Chapter 16

January 27, 2005 11:56 L24-CH16 Sheet number 1 Page number 693 black

CHAPTER 16
Topics in Vector Calculus

EXERCISE SET 16.1
1. (a) III because the vector field is independent of y and the direction is that of the negative x-axis
for negative x, and positive for positive
(b) IV, because the y-component is constant, and the x-component varies periodically with x

2. (a) I, since the vector field is constant
(b) II, since the vector field points away from the origin

3. (a) true (b) true (c) true

4. (a) false, the lengths are equal to 1 (b) false, the y-component is then zero
(c) false, the x-component is then zero
y y y
5. 6. 7.

x x

x

y y y
8. 9. 10.

x x x

y x
11. (a) ∇φ = φx i + φy j = i+ j = F, so F is conservative for all x, y
1 + x2 y 2 1 + x2 y 2
(b) ∇φ = φx i + φy j = 2xi − 6yj + 8zk = F so F is conservative for all x, y

12. (a) ∇φ = φx i + φy j = (6xy − y 3 )i + (4y + 3x2 − 3xy 2 )j = F, so F is conservative for all x, y
(b) ∇φ = φx i + φy j + φz k = (sin z + y cos x)i + (sin x + z cos y)j + (x cos z + sin y)k = F, so F is
conservative for all x, y

13. div F = 2x + y, curl F = zi

14. div F = z 3 + 8y 3 x2 + 10zy, curl F = 5z 2 i + 3xz 2 j + 4xy 4 k

15. div F = 0, curl F = (40x2 z 4 − 12xy 3 )i + (14y 3 z + 3y 4 )j − (16xz 5 + 21y 2 z 2 )k

16. div F = yexy + sin y + 2 sin z cos z, curl F = −xexy k

693


694 Chapter 16

2
17. div F = , curl F = 0
x2 + y2 + z2

1 x z
18. div F = + xzexyz + 2 2
, curl F = −xyexyz i + 2 j + yzexyz k
x x +z x + z2

19. ∇ · (F × G) = ∇ · (−(z + 4y 2 )i + (4xy + 2xz)j + (2xy − x)k) = 4x

20. ∇ · (F × G) = ∇ · ((x2 yz 2 − x2 y 2 )i − xy 2 z 2 j + xy 2 zk) = −xy 2

21. ∇ · (∇ × F) = ∇ · (− sin(x − y)k) = 0

22. ∇ · (∇ × F) = ∇ · (−zeyz i + xexz j + 3ey k) = 0

23. ∇ × (∇ × F) = ∇ × (xzi − yzj + yk) = (1 + y)i + xj

24. ∇ × (∇ × F) = ∇ × ((x + 3y)i − yj − 2xyk) = −2xi + 2yj − 3k

∂f ∂g ∂h
27. Let F = f i + gj + hk ; div (kF) = k +k +k = k div F
∂x ∂y ∂z

∂h ∂g ∂f ∂h ∂g ∂f
28. Let F = f i + gj + hk ; curl (kF) = k − i+k − j+k − k = k curl F
∂y ∂z ∂z ∂x ∂x ∂y

29. Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
∂f ∂P ∂g ∂Q ∂h ∂R
div (F + G) = + + + + +
∂x ∂x ∂y ∂y ∂z ∂z
∂f ∂g ∂h ∂P ∂Q ∂R
= + + + + + = div F + div G
∂x ∂y ∂z ∂x ∂y ∂z

30. Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
∂ ∂ ∂ ∂
curl (F + G) = (h + R) − (g + Q) i + (f + P ) − (h + R) j
∂y ∂z ∂z ∂x
∂ ∂
(g + Q) −
+ (f + P ) k;
∂x ∂y
expand and rearrange terms to get curl F + curl G.

31. Let F = f i + gj + hk ;
∂f ∂φ ∂g ∂φ ∂h ∂φ
div (φF) = φ + f + φ + g + φ + h
∂x ∂x ∂y ∂y ∂z ∂z
∂f ∂g ∂h ∂φ ∂φ ∂φ
=φ + + + f+ g+ h
∂x ∂y ∂z ∂x ∂y ∂z
= φ div F + ∇φ · F

32. Let F = f i + gj + hk ;
∂ ∂ ∂ ∂ ∂ ∂
curl (φF) = (φh) − (φg) i+ (φf ) − (φh) j+ (φg) − (φf ) k; use the product
∂y ∂z ∂z ∂x ∂x ∂y
rule to expand each of the partial derivatives, rearrange to get φ curl F + ∇φ × F


Exercise Set 16.1 695

33. Let F = f i + gj + hk ;
∂ ∂h ∂g ∂ ∂f ∂h ∂ ∂g ∂f
div(curl F) = − + − + −
∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y
∂2h ∂2g ∂2f ∂2h ∂2g ∂2f
= − + − + − = 0,
∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y
assuming equality of mixed second partial derivatives

∂2φ ∂2φ ∂2φ ∂2φ ∂2φ ∂2φ
34. curl (∇φ) = − i+ − j+ − k = 0, assuming equality
∂y∂z ∂z∂y ∂z∂x ∂x∂z ∂x∂y ∂y∂x
of mixed second partial derivatives

35. ∇ · (kF) = k∇ · F, ∇ · (F + G) = ∇ · F + ∇ · G, ∇ · (φF) = φ∇ · F + ∇φ · F, ∇ · (∇ × F) = 0

36. ∇ × (kF) = k∇ × F, ∇ × (F + G) = ∇ × F + ∇ × G, ∇ × (φF) = φ∇ × F + ∇φ × F, ∇ × (∇φ) = 0

37. (a) curl r = 0i + 0j + 0k = 0
x y z r
(b) ∇ r = ∇ x2 + y 2 + z 2 = i+ j+ k=
x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z2 r

38. (a) div r = 1 + 1 + 1 = 3
1 xi + yj + zk r
(b) ∇ = ∇(x2 + y 2 + z 2 )−1/2 = − 2 =−
r (x + y 2 + z 2 )3/2 r 3

