

r=

1/r

1/R.

This relationship shows that ‖ ~B ‖ varies inversely with the radial distance, r.

55. (a) The distance from any point to the origin is given by√x2 + y2 + z2, so the denominator is simply r3. The field can

then be rewritten in components as ( kxr3~i + ky

r3~j + kz

r3~k ). Its magnitude is thus:

‖~v ‖ =

√K2

x2 + y2 + z2

r6=

√K2

r4=K

r2

which is only a function of the distance from the origin. It is clear that the vector field points away from the origin forall points (x, y, z), because it is the radius vector x~i + y~j + z~k , multiplied by the positive scalar K/r3. Suppose(x, y, z) 6= (0, 0, 0), then

div~v (x, y, z) =∂

∂x

(Kx

r3

)+

∂

∂y

(Ky

r3

)+∂

∂z

(Kz

r3

)

= K

(−2x2 + y2 + z2

r5+x2 − 2y2 + z2

r5+x2 + y2 − 2z2

r5

)= 0.

Hence, indeed ~v is a point source at the origin.(b) The dependence of ~v on r, the distance from the origin, is shown in part (a).(c) The flux through a sphere centered at the origin is calculated as:

Flux =

∫

S

~v · d ~A

Since the vector field’s magnitude is a function only of the distance from the origin, it will be constant over thesurface of a sphere centered at the origin. Furthermore, since it is pointed away from the origin, ~v · d ~A will besimply ‖~v ‖ · ‖d ~A ‖. Thus

Total flux out of a sphere of radius r = ‖~v r‖ · ‖ ~A sphere‖ =K

r2

4

3πr2 =

4

3πK.

So the flux does not even depend upon r since the rate at which the area of the sphere is increasing is exactly equalto the rate at which the magnitude of the field is decreasing.

(d) This is best handled by observing (as in part (a)) that the divergence of the vector field at any point besides the originis zero. Since the divergence anywhere but the origin is zero, the net flux through any closed surface not enclosingthe origin must also be zero.


56. Calculating the gradient gives us

grad ~φ =

2x~i + 2y~j + 2z~k for x2 + y2 + z2 ≤ b2

4b3x

4(x2 + y2 + z2)3/2~i +

b3y

4(x2 + y2 + z2)3/2~j +

b3z

4(x2 + y2 + z2)3/2~k for

b2

4≤ x2 + y2 + z2

The electric field, ~E , is the negative of gradφ. Both vector fields are radial. The magnitude of the vector field in the regionx2 + y2 + z2 ≤ b2/4 is proportional to ‖~r ‖ while the magnitude of the vector field in the region b2/4 ≤ x2 + y2 + z2

is proportional to 1/‖~r ‖2, or the inverse square. Thus, for x2 + y2 + z2 ≤ b2/4,

Charge distribution = ρ =1

4πdiv ~E = −2 + 2 + 2

4π= − 3

2π

For b2/4 ≤ x2 + y2 + z2,

Charge distribution = ρ =1

4πdiv ~E

= − b3

4(x2 + y2 + z2)3/2+b3

4

3

2

2x · x(x2 + y2 + z2)5/2

− b3

4(x2 + y2 + z2)3/2+b3

4

3

2

2y · y(x2 + y2 + z2)5/2

− b3

4(x2 + y2 + z2)3/2+b3

4

3

2

2z · z(x2 + y2 + z2)5/2

= − b3

4

x2 + y2 + z2 − 3x2 + x2 + y2 + z2 − 3y2 + x2 + y2 + z2 − 3z2

(x2 + y2z2)5/2

= 0

The charge density is constant: −3/(2π) for x2 + y2 + z2 ≤ b2/4 and zero further away.

57. (a) The path along which we integrate is shown in Figure 20.28.

(a, 0)C1

(a, b)

C2

x

y

Figure 20.28

The path C1 is given by

C1 :

{x = t 0 ≤ t ≤ ay = 0

and path C2 is given by

C2 :

{x = a

y = t 0 ≤ t ≤ b .

We integrate along the path C = C1 + C2. Then,∫

C

~E · d~r =

∫

C1

~E · d~r +

∫

C2

~E · d~r

=

∫ a

0

5t2~j ·~i dt+

∫ b

0

(10at~i + (5a2 − 5t2)~j ) ·~j dt

= 5a2b− 5

3b3


(b) The path along which we integrate is shown in Figure 20.29.

(0, b)

C1

(a, b)C2

y

x

Figure 20.29

The path C1 is given by

C1 :

{x = 0

y = t 0 ≤ t ≤ band Path C2 is given by

C2 :

{x = t 0 ≤ t ≤ ay = b

.

We integrate along the path C = C1 + C2. Then,∫

C

~E · d~r =

∫

C1

~E · d~r +

∫

C2

~E · d~r

=

∫ b

0

−5t2~j ·~j dt+

∫ a

0

(10bt~i + (5t2 − 5b2)~j ) ·~i dt

= −5

3b3 + 5a2b

(c) Notice that the line integrals along both paths in part (a) and (b) are equal. Thus ~E could be path independent. Thisproperty is a property of electric fields.

(d) Using the calculations of the parts (a) and (b), with the point (a, b) replaced by (x, y), we see that the electricpotential, φ, at (x, y) is given by

φ =5

3y3 − 5x2y.

Then, taking the gradient gives

gradφ = φx~i + φy~j

= −10xy~i + (5

3(3y2)− 5x2)~j

= −10xy~i + (5y2 − 5x2)~j

= − ~E .

Thus, we have confirmed that ~E = − gradφ.

58. (a) Differentiating div ~E = 4πρ with respect to time gives

∂

∂t

(div ~E

)=

∂

∂t

(∂E1

∂x+∂E2

∂y+∂E3

∂z

)= 4π

∂ρ

∂t.

Since, for example, ∂∂t

(∂E1∂x

)= ∂

∂x

(∂E1∂t

), we can rewrite this as

∂

∂x

(∂E1

∂t

)+

∂

∂y

(∂E2

∂t

)+

∂

∂z

(∂E3

∂t

)= 4π

∂ρ

∂t.

So we have shown that

div

(∂ ~E

∂t

)= 4π

∂ρ

∂t.


Now consider the equation

curl ~B − 1

c

∂ ~E

∂t=

4π

c~J

and take the divergence of both sides:

div curl ~B − 1

cdiv

(∂ ~E

∂t

)=

4π

cdiv ~J .

Since div curl ~B = 0, by Problem 25 on page 1366, we have

−1

cdiv

(∂ ~E

∂t

)=

4π

cdiv ~J .

Thus−1

c

(4π∂ρ

∂t

)=

4π

cdiv ~J ,

so−∂ρ∂t

= div ~J ,

or∂ρ

∂t+ div ~J = 0.

(b) The equation derived in part (a) says that the rate of change with time of the charge density at a point is the negativeof the divergence of the current density at that point.

Why is this reasonable? Suppose div ~J < 0 at some point, that is, there is a current “sink” there. This means thatcurrent is “piling up” at this point – in other words, the charge is “piling up” there. Thus we would expect ∂ρ/∂t > 0there. Similarly, if div ~J > 0, at some point, there is a current source there. This means that current is being “created”near that point, which means that charge density is decreasing there. Thus we would expect ∂ρ/∂t < 0 there.

(c) The equation is called the charge conservation equation because it reflects the fact that charge is neither created nordestroyed. If div ~J is negative at some point, there is a net influx of current into a small surface around the point,so the charge density must be increasing there. If div ~J is positive, there is a net outflow of current through a smallsurface around the point, so the charge density must be decreasing there.

59. (a) Cross-section in the xy-plane of the vector field with K = 1 is in Figure 20.30. Cross-section in the xy-plane of thevector field with K = −1 is in Figure 20.31.

x

y

Figure 20.30

x

y

Figure 20.31

(b) ‖~v ‖ = |K| · ‖ − y~i + x~j ‖/(x2 + y2) = |K| · r/r2 = |K|/r.(c) For (x, y, z) 6= (0, 0, z),

div~v =∂

∂x

(−Kyx2 + y2

)+

∂

∂y

(Kx

x2 + y2

)= K

(−y)(−2x)

(x2 + y2)2+K

x(−2y)

(x2 + y2)2= 0


(d) In order for curl~v to make sense, ~v needs to have 3 components. Write

~v =K

x2 + y2(−y~i + x~j + 0~k ),

Hence, for (x, y, z) 6= (0, 0, z),

curl~v =

(∂

∂x

(Kx

x2 + y2

)− ∂

∂y

(Ky

x2 + y2

))~k

=

(K

y2 − x2

(x2 + y2)2+K

x2 − y2

(x2 + y2)2

)~k

= ~0 .