∂r ∂r ∂r f (r)
39. (a) ∇f (r) = f (r) i + f (r) j + f (r) k = f (r)∇r = r
∂x ∂y ∂z r
f (r)
(b) div[f (r)r] = f (r)div r + ∇f (r) · r = 3f (r) + r · r = 3f (r) + rf (r)
r

f (r)
40. (a) curl[f (r)r] = f (r)curl r + ∇f (r) × r = f (r)0 + r×r=0+0=0
r
f (r) f (r) f (r)
(b) ∇2 f (r) = div[∇f (r)] = div r = div r + ∇ ·r
r r r

f (r) rf (r) − f (r) f (r)
=3 + r·r=2 + f (r)
r r3 r

41. f (r) = 1/r3 , f (r) = −3/r4 , div(r/r3 ) = 3(1/r3 ) + r(−3/r4 ) = 0

42. Multiply 3f (r) + rf (r) = 0 through by r2 to obtain 3r2 f (r) + r3 f (r) = 0,
d[r3 f (r)]/dr = 0, r3 f (r) = C, f (r) = C/r3 , so F = Cr/r3 (an inverse-square ﬁeld).

43. (a) At the point (x, y) the slope of the line along which the vector −yi + xj lies is −x/y; the
slope of the tangent line to C at (x, y) is dy/dx, so dy/dx = −x/y.
(b) ydy = −xdx, y 2 /2 = −x2 /2 + K1 , x2 + y 2 = K


696 Chapter 16

44. dy/dx = x, y = x2 /2 + K 45. dy/dx = 1/x, y = ln x + K
y y

x

x

46. dy/dx = −y/x, (1/y)dy = (−1/x)dx, ln y = − ln x + K1 , y
y = eK1 e− ln x = K/x

x

EXERCISE SET 16.2
1
1. (a) dy = 1 because s = y is arclength measured from (0, 0)
0

(b) 0, because sin xy = 0 along C

2. (a) ds = length of line segment = 2 (b) 0, because x is constant and dx = 0
C

3. Since F and r are parallel, F · r = F r , and since F is constant,
√ 4 √
F · dr = d(F · r) = d( F r ) = 2 2dt = 16
−4
C C

4. F · r = 0, since F is perpendicular to the curve.
C

5. By inspection the tangent vector in part (a) is given by T = j, so F · T = F · j = sin x on C. But

x = −π/2 on C, thus sin x = −1, F · T = −1 and F · dr = (−1)ds.
C C

6. (a) Let α be the angle between F and T. Since F = 1, cos α = F T cos α = F · T, and

F · T ds = cos α(s) ds. From Figure 16.2.12(b) it is apparent that α is close to zero on
C C

most of the parabola, thus cos α ≈ 1 though cos α ≤ 1. Hence cos α(s) ds ≤ ds and
C C
the ﬁrst integral is close to the second.



(b) From Example 8(b) cos α ds = F · dr ≈ 5.83629, and
C C
2
ds = 1 + (2t)2 dt ≈ 6.125726619.
C −1

11 √ √
2 2 1
dx dy 1 4
7. (a) ds = + dt, so (2t − 3t2 ) 4 + 36t2 dt = − 10 − ln( 10 − 3) −
dt dt 0 108 36 27
1 1
1
(b) (2t − 3t2 )2 dt = 0 (c) (2t − 3t2 )6t dt = −
0 0 2

1 1
864 54
8. (a) t(3t2 )(6t3 )2 1 + 36t2 + 324t4 dt = (b) t(3t2 )(6t3 )2 dt =
0 5 0 5
1 1
648
(c) t(3t2 )(6t3 )2 6t dt = (d) t(3t2 )(6t3 )2 18t2 dt = 162
0 11 0

1
9. (a) C : x = t, y = t, 0 ≤ t ≤ 1; 6t dt = 3
0
1
(b) C : x = t, y = t2 , 0 ≤ t ≤ 1; (3t + 6t2 − 2t3 )dt = 3
0

(c) C : x = t, y = sin(πt/2), 0 ≤ t ≤ 1;
1
[3t + 2 sin(πt/2) + πt cos(πt/2) − (π/2) sin(πt/2) cos(πt/2)]dt = 3
0
1
(d) C : x = t3 , y = t, 0 ≤ t ≤ 1; (9t5 + 8t3 − t)dt = 3
0

1
1
10. (a) C : x = t, y = t, z = t, 0 ≤ t ≤ 1; (t + t − t) dt =
0 2
1
1
(b) C : x = t, y = t2 , z = t3 , 0 ≤ t ≤ 1; (t2 + t3 (2t) − t(3t2 )) dt = −
0 60
1
π 2
(c) C : x = cos πt, y = sin πt, z = t, 0 ≤ t ≤ 1; (−π sin2 πt + πt cos πt − cos πt) dt = − −
0 2 π
√
3
1+t 3 √ 1
1 + 2t √
11. dt = (1 + t)−1/2 dt = 2 12. 5 dt = 5(π/4 + ln 2)
0 1+t 0 0 1 + t2
1 1
13. 3(t2 )(t2 )(2t3 /3)(1 + 2t2 ) dt = 2 t7 (1 + 2t2 ) dt = 13/20
0 0
√
5 2π √ π/4
14. e−t dt = 5(1 − e−2π )/4 15. (8 cos2 t−16 sin2 t−20 sin t cos t)dt = 1−π
4 0 0

1
2 2
16. t − t5/3 + t2/3 dt = 6/5
−1 3 3
3
1
17. C : x = (3 − t)2 /3, y = 3 − t, 0 ≤ t ≤ 3; (3 − t)2 dt = 3
0 3
1
2 2/3 2 1/3
18. C : x = t2/3 , y = t, −1 ≤ t ≤ 1; t − t + t7/3 dt = 4/5
−1 3 3