(e) On the circle of radius R, the magnitude of ~v is constant; ‖~v ‖ = |K|/R. If K > 0, ~v and ∆~r are parallel and inthe same direction so that ∫

C

~v · d~r =K

R· (Length of circle) =

K

R· 2πR = 2πK.

If K < 0, |K| = −K, and ~v and ∆~r are in opposite directions so that∫

C

~v · d~r =−|K|R· (Length of circle) =

K

R· 2πR = 2πK.

(f) Since ~v (and hence curl~v ) is undefined on the z-axis, Stokes’ Theorem does not apply on the circle (or on any curveencircling the z-axis).

CAS Challenge Problems

60. (a) curl ~F (1, 2, 1) = −6~i − 8~j + ~k .(b)∫Ca

~F · d~r = − 32(4a2 + a4)π.

lima→0

∫Ca

~F · d~rπa2

= −6.

This is the~i component of curl ~F (1, 2, 1) according to the circulation density formula.(c)∫Da

~F · d~r = −8a2π.

lima→0

∫Ca

~F · d~rπa2

= −8.

This is the ~j component of curl ~F (1, 2, 1).(d)∫Ea

~F · d~r = − 14(−4a2 + 3a4)π.

lima→0

∫Ca

~F · d~rπa2

= 1.

This is the ~k component of curl ~F (1, 2, 1).(e) The curvesCa are circles of radius a around the point (1, 2, 1) facing in the positive x-direction, so the limit as a→ 0

of the circulation around them is the ~i -component of the curl, according to the geometric definition. Similarly, thecurvesDa are circles facing in the y-direction and the curves Ea are circles facing in the z-direction, so they give the~j and ~k components of the curl.

61. (a) LetW be the region enclosed by the sphere. We have div ~F = 2ax+bz+2cy+p+q so by the Divergence Theorem∫S~F ·d ~A =

∫W

(2ax+bz+2cy+p+q) dV . Now∫Wx dV =

∫Wy dV =

∫Wz dV = 0, becauseW is symmetric

about the origin and x, y, and z are odd functions. So∫W

(2ax+bz+2cy+p+q) dV =∫B

(p+q)dV = 4(p+q)πR3

3.

(b) Using spherical coordinates, we calculate the flux integral directly as∫ 2π

0

∫ π

0

((bR2 cos(θ) cos(φ) sin(φ) + aR2cos(θ)2sin(φ)2)~i

+(pR sin(θ) sin(φ) + cR2sin(θ)2sin(φ)2)~j + (qR cos(φ)

+rR3cos(θ)3sin(φ)3)~k ) · (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )R2 sinφ dφdθ =4(p+ q)πR3

3.

Rather than entering this integral directly into your CAS, it is better to define the vector field and parameterizationseparately and enter the formula for flux integral through a sphere.

CHECK YOUR UNDERSTANDING 1407

62. (a) div ~F (1, 1, 1) = 5.(b) On a small sphere centered at (1, 1, 1) the divergence is approximately constant and equal to its value at (1, 1, 1),

namely 5. Geometrically the divergence is the flux density, so the total flux is approximately the divergence multipliedby the volume, so

∫S~F · d ~A ≈ 5( volume of S) = 20

3π(0.1)3 ≈ 0.02094...

(c) The area element on the sphere is d ~A = (sinφ cos θ~i + sinφ sin θ~j + cosφ~k )a2 sinφ dφdθ. Using a CAS tocalculate the flux integral we get

∫S~F · d ~A = (20πa3/3) + (4πa5/5), which is .02097 when a = .1. This is close

to the approximation we found in part (b). The limit

lima→0

∫Sa

~F · d ~AVolume inside Sa

= lima→0

(20πa3/3) + (4πa5/5)

(4/3)πa3= lima→0

(5 +3a2

5) = 5.

This agrees with the value of the divergence we found in part (a). This makes sense, because the limit is just thegeometric definition of divergence.

CHECK YOUR UNDERSTANDING

1. True. By Stokes’ Theorem, the circulation of ~F around C is the flux of curl ~F through the flat disc S in the xy-planeenclosed by the circle. An area element for S is d ~A = ±~k dA, where the sign depends on the orientation of the circle.Since curl ~F is perpendicular to the z-axis, curl ~F · d ~A = ±(curl ~F · ~k )dA = 0, so the flux of curl ~F through S iszero, hence the circulation of ~F around C is zero.

2. True.

div(~F + ~G ) =∂(F1 +G1)

∂x+∂(F2 +G2)

∂y+∂(F3 +G3)

∂z

=∂F1

∂x+∂F2

∂y+∂F3

∂z+∂G1

∂x+∂G2

∂y+∂G3

∂z

= div ~F + div ~G .

3. False. Let’s compare the x–components of each side of the equation. The x–component of grad( ~F · ~G ) is given by

(grad( ~F · ~G ))1 =∂(F1G1 + F2G2 + F3G3)

∂x

=∂F1

∂xG1 + F1

∂G1

∂x+∂F2

∂xG2 + F2

∂G2

∂x+∂F3

∂xG3 + F3

∂G3

∂x.

However, the x–component of ~F · (div ~G ) + (div ~F ) · ~G is

(~F (div ~G ) + (div ~F ) ~G )1 = F1(div ~G ) + (div ~F )G1

= F1

(∂G1

∂x+∂G2

∂y+∂G3

∂z

)+

(∂F1

∂x+∂F2

∂y+∂F3

∂z

)G1.

These two x–components are different and therefore

grad( ~F · ~G ) 6= ~F (div ~G ) + (div ~F ) ~G .

4. True. We calculate the x–components for each side of the equation:

(curl(f ~G ))1 =∂(fG3)

∂y− ∂(fG2)

∂z

=∂f

∂yG3 + f

∂G3

∂y− ∂f

∂zG2 − f ∂G2

∂z

= (∂f

∂yG3 − ∂f

∂zG2) + f(

∂G3

∂y− ∂G2

∂z)

= ((grad f)× ~G )1 + (f(curl ~G ))1.

Computations for the other two components are similar, so

curl(f ~G ) = (grad f)× ~G + f · (curl ~G ).


5. True. div ~F is a scalar whose value depends on the point at which it is calculated.

6. False. You need a closed surface to use the divergence theorem.

7. True. ~F is rotating around the y axis, so by the right hand rule curl ~F has a positive y component. Therefore taking thedot product of curl ~F and ~j will give a positive number.

8. False. As a counterexample, consider ~F =~i and f(x, y, z) = x. Then div(f ~F ) = divx~i = 1, and fdiv ~F = x ·0 = 0.

9. False. As a counterexample, consider ~F = 2x~i+2y~j +2z~k . Then ~F = grad(x2+y2+z2), and div ~F = 2+2+2 6= 0.

10. False. As a counterexample, consider ~F = x2~i Then div ~F = 2x, and grad 2x = 2~i 6= ~0 .

11. False. Since ~F can be written ~F (x, y, z) = x~i + y~j + z~k , the divergence of ~F is 3.

12. True. Here is a way of constructing a vector field ~F . The idea is to think of f as a function of x (with y and z constant)and take the antiderivative. We define

g(x, y, z) =

∫ x

0

f(t, y, z) dt.

By the Fundamental Theorem of one-variable calculus, we know∂g

∂x= f. So, if f is given, the vector field ~F =

g(x, y, z)~i has div ~F = f.

13. False. As a counterexample, note that ~F =~i and ~G = ~j both have divergence zero, but are not the same vector fields.

14. False. Neither side of this equation makes sense: div ~F is a scalar, and we cannot take the flux integral of a scalar. On theother side, ~F is a vector field, and we cannot take the triple integral of a vector field.

15. True. By the Divergence theorem,∫S~F · d ~A = −

∫W

div ~F dV , where W is the solid interior of S and the negativesign is due to the inward orientation of S. Since div ~F = 0, we have

∫S~F · d ~A = 0.