698 Chapter 16

π/2
19. C : x = cos t, y = sin t, 0 ≤ t ≤ π/2; (− sin t − cos2 t)dt = −1 − π/4
0

1
20. C : x = 3 − t, y = 4 − 3t, 0 ≤ t ≤ 1; (−37 + 41t − 9t2 )dt = −39/2
0

1
21. (−3)e3t dt = 1 − e3
0

π/2
π6
22. (sin2 t cos t − sin2 t cos t + t4 (2t)) dt =
0 192

ln 2
63 √ 1 √ 1 1√
23. (a) e3t + e−3t e2t + e−2t dt = 17 + ln(4 + 17) − tanh−1 17
0 64 4 4 17
π/2
1 3 1 π/2 1 6
(b) et sin t cos t − (sin t − t) sin t + (1 + t2 ) dt = π + e + π+
0 24 5 4 5

π/2
24. (a) cos21 t sin9 t (−3 cos2 t sin t)2 + (3 sin2 t cos t)2 dt
0
π/2
61,047
= 3 cos22 t sin10 t dt = π
0 4,294,967,296
e 2
1 1
(b) t5 ln t + 7t2 (2t) + t4 (ln t) 1 + (2t)2 + dt ≈ 1177.660136
1 t t

25. (a) C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1
C2 : (1, 0) to (0, 1); x = 1 − t, y = t, 0 ≤ t ≤ 1
C3 : (0, 1) to (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1
1 1 1
(0)dt + (−1)dt + (0)dt = −1
0 0 0

(b) C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1
C2 : (1, 0) to (1, 1); x = 1, y = t, 0 ≤ t ≤ 1
C3 : (1, 1) to (0, 1); x = 1 − t, y = 1, 0 ≤ t ≤ 1
C4 : (0, 1) to (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1
1 1 1 1
(0)dt + (−1)dt + (−1)dt + (0)dt = −2
0 0 0 0

26. (a) C1 : (0, 0) to (1, 1); x = t, y = t, 0 ≤ t ≤ 1
C2 : (1, 1) to (2, 0); x = 1 + t, y = 1 − t, 0 ≤ t ≤ 1
C3 : (2, 0) to (0, 0); x = 2 − 2t, y = 0, 0 ≤ t ≤ 1
1 1 1
(0)dt + 2dt + (0)dt = 2
0 0 0

(b) C1 : (−5, 0) to (5, 0); x = t, y = 0, −5 ≤ t ≤ 5
C2 : x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π
5 π
(0)dt + (−25)dt = −25π
−5 0



1 1
1
27. C1 : x = t, y = z = 0, 0 ≤ t ≤ 1, 0 dt = 0; C2 : x = 1, y = t, z = 0, 0 ≤ t ≤ 1, (−t) dt = −
0 0 2
1
1 5
C3 : x = 1, y = 1, z = t, 0 ≤ t ≤ 1, 3 dt = 3; x2 z dx − yx2 dy + 3 dz = 0 − + 3 =
0 C 2 2

28. C1 : (0, 0, 0) to (1, 1, 0); x = t, y = t, z = 0, 0 ≤ t ≤ 1
C2 : (1, 1, 0) to (1, 1, 1); x = 1, y = 1, z = t, 0 ≤ t ≤ 1
C3 : (1, 1, 1) to (0, 0, 0); x = 1 − t, y = 1 − t, z = 1 − t, 0 ≤ t ≤ 1
1 1 1
(−t3 )dt + 3 dt + −3dt = −1/4
0 0 0

π 1
29. (0)dt = 0 30. (e2t − 4e−t )dt = e2 /2 + 4e−1 − 9/2
0 0

1 π/2
31. e−t dt = 1 − e−1 32. (7 sin2 t cos t + 3 sin t cos t)dt = 23/6
0 0

33. Represent the circular arc by x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π/2.
√ π/2 √ √
√
x yds = 9 3 sin t cos t dt = 6 3
C 0

34. δ(x, y) = k x2 + y 2 where k is the constant of proportionality,
1 √ √ 1 √
k x2 + y 2 ds = k et ( 2et ) dt = 2k e2t dt = (e2 − 1)k/ 2
C 0 0

π/2
kx cos t
35. ds = 15k dt = 5k tan−1 3
C 1 + y2 0 1 + 9 sin2 t

36. δ(x, y, z) = kz where k is the constant of proportionality,
4 √
k z ds = k(4 t)(2 + 1/t) dt = 136k/3
C 1

1
37. C : x = t2 , y = t, 0 ≤ t ≤ 1; W = 3t4 dt = 3/5
0

3 1
38. W = (t2 + 1 − 1/t3 + 1/t)dt = 92/9 + ln 3 39. W = (t3 + 5t6 )dt = 27/28
1 0

40. C1 : (0, 0, 0) to (1, 3, 1); x = t, y = 3t, z = t, 0 ≤ t ≤ 1
C2 : (1, 3, 1) to (2, −1, 4); x = 1 + t, y = 3 − 4t, z = 1 + 3t, 0 ≤ t ≤ 1
1 1
W = (4t + 8t2 )dt + (−11 − 17t − 11t2 )dt = −37/2
0 0

41. C : x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/2
π/2
1
− sin t + cos t dt = 3/4
0 4


700 Chapter 16

42. C1 : (0, 3) to (6, 3); x = 6t, y = 3, 0 ≤ t ≤ 1
C2 : (6, 3) to (6, 0); x = 6, y = 3 − 3t, 0 ≤ t ≤ 1
1
6 1
−12 1 2
dt + dt = tan−1 2 − tan−1 (1/2)
0 36t2 + 9 0 36 + 9(1 − t)2 3 3

43. Represent the parabola by x = t, y = t2 , 0 ≤ t ≤ 2.
2 √
3x ds = 3t 1 + 4t2 dt = (17 17 − 1)/4
C 0

44. Represent the semicircle by x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π.
π
x2 y ds = 16 cos2 t sin t dt = 32/3
C 0

45. (a) 2πrh = 2π(1)2 = 4π (b) S = z(t) dt
C
2π
(c) C : x = cos t, y = sin t, 0 ≤ t ≤ 2π; S = (2 + (1/2) sin 3t) dt = 4π
0