16. True. By the Divergence theorem,∫S~F · d ~A =

∫W

div ~F dV , where W is the solid interior of S. Since div ~F = 1,we have

∫S~F · d ~A =

∫W

1 dV which is equal to the volume enclosed by S.

17. True. The Divergence theorem says that∫W

div ~F dV =∫S~F · d ~A , where S is the outward oriented boundary of W .

In this case, the boundary of W consists of the surfaces S1 and S2. To give this boundary surface a consistent outwardorientation, we use a normal vector on S1 that points toward the origin, and a normal on S2 that points away from theorigin. Thus

∫W

div ~F dV =∫S2

~F · d ~A +∫S1

~F · d ~A , with S2 oriented outward and S1 oriented inward. Reversing

the orientation on S1 so that both spheres are oriented outward yields∫W

div ~F dV =∫S2

~F · d ~A −∫S1

~F · d ~A .18. False. The boundary of the cube W consists of six squares, so the Divergence theorem requires adding the flux integrals

over the four remaining sides.

19. True. The boundary of the cube W consists of six squares, but four of them are parallel to the xz or yz-planes and socontribute zero flux for this particular vector field. The only two surfaces of the boundary with nonzero flux are S1 andS2, which are parallel to the xy-plane.

20. False. The Divergence theorem tells us in this case that∫S~F · d ~A =

∫W

div ~F dV = 0, where W is the solid ball withboundary S. But this does not necessarily mean that div ~F = 0 at all points in W . The divergence of ~F can be positiveat some points in W and negative at other points in W , yielding a triple integral of zero. For example, let ~F = x2~i .The flux of this vector field through the sphere is 0. (The flux out for x > 0 cancels the flux in for x < 0.) However,div ~F = 2x, which is not 0.

21. False. The left-hand side of the equation, div(grad f), is a scalar function and the right hand side, grad(divF ), is avector. There cannot be an equality between a scalar and a vector.

22. True. Let D be the disk that forms the bottom of the cylinder, x2 + y2 ≤ 1, z = 0, oriented downward. Then the surfaceconsisting of Sh andD is closed and oriented outward, so the Divergence Theorem says

∫Sh+D

~F ·d ~A =∫W

div ~F dV ,

where W is the solid interior of the cylinder. Since div ~F = 0, we have∫Sh+D

~F · d ~A = 0. Writing the flux integral

as the sum of integrals over Sh and D gives∫Sh

~F · d ~A +∫D~F · d ~A = 0, so

∫Sh

~F · d ~A = −∫D~F · d ~A . The flux

integral∫D~F · d ~A does not depend on the height h, so

∫Sh

~F · d ~A is independent of h.

23. True. The circulation density is obtained by dividing the circulation around a circle C (a scalar) by the area enclosed byC (also a scalar), in the limit as the area tends to zero.

24. False. As a counterexample, any constant vector field ~F = a~i + b~j + c~k has div ~F = 0 and curl ~F = ~0 .

25. True. Writing ~F = F1~i +F2

~j +F3~k and ~G = G1

~i +G2~j +G3

~k , we have ~F + ~G = (F1 +G1)~i + (F2 +G2)~j +

CHECK YOUR UNDERSTANDING 1409

(F3 +G3)~k . Then the~i component of curl( ~F + ~G ) is

∂(F3 +G3)

∂y− ∂(F2 +G2)

∂z=∂F3

∂y− ∂F2

∂z+∂G3

∂y− ∂G2

∂z

which is the ~i component of curl ~F plus the ~i component of curl ~G . The ~j and ~k components work out in a similarmanner.

26. False. The left-hand side of the equation does not make sense. The quantity ( ~F · ~G ) is a scalar, so we cannot computethe curl of it.

27. True. For example, take ~F = y~k .

28. True. The curl is curl ~F = 0~i + 0~j + ( ∂∂x

( xx2+y2 )− ∂

∂y( −yx2+y2 ))~k = ( y2−x2

(x2+y2)2+ x2−y2

(x2+y2)2)~k = ~0 . The divergence

is divF = 2xy(x2+y2)2

+ −2xy(x2+y2)2

= 0.

29. False. For example, take ~F = z~i +x~j . Then curl ~F = ~j +~k , which is not perpendicular to ~F , since (z~i +x~j ) · (~j +~k ) = x 6= 0.

30. False. For example, take ~F = z~i and ~G = x~j . Then ~F × ~G = xz~k , and curl( ~F × ~G ) = −z~j . However,(curl~F )× (curl ~G ) = ~j × ~k =~i .

31. False. To see why, write ~F = F1~i + F2

~j + F3~k . Then curl ~F = x~i gives

∂F3

∂y− ∂F2

∂z= x;

∂F1

∂z− ∂F3

∂x= 0;

∂F2

∂x− ∂F1

∂y= 0.

Now take the partial ∂∂x

of the first equation, ∂∂y

of the second and ∂∂z

of the third. This gives

∂2F3

∂x∂y− ∂2F2

∂x∂z= 1;

∂2F1

∂y∂z− ∂2F3

∂y∂x= 0;

∂2F2

∂z∂x− ∂2F1

∂z∂y= 0.

Assuming the continuity of the second order partials, the equality of mixed partials in the second and third equationsshows that ∂

2F3∂x∂y

= ∂2F2∂x∂z

, which contradicts the first equation. Thus there cannot be a vector field ~F with curl ~F = x~i .

32. True. On a small patch of S that includes the boundary circle, the positive normal is outward. Letting the thumb of theright hand point in this direction makes the fingers curl in the counterclockwise direction.

33. False. The curl needs to be in the flux integral, not the line integral, for a correct statement of Stokes’ theorem:∫C~F ·d~r =∫

Scurl ~F · d ~A .

34. True. By Stokes’ theorem, both flux integrals are equal to the line integral∫C~F · d~r , where C is the circle x2 + y2 = 1,

oriented counterclockwise when viewed from the positive z-axis.

35. True. By Stokes’ theorem, the flux integral is equal to the line integral∫C~F · d~r , where C is the boundary of the closed

sphere. Since the sphere has no boundary curve, the line integral is zero. Alternatively, the closed sphere can be dividedinto two hemispheres S1 (the top half with upward orientation) and S2 (the bottom half with downward orientation.) ThenS1 and S2 both have the circle C (x2 + y2 = 1, z = 0) as their common boundary, except that for S1, C is orientedcounterclockwise when viewed from above, and as the boundary of S2, C is oriented clockwise. Then, using Stokes’theorem on S1 and S2 gives

∫S

curl~F · d ~A =∫S1

curl~F · d ~A +∫S2

curl~F · d ~A =∫C~F · d~r −

∫C~F · d~r = 0.

The same result can be obtained using the Divergence theorem and the fact that div curl ~F = 0.

36. True. Let D be the interior disk of C, oriented by the right hand rule. By Stokes’ theorem,∫C~F · d~r =

∫D

curl~F · d ~A ,and∫C~G · d~r =

∫D

curl ~G · d ~A . Since curl ~F = curl ~G , we have∫C~F · d~r =

∫C~G · d~r .

37. True. Let S be the rectangular region inside C, oriented by the right hand rule. By Stokes’ theorem,∫C~F · d~r =∫

Scurl~F · d ~A = 0.

38. True. By Stokes’ theorem,∫S

curl ~F · d ~A =∫C~F · d~r , where C is one or more closed curves that form the boundary

of S. Since ~F is a gradient field, its line integral over any closed curve will be zero.

39. False. Using the right-hand rule gives C1 oriented clockwise and C2 oriented counterclockwise when viewed from thepositive y-axis.

40. False. The condition that∫C~F · d~r = 0 implies, by Stokes’ theorem, that

∫S

curl ~F · d ~A = 0. However, curl ~F neednot be ~0 for this to occur. For example, let ~F = y~i , and let S be the upper unit hemisphere x2 + y2 + z2 = 1, z ≥ 0oriented upward. Then C is the circle x2 + y2 = 1, z = 0 oriented counterclockwise when viewed from above. The lineintegral

∫Cy~i · d~r = 0, but curl ~F = −~k .