46. C : x = a cos t, y = −a sin t, 0 ≤ t ≤ 2π,
x dy − y dx 2π
−a2 cos2 t − a2 sin2 t 2π
= dt = dt = 2π
C x2 + y 2 0 a2 0

1
47. W = F · dr = (λt2 (1 − t), t − λt(1 − t)) · (1, λ − 2λt) dt = −λ/12, W = 1 when λ = −12
C 0

1 3
48. The force exerted by the farmer is F = 150 + 20 − z k= 170 − t k, so
10 4π
60
1 1
F · dr = 170 −z dz, and W = 170 − z dz = 10,020. Note that the functions
10 0 10
x(z), y(z) are irrelevant.
tk
49. (a) From (8), ∆sk = r (t) dt, thus m∆tk ≤ ∆sk ≤ M ∆tk for all k. Obviously
tk−1
∆sk ≤ M (max∆tk ), and since the right side of this inequality is independent of k, it follows
that max∆sk ≤ M (max∆tk ). Similarly m(max∆tk ) ≤ max∆sk .
1
(b) This follows from max∆tk ≤ max∆sk and max∆sk ≤ M max∆tk .
m

EXERCISE SET 16.3
1. ∂x/∂y = 0 = ∂y/∂x, conservative so ∂φ/∂x = x and ∂φ/∂y = y, φ = x2 /2 + k(y), k (y) = y,
k(y) = y 2 /2 + K, φ = x2 /2 + y 2 /2 + K

2. ∂(3y 2 )/∂y = 6y = ∂(6xy)/∂x, conservative so ∂φ/∂x = 3y 2 and ∂φ/∂y = 6xy,
φ = 3xy 2 + k(y), 6xy + k (y) = 6xy, k (y) = 0, k(y) = K, φ = 3xy 2 + K

3. ∂(x2 y)/∂y = x2 and ∂(5xy 2 )/∂x = 5y 2 , not conservative



4. ∂(ex cos y)/∂y = −ex sin y = ∂(−ex sin y)/∂x, conservative so ∂φ/∂x = ex cos y and
∂φ/∂y = −ex sin y, φ = ex cos y + k(y), −ex sin y + k (y) = −ex sin y,
k (y) = 0, k(y) = K, φ = ex cos y + K

5. ∂(cos y + y cos x)/∂y = − sin y + cos x = ∂(sin x − x sin y)/∂x, conservative so
∂φ/∂x = cos y + y cos x and ∂φ/∂y = sin x − x sin y, φ = x cos y + y sin x + k(y),
−x sin y + sin x + k (y) = sin x − x sin y, k (y) = 0, k(y) = K, φ = x cos y + y sin x + K

6. ∂(x ln y)/∂y = x/y and ∂(y ln x)/∂x = y/x, not conservative

7. (a) ∂(y 2 )/∂y = 2y = ∂(2xy)/∂x, independent of path
1
(b) C : x = −1 + 2t, y = 2 + t, 0 ≤ t ≤ 1; (4 + 14t + 6t2 )dt = 13
0

(c) ∂φ/∂x = y 2 and ∂φ/∂y = 2xy, φ = xy 2 + k(y), 2xy + k (y) = 2xy, k (y) = 0, k(y) = K,
φ = xy 2 + K. Let K = 0 to get φ(1, 3) − φ(−1, 2) = 9 − (−4) = 13

8. (a) ∂(y sin x)/∂y = sin x = ∂(− cos x)/∂x, independent of path
1
(b) C1 : x = πt, y = 1 − 2t, 0 ≤ t ≤ 1; (π sin πt − 2πt sin πt + 2 cos πt)dt = 0
0

(c) ∂φ/∂x = y sin x and ∂φ/∂y = − cos x, φ = −y cos x + k(y), − cos x + k (y) = − cos x,
k (y) = 0, k(y) = K, φ = −y cos x+K. Let K = 0 to get φ(π, −1)−φ(0, 1) = (−1)−(−1) = 0

9. ∂(3y)/∂y = 3 = ∂(3x)/∂x, φ = 3xy, φ(4, 0) − φ(1, 2) = −6

10. ∂(ex sin y)/∂y = ex cos y = ∂(ex cos y)/∂x, φ = ex sin y, φ(1, π/2) − φ(0, 0) = e

11. ∂(2xey )/∂y = 2xey = ∂(x2 ey )/∂x, φ = x2 ey , φ(3, 2) − φ(0, 0) = 9e2

12. ∂(3x − y + 1)/∂y = −1 = ∂[−(x + 4y + 2)]/∂x,
φ = 3x2 /2 − xy + x − 2y 2 − 2y, φ(0, 1) − φ(−1, 2) = 11/2

13. ∂(2xy 3 )/∂y = 6xy 2 = ∂(3x2 y 2 )/∂x, φ = x2 y 3 , φ(−1, 0) − φ(2, −2) = 32

14. ∂(ex ln y − ey /x)/∂y = ex /y − ey /x = ∂(ex /y − ey ln x)/∂x,
φ = ex ln y − ey ln x, φ(3, 3) − φ(1, 1) = 0

15. φ = x2 y 2 /2, W = φ(0, 0) − φ(1, 1) = −1/2 16. φ = x2 y 3 , W = φ(4, 1) − φ(−3, 0) = 16

17. φ = exy , W = φ(2, 0) − φ(−1, 1) = 1 − e−1

18. φ = e−y sin x, W = φ(−π/2, 0) − φ(π/2, 1) = −1 − 1/e

19. ∂(ey + yex )/∂y = ey + ex = ∂(xey + ex )/∂x so F is conservative, φ(x, y) = xey + yex so

F · dr = φ(0, ln 2) − φ(1, 0) = ln 2 − 1
C

20. ∂(2xy)/∂y = 2x = ∂(x2 + cos y)/∂x so F is conservative, φ(x, y) = x2 y + sin y so

F · dr = φ(π, π/2) − φ(0, 0) = π 3 /2 + 1
C


702 Chapter 16

21. F · dr = [(ey + yex )i + (xey + ex )j] · [(π/2) cos(πt/2)i + (1/t)j]dt
π
= cos(πt/2)(ey + yex ) + (xey + ex )/t dt,
2
2
π 1
so F · dr = cos(πt/2) t + (ln t)esin(πt/2) + sin(πt/2) + esin(πt/2) dt = ln 2 − 1
C 1 2 t

22. F · dr = 2t2 cos(t/3) + [t2 + cos(t cos(t/3))](cos(t/3) − (t/3) sin(t/3)) dt, so
π
F · dr = 2t2 cos(t/3) + [t2 + cos(t cos(t/3))](cos(t/3) − (t/3) sin(t/3)) dt = 1 + π 3 /2
C 0

23. No; a closed loop can be found whose tangent everywhere makes an angle < π with the vector
field, so the line integral F · dr > 0, and by Theorem 16.3.2 the vector field is not conservative.
C

24. The vector field is constant, say F = ai + bj, so let φ(x, y) = ax + by and F is conservative.

25. Let r(t) be a parametrization of the circle C. Then by Theorem 16.3.2(b),

Fdr = F · r (t) dt = 0. Let h(t) = F(x, y) · r (t). Then h is continuous. We must find two
C C
points at which h = 0. If h(t) = 0 everywhere on the circle, then we are done; otherwise there are
points at which h is nonzero, say h(t1 ) > 0. Then there is a small interval around t1 on which the
integral of h is positive.
(Let = h(t1 )/2. Since h(t) is continuous there exists δ > 0 such that for |t − t1 | < δ, h(t) > /2.
t1 +δ
Then h(t) dt ≥ (2δ) /2 > 0.)
t1 −δ

Since h = 0, there are points on the circle where h < 0, say h(t2 ) < 0. Now consider the
C
parametrization h(θ), 0 ≤ θ ≤ 2π. Let θ1 < θ2 correspond to the points above where h > 0 and
h < 0. Then by the Intermediate Value Theorem on [θ1 , θ2 ] there must be a point where h = 0,
say h(θ3 ) = 0, θ1 < θ3 < θ2.
To find a second point where h = 0, assume that h is a periodic function with period 2π (if need be,
extend the definition of h). Then h(t2 − 2π) = h(t2 ) < 0. Apply the Intermediate Value Theorem
on [t2 − 2π, t1 ] to find an additional point θ4 at which h = 0. The reader should prove that θ3 and
θ4 do indeed correspond to distinct points on the circle.

26. The function F · r (t) is not necessarily continuous since the tangent to the square has obvious
discontinuities. For a counterexample to the result, let the square have vertices at (0, 0), (0, 1),
(1, 1), (1, 0). Let Φ(x, y) = xy + x + y and let F = ∇Φ = (y + 1)i + (x + 1)j. Then F is conservative
, but on the bottom side of the square, where y = 0, F · r = −F · j = −x − 1 ≤ 1 < 0. On the top
edge F · r = F · j = x + 1 ≥ 1 > 0. Similarly for the other two sides of the square. Thus at no
point is F · r = 0.

∂φ ∂φ ∂φ ∂φ ∂φ ∂φ
27. If F is conservative, then F = ∇φ = i+ j+ k and hence f = ,g = , and h = .
∂x ∂y ∂z ∂x ∂y ∂z
∂f ∂2φ ∂g ∂ 2 φ ∂f ∂2φ ∂h ∂ 2 φ ∂g ∂2φ ∂h ∂2φ
Thus = and = , = and = , = and = .
∂y ∂y∂x ∂x ∂x∂y ∂z ∂z∂x ∂x ∂x∂z ∂z ∂z∂y ∂y ∂y∂z
The result follows from the equality of mixed second partial derivatives.



28. Let f (x, y, z) = yz, g(x, y, z) = xz, h(x, y, z) = yx2 , then ∂f /∂z = y, ∂h/∂x = 2xy = ∂f /∂z, thus

by Exercise 27, F = f i+gj+hk is not conservative, and by Theorem 16.3.2, yz dx+xz dy+yx2 dz
C
is not independent of the path.

∂
29. (h(x)[x sin y + y cos y]) = h(x)[x cos y − y sin y + cos y]
∂y
∂
(h(x)[x cos y − y sin y]) = h(x) cos y + h (x)[x cos y − y sin y],
∂x
equate these two partial derivatives to get (x cos y − y sin y)(h (x) − h(x)) = 0 which holds for all
x and y if h (x) = h(x), h(x) = Cex where C is an arbitrary constant.

∂ cx 3cxy ∂ cy
30. (a) 2 + y 2 )3/2
=− 2 2 )−5/2
= 2 + y 2 )3/2
when (x, y) = (0, 0),
∂y (x (x + y ∂x (x
so by Theorem 16.3.3, F is conservative. Set ∂φ/∂x = cx/(x2 + y 2 )−3/2 ,
then φ(x, y) = −c(x2 + y 2 )−1/2 + k(y), ∂φ/∂y = cy/(x2 + y 2 )−3/2 + k (y), so k (y) = 0.
c
Thus φ(x, y) = − 2 is a potential function.
(x + y 2 )1/2
(b) curl F = 0 is similar to Part (a), so F is conservative. Let
cx
φ(x, y, z) = dx = −c(x2 + y 2 + z 2 )−1/2 + k(y, z). As in Part (a),
(x2 + y 2 + z 2 )3/2
∂k/∂y = ∂k/∂z = 0, so φ(x, y, z) = −c/(x2 + y 2 + z 2 )1/2 is a potential function for F.