41. False. The condition that∫S

curl~F ·d ~A = 0 implies, by Stokes’ theorem, that∫C~F ·d~r = 0. However, ~F need not be a

gradient field for this to occur. For example, let ~F = x~k , and let S be the upper unit hemisphere x2 +y2 +z2 = 1, z ≥ 0oriented upward. Then C is the circle x2 + y2 = 1, z = 0 oriented counterclockwise when viewed from above. The lineintegral

∫Cx~k · d~r = 0, since the field ~F is everywhere perpendicular to C. The curl of ~F is the constant field −~j ,

so ~F is not a gradient field. Yet we have∫S−~j · d ~A = 0, since the constant field −~j flows in, and then out of the

hemisphere S.

PROJECTS FOR CHAPTER TWENTY

1. (a) Since ~e ρ is a unit vector pointing radially away from the origin, ~e ρ = ~r /‖~r ‖ = ~r /ρ. Thus, we have

~F =f(ρ)

ρ~r =

f(ρ)

ρx~i +

f(ρ)

ρy~j +

f(ρ)

ρz~k .

Let g(ρ) = f(ρ)/ρ. Since ρ = ‖~r ‖ =√x2 + y2 + z2, and ∂ρ/∂x = 1

2 (x2 + y2 + z2)−1/2 · 2x =

x/√x2 + y2 + z2, using the chain rule, we have

∂

∂xg(ρ) =

d

dρg(ρ) · ∂ρ

∂x=

x√x2 + y2 + z2

g′(ρ)

∂

∂yg(ρ) =

y√x2 + y2 + z2

g′(ρ)

∂

∂zg(ρ) =

z√x2 + y2 + z2

g′(ρ).

So

div ~F =∂

∂x(g(ρ)x) +

∂

∂y(g(ρ)y) +

∂

∂z(g(ρ)z)

=

(x∂

∂xg(ρ) + g(ρ)

)+

(y∂

∂yg(ρ) + g(ρ)

)+

(z∂

∂zg(ρ) + g(ρ)

)

=x2

√x2 + y2 + z2

g′(ρ) +y2

√x2 + y2 + z2

g′(ρ) +z2

√x2 + y2 + z2

g′(ρ) + 3g(ρ)

=x2 + y2 + z2

√x2 + y2 + z2

g′(ρ) + 3g(ρ) =ρ2

ρg′(ρ) + 3g(ρ) = ρg′(ρ) + 3g(ρ)

= ρ

(f ′(ρ)

ρ− f(ρ)

ρ2

)+

3f(ρ)

ρ= f ′(ρ) + 2

f(ρ)

ρ.

On the other hand,

1

ρ2

d

dρ(ρ2f(ρ)) =

1

ρ2(2ρf(ρ) + ρ2f ′(ρ)) = 2

f(ρ)

ρ+ f ′(ρ).

Thus,

div ~F =1

ρ2(ρ2f(p)).

(b) Applying the formula from part (a), we have

div ~F =1

ρ2

d

dρ(ρ2f(ρ)) = 0.

Since ρ 6= 0, multiplying by ρ2 gives us (d/dρ)(ρ2f(ρ)) = 0, hence ρ2f(ρ) is a constant, say k. Sof(ρ) = k/ρ2, and hence

~F =k

ρ2~e ρ.

PROJECTS FOR CHAPTER TWENTY 1411

(c) Suppose the sphere, S, has radius R. Since ~F has constant magnitude on S and points perpendicularly toS, we have ~F · d ~A = f(R) dA, so the flux of ~F out of S is

∫S~F · d ~A =

∫Sf(R) dA = f(R)

∫SdA =

f(R) · Area of S = 4πR2f(R). On the other hand, using spherical coordinates, part (a) tells us that:

div ~F =1

ρ2

d

dρ

(ρ2f(ρ)

),

so if W is the region enclosed by S then∫

W

div ~F dV =

∫ R

0

∫ π

0

∫ 2π

0

1

ρ2

d

dρ(ρ2f(ρ)) ρ2 sinφdθ dφ dρ

= 4π

∫ R

0

d

dρ

(ρ2f(ρ)

)dρ = 4πρ2f(ρ)

∣∣∣∣R

0

= 4πR2f(R).

The fact that∫W

div ~F dV equals the flux integral,∫S~F · d ~A , confirms the Divergence Theorem.

(d) Since we are assuming ~E is spherically symmetric, we can write

~E = E(ρ)~e ρ.