Q
1 1
31. (a) See Exercise 30, c = 1; W = F · dr = φ(3, 2, 1) − φ(1, 1, 2) = − √ + √
P 14 6
1 1
(b) C begins at P (1, 1, 2) and ends at Q(3, 2, 1) so the answer is again W = − √ + √ .
14 6
(c) The circle is not specified, but cannot pass through (0, 0, 0), so Φ is continuous and differ-
entiable on the circle. Start at any point P on the circle and return to P , so the work is
Φ(P ) − Φ(P ) = 0.
C begins at, say, (3, 0) and ends at the same point so W = 0.

dx dy
32. (a) F · dr = y −x dt for points on the circle x2 + y 2 = 1, so
dt dt
π
C1 : x = cos t, y = sin t, 0 ≤ t ≤ π, F · dr = (− sin2 t − cos2 t) dt = −π
C1 0
π
C2 : x = cos t, y = − sin t, 0 ≤ t ≤ π, F · dr = (sin2 t + cos2 t) dt = π
C2 0

∂f x2 − y 2 ∂g y 2 − x2 ∂f
(b) = 2 ,
2 )2 ∂x
=− 2 =
∂y (x + y (x + y 2 )2 ∂y
(c) The circle about the origin of radius 1, which is formed by traversing C1 and then traversing
C2 in the reverse direction, does not lie in an open simply connected region inside which F
is continuous, since F is not defined at the origin, nor can it be defined there in such a way
as to make the resulting function continuous there.


704 Chapter 16

33. If C is composed of smooth curves C1 , C2 , . . . , Cn and curve Ci extends from (xi−1 , yi−1 ) to (xi , yi )
n n
then F · dr = F · dr = [φ(xi , yi ) − φ(xi−1 , yi−1 )] = φ(xn , yn ) − φ(x0 , y0 )
C i=1 Ci i=1
where (x0 , y0 ) and (xn , yn ) are the endpoints of C.

34. F · dr + F · dr = 0, but F · dr = − F · dr so F · dr = F · dr, thus
C1 −C2 −C2 C2 C1 C2

F · dr is independent of path.
C

35. Let C1 be an arbitrary piecewise smooth curve from (a, b) to a point (x, y1 ) in the disk, and C2
the vertical line segment from (x, y1 ) to (x, y). Then
(x,y1 )
φ(x, y) = F · dr + F · dr = F · dr + F · dr.
C1 C2 (a,b) C2

The ﬁrst term does not depend on y;
∂φ ∂ ∂
hence = F · dr = f (x, y)dx + g(x, y)dy.
∂y ∂y C2 ∂y C2
∂φ ∂
However, the line integral with respect to x is zero along C2 , so = g(x, y) dy.
∂y ∂y C2
y
∂φ ∂
Express C2 as x = x, y = t where t varies from y1 to y, then = g(x, t) dt = g(x, y).
∂y ∂y y1

EXERCISE SET 16.4
1 1
1. (2x − 2y)dA = (2x − 2y)dy dx = 0; for the line integral, on x = 0, y 2 dx = 0, x2 dy = 0;
0 0
R
on y = 0, y 2 dx = x2 dy = 0; on x = 1, y 2 dx + x2 dy = dy; and on y = 1, y 2 dx + x2 dy = dx,
1 0
hence y 2 dx + x2 dy = dy + dx = 1 − 1 = 0
0 1
C

2. (1 − 1)dA = 0; for the line integral let x = cos t, y = sin t,
R
2π
y dx + x dy = (− sin2 t + cos2 t)dt = 0
0
C

4 2 2π 3
3. (2y − 3x)dy dx = 0 4. (1 + 2r sin θ)r dr dθ = 9π
−2 1 0 0

π/2 π/2
5. (−y cos x + x sin y)dy dx = 0 6. (sec2 x − tan2 x)dA = dA = π
0 0
R R
1 x
7. [1 − (−1)]dA = 2 dA = 8π 8. (2x − 2y)dy dx = 1/30
0 x2
R R



y 1
9. − − dA = − dA = −4
1+y 1+y
R R

π/2 4
10. (−r2 )r dr dθ = −32π
0 0

y2 1
11. − 2
− dA = − dA = −1
1+y 1 + y2
R R

√
1 x
12. (cos x cos y − cos x cos y)dA = 0 13. (y 2 − x2 )dy dx = 0
0 x2
R
2 2x 2 2x
14. (a) (−6x + 2y)dy dx = −56/15 (b) 6y dy dx = 64/5
0 x2 0 x2

15. (a) C : x = cos t, y = sin t, 0 ≤ t ≤ 2π;
2π
= esin t (− sin t) + sin t cos tecos t dt ≈ −3.550999378;
C 0

∂ ∂ y
(yex ) − e dA = [yex − ey ] dA
∂x ∂y
R R
2π 1
= r sin θer cos θ − er sin θ r dr dθ ≈ −3.550999378
0 0
1
2
(b) C1 : x = t, y = t2 , 0 ≤ t ≤ 1; [ey dx + yex dy] = et + 2t3 et dt ≈ 2.589524432
0
C1
1
2 e+3
C2 : x = t2 , y = t, 0 ≤ t ≤ 1; [ey dx + yex dy] = 2tet + tet dt = ≈ 2.859140914
0 2
C2
√
1 x
− ≈ −0.269616482; = [yex − ey ] dy dx ≈ −0.269616482
0 x2
C1 C2 R

2π 2π
16. (a) x dy = ab cos2 t dt = πab (b) −y dx = ab sin2 t dt = πab
C 0 C 0

2π
1 1
17. A = −y dx + x dy = (3a2 sin4 φ cos2 φ + 3a2 cos4 φ sin2 φ)dφ
2 C 2 0
2π 2π
3 2 3 2
= a sin2 φ cos2 φ dφ = a sin2 2φ dφ = 3πa2 /8
2 0 8 0

18. C1 : (0, 0) to (a, 0); x = at, y = 0, 0≤t≤1
C2 : (a, 0) to (0, b); x = a − at, y = bt, 0≤t≤1
C3 : (0, b) to (0, 0); x = 0, y = b − bt, 0 ≤ t ≤ 1
1 1 1
1
A= x dy = (0)dt + ab(1 − t)dt + (0)dt = ab
C 0 0 0 2


706 Chapter 16

19. C1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1
C2 : (a, 0) to (a cos t0 , b sin t0 ); x = a cos t, y = b sin t, 0 ≤ t ≤ t0
C3 : (a cos t0 , b sin t0 ) to (0, 0); x = −a(cos t0 )t, y = −b(sin t0 )t, −1 ≤ t ≤ 0
1 t0 0
1 1 1 1 1
A= −y dx + x dy = (0) dt + ab dt + (0) dt = ab t0
2 C 2 0 2 0 2 −1 2

20. C1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1
C2 : (a, 0) to (a cosh t0 , b sinh t0 ); x = a cosh t, y = b sinh t, 0 ≤ t ≤ t0
C3 : (a cosh t0 , b sinh t0 ) to (0, 0); x = −a(cosh t0 )t, y = −b(sinh t0 )t, −1 ≤ t ≤ 0
1 t0 0
1 1 1 1 1
A= −y dx + x dy = (0) dt + ab dt + (0) dt = ab t0
2 C 2 0 2 0 2 −1 2

21. curl F(x, y, z) = −gz i + fz j + (gx − fy )k = fz j(gx − fy )k, since f and g are independent of z. Thus

curl F · k dA = (fx − gy ) dA = f (x, y) dx + g(x, y) dy = F · dr by Green’s Theorem.
C C
R R

22. The boundary of the region R in Figure Ex-22 is C = C1 − C2 , so by Green’s Theorem,
F · dr − F · dr = F · dr = F · dr = 0 , since fy = gx . Thus = .
C1 C2 C1 −C2 C C1 C2

23. Let C1 denote the graph of g(x) from left to right, and C2 the graph of f (x) from left to right. On
the vertical sides x = const, and so dx = 0 there. Thus the area between the curves is

A(R) = dA = − y dx = − g(x) dx + f (x) dx
C C1 C2
R
b b b
=− g(x) dx + f (x) dx = (f (x) − g(x)) dx
a a a

24. Let A(R1 ) denote the area of the region R1 bounded by C and the lines y = y0 , y = y1 and the
y-axis. Then by Formula (6) A(R1 ) = C x dy, since the integrals on the top and bottom are zero
(dy = 0 there), and x = 0 on the y-axis. Similarly, A(R2 ) = −C y dx = − C y dx, where R2 is the
region bounded by C, x = x0 , x = x1 and the x-axis.
(a) R1 (b) R2

(c) y dx + x dy = A(R1 ) + A(R2 ) = x1 y1 − x0 y0
C

(d) Let φ(x, y) = xy. Then ∇φ · dr = y dx + x dy and thus by the Fundamental Theorem

y dx + x dy = φ(x1 , y1 ) − φ(x0 , y0 ) = x1 y1 − x0 y0 .
C
t1 t1
dy dx
(e) x(t) dt = x(t1 )y(t1 ) − x(t0 )y(t0 ) − y(t) dt which is equivalent to
t0 dt t0 dt

y dx + x dy = x1 y1 − x0 y0
C

π 5
25. W = y dA = r2 sin θ dr dθ = 250/3
0 0
R



26. We cannot apply Green’s Theorem on the region enclosed by the closed curve C, since F does not
have ﬁrst order partial derivatives at the origin. However, the curve Cx0 , consisting of
y = x3 /4, x0 ≤ x ≤ 2; x = 2, x3 /4 ≤ y ≤ 2; and y = x3 /4, x0 ≤ x ≤ 2 encloses a region Rx0 in
0 0
which Green’s Theorem does hold, and

W = F · dr = lim F · dr = lim ∇ · F dA
x0 →0+ x0 →0+
C Cx0 Rx0
3
2 x /4
1 −1/2 1 −1/2
= lim x − y dy dx
x0 →0+ x0 x3 /4
0
2 2
√
18 √ 2 3 3/2 3 7/2 3 5/2 18 √
= lim − 2− x0 + x0 + x0 − x0 =− 2
x0 →0+ 35 4 14 10 35

2π a(1+cos θ)
27. y dx − x dy = (−2)dA = −2 r dr dθ = −3πa2
C 0 0
R

1 1 2
28. x =
¯ x dA, but x dy = x dA from Green’s Theorem so
A C 2
R R

1 1 2 1 1
x=
¯ x dy = x2 dy. Similarly, y = −
¯ y 2 dx.
A C 2 2A C 2A C

1 x 1
1 3
29. A = dy dx = ; C1 : x = t, y = t3 , 0 ≤ t ≤ 1, x2 dy = t2 (3t2 ) dt =
0 x3 4 C1 0 5
1
1 3 1 4 8
C2 : x = t, y = t, 0 ≤ t ≤ 1; x2 dy = t2 dt = , x2 dy = − = − = ,x =
¯
C2 0 3 C C1 C2 5 3 15 15

1 1
1 1 4 8 8 8
y 2 dx = t6 dt − t2 dt = − = − ,y =
¯ , centroid ,
C 0 0 7 3 21 21 15 21

a2
30. A = ; C1 : x = t, y = 0, 0 ≤ t ≤ a, C2 : x = a − t, y = t, 0 ≤ t ≤ a; C3 : x = 0, y = a − t, 0 ≤ t ≤ a;
2
a
a3 a3 a
x2 dy = 0, x2 dy = (a − t)2 dt = , x2 dy = 0, x2 dy = + + = , x = ; ¯
0 3 3 3
C1 C2 C3 C C1 C2 C3
a 3
a a a a
y 2 dx = 0 − t2 dt + 0 = − , y = , centroid
¯ ,
0 3 3 3 3
C

31. x = 0 from the symmetry of the region,
¯
√
C1 : (a, 0) to (−a, 0) along y = a2 − x2 ; x = a cos t, y = a sin t, 0 ≤ t ≤ π
C2 : (−a, 0) to (a, 0); x = t, y = 0, −a ≤ t ≤ a
π a
1
A = πa2 /2, y= −
¯ −a3 sin3 t dt + (0)dt
2A 0 −a

1 4a3 4a 4a
=− − = ; centroid 0,
πa2 3 3π 3π


708 Chapter 16

ab
32. A = ; C1 : x = t, y = 0, 0 ≤ t ≤ a, C2 : x = a, y = t, 0 ≤ t ≤ b;
2
C3 : x = a − at, y = b − bt, 0 ≤ t ≤ 1;
b 1
ba2
x2 dy = 0, x2 dy = a2 dt = ba2 , x2 dy = a2 (1 − t)2 (−b) dt = − ,
C1 C2 0 C3 0 3
2
2ba 2a
x2 dy = + + = , x=
¯ ;
C C1 C2 C3 3 3
1
ab2 b 2a b
y 2 dx = 0 + 0 − ab2 (1 − t)2 dt = − , y = , centroid
¯ ,
C 0 3 3 3 3

33. From Green’s Theorem, the given integral equals (1−x2 −y 2 )dA where R is the region enclosed
R
by C. The value of this integral is maximum if the integration extends over the largest region for
which the integrand 1 − x2 − y 2 is nonnegative so we want 1 − x2 − y 2 ≥ 0, x2 + y 2 ≤ 1. The
largest region is that bounded by the circle x2 + y 2 = 1 which is the desired curve C.