By part (a) and Gauss’s Law,

div ~E =1

ρ2

d

dρ(ρ2E(ρ)) =

{δ0 ρ ≤ a0 ρ > a

Suppose ρ ≤ a. Then, since ρ 6= 0,d

dρ(ρ2E(ρ)) = δ0ρ

2.

Therefore

ρ2E(ρ) =

∫δ0ρ

2dρ =δ0ρ

3

3+ C,

and soE(ρ) =

δ0ρ

3+C

ρ2.

Since ~E , and hence E, is assumed to be continuous, we must have C = 0, so E(ρ) = δ0ρ/3.Now suppose that ρ > a. Then

d

dρ(ρ2E(ρ)) = 0,

thereforeρ2E(ρ) = k,

and soE(ρ) =

k

ρ2.

Since E is assumed to be continuous,

limρ→a+

E(ρ) = limρ→a−

E(ρ).

Now,

limρ→a+

E(ρ) =k

a2

limρ→a−

E(ρ) =δ0a

3


so k = δ0a3/3. Thus

E(ρ) =

δ0ρ

3ρ ≤ a

δ0a3

3ρ > a

2. (a) Since

~e r =x~i + y~j

r,

we have~F =

f(r)

rx~i +

f(r)

ry~j .

Let g(r) = f(r)/r. Since r =√x2 + y2, we have

∂

∂xg(r) =

x√x2 + y2

g′(r)

∂

∂yg(r) =

y√x2 + y2

g′(r)

So

div ~F =∂

∂x(g(r)x) +

∂

∂y(g(r)y)

= (x∂

∂xg(r) + g(r)) + (y

∂

∂yg(r) + g(r))

=x2

√x2 + y2

g′(r) +y2

√x2 + y2

g′(r) + 2g(r)

=x2 + y2

√x2 + y2

g′(r) + 2g(r) =r2

rg′(r) + 2g(r) = rg′(r) + 2g(r)

= r(f ′(r)r− f(r)

r2) +

2f(r)

r= f ′(r) +

f(r)

r.

On the other hand,1

r

d

dr(rf(r)) =

1

r(f(r) + rf ′(r)) =

f(r)

r+ f ′(r).

(b) Applying the formula from part (a), we have

div ~F =1

r

d

dr(rf(r)) = 0.

Multiplying both sides by r we have (d/dr)(rf(r)) = 0, hence rf(r) is a constant, k. So f(r) = k/r,and

~F =k

r~e r.

(c) Since ~F is perpendicular to the sides of S, the flux out of the top and bottom of the cylinder is 0. On theside of the cylinder, ~F · d ~A = f(R) dA, so

Flux of ~F out of S =

∫

S

f(R) dA = f(R) · Area of side = 2πRhf(R).

By part (a), in cylindrical coordinates,

div ~F =1

r

d

dr(rf(r)) ,

PROJECTS FOR CHAPTER TWENTY 1413

so if W is the region enclosed by S then

∫

W

div ~F dV =

∫ R

0

∫ h

0

∫ 2π

0

1

r2

d

dr

(r2f(r)

)r dθ dz dr

= 2πh

∫ R

0

d

dr(rf(r)) dr = 2πh rf(r)|R0 = 2πhRf(R) = 2πRhf(R).

Since this agrees with the expression for the flux integral, it confirms the Divergence Theorem.(d) Since ~E is cylindrically symmetric,

~E = E(r)~e r.

By part (a) and Gauss’s Law,

div ~E =1

r

d

dr(rE(r)) =

{δ0 r ≤ a0 r > a

Suppose r ≤ a. Thend

dr(rE(r)) = δ0r,

therefore

rE(r) =δ0r

2

2+ C,

and soE(r) =

δ0r

2+C

r.

Since E is assumed to be continuous, we must have C = 0, so E(r) = δ0r/2.Now suppose that r > a. Then

d

dr(rE(r)) = 0, therefore

rE(r) = k, and so

E(r) =k

r.

Since E is assumed to be continuous,

limr→a+

E(r) = limr→a−

E(r).

Now,

limr→a+

E(r) =k

a

limr→a−

E(r) =δ0a

2

so k = δ0a2/2. Thus

E(r) =

{δ0r2 r ≤ aδ0a

2

2 r > a

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