34. (a) C : x = a + (c − a)t, y = b + (d − b)t, 0 ≤ t ≤ 1
1
−y dx + x dy = (ad − bc)dt = ad − bc
C 0

(b) Let C1 , C2 , and C3 be the line segments from (x1 , y1 ) to (x2 , y2 ), (x2 , y2 ) to (x3 , y3 ), and
(x3 , y3 ) to (x1 , y1 ), then if C is the entire boundary consisting of C1 , C2 , and C3
3
1 1
A= −y dx + x dy = −y dx + x dy
2 C 2 i=1 Ci
1
= [(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (x3 y1 − x1 y3 )]
2
1
(c) A = [(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + · · · + (xn y1 − x1 yn )]
2
1
(d) A = [(0 − 0) + (6 + 8) + (0 + 2) + (0 − 0)] = 8
2

35. F · dr = (x2 + y) dx + (4x − cos y) dy = 3 dA = 3(25 − 2) = 69
C C
R

36. F · dr = (e−x + 3y) dx + x dy = −2 dA = −2[π(4)2 − π(2)2 ] = −24π
C C
R

EXERCISE SET 16.5
1. R is the annular region between x2 + y 2 = 1 and x2 + y 2 = 4;

x2 y2
z 2 dS = (x2 + y 2 ) + 2 + 1 dA
x2 +y 2 x + y2
σ R
√ √ 2π 2
15 √
= 2 (x2 + y 2 )dA = 2 r3 dr dθ = π 2.
0 1 2
R



2. z = 1 − x − y, R is the triangular region enclosed by x + y = 1, x = 0 and y = 0;
√
√ √ 1 1−x
3
xy dS = xy 3 dA = 3 xy dy dx = .
0 0 24
σ R

3. Let r(u, v) = cos ui + vj + sin uk, 0 ≤ u ≤ π, 0 ≤ v ≤ 1. Then ru = − sin ui + cos uk, rv = j,
1 π
ru × rv = − cos ui − sin uk, ru × rv = 1, x2 y dS = v cos2 u du dv = π/4
0 0
σ

4. z = 4 − x2 − y 2 , R is the circular region enclosed by x2 + y 2 = 3;

x2 y2
(x2 + y 2 )z dS = (x2 + y 2 ) 4 − x2 − y 2 + + 1 dA
4−x2 − y2 4 − x2 − y 2
σ R
√
2π 3
= 2(x2 + y 2 )dA = 2 r3 dr dθ = 9π.
0 0
R

5. If we use the projection of σ onto the xz-plane then y = 1 − x and R is the rectangular region in
the xz-plane enclosed by x = 0, x = 1, z = 0 and z = 1;
√ √ 1 1 √
(x − y − z)dS = (2x − 1 − z) 2dA = 2 (2x − 1 − z)dz dx = − 2/2
0 0
σ R

6. R is the triangular region enclosed by 2x + 3y = 6, x = 0, and y = 0;
√ √ 3 (6−2x)/3 √
(x + y)dS = (x + y) 14 dA = 14 (x + y)dy dx = 5 14.
0 0
σ R

7. There are six surfaces, parametrized by projecting onto planes:
σ1 : z = 0; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (onto xy-plane), σ2 : x = 0; 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 (onto yz-plane),
σ3 : y = 0; 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 (onto xz-plane), σ4 : z = 1; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (onto xy-plane),
σ5 : x = 1; 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 (onto yz-plane), σ6 : y = 1; 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 (onto xz-plane).

By symmetry the integrals over σ1 , σ2 and σ3 are equal, as are those over σ4 , σ5 and σ6 , and
1 1 1 1
(x + y + z)dS = (x + y)dx dy = 1; (x + y + z)dS = (x + y + 1)dx dy = 2,
0 0 0 0
σ1 σ4

thus, (x + y + z)dS = 3 · 1 + 3 · 2 = 9.
σ

8. Let r(φ, θ) = a sin φ cos θ i + a sin φ sin θ j + a cos φ k,
0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π; rφ × rθ = a2 sin φ, x2 + y 2 = a2 sin2 φ
2π π
8 4
f (x, y, z) = a4 sin3 φ dφ dθ = πa
0 0 3
σ


710 Chapter 16

9. (a) The integral is improper because the function z(x, y) is not differentiable when x2 + y 2 = 1.
(b) Fix r0 , 0 < r0 < 1. Then z + 1 = 1 − x2 − y 2 + 1, and
x2 y2
(z + 1) dS = ( 1 − ‘x2 − y 2 + 1) 1 + + dx dy
1−x2 − y2 1 − x2 − y 2
σr0 σr 0
2π r0
1 1 2
= ( 1 − r2 + 1) √ r dr dθ = 2π 1 + r0 − 1 − r0 , and, after passing to
2
0 0 1−r 2 2

the limit as r0 → 1, (z + 1) dS = 3π
σ

(c) Let r(φ, θ) = sin φ cos θi + sin φ sin θj + cos φk, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2; rφ × rθ = sin φ,
2π π/2
(1 + cos φ) dS = (1 + cos φ) sin φ dφ dθ
0 0
σ
π/2
= 2π (1 + cos φ) sin φ dφ = 3π
0

10. (a) The function z(x, y) is not differentiable at the origin (in fact it’s partial derivatives are
unbounded there).
(b) R is the circular region enclosed by x2 + y 2 = 1;

x2 y2
x2 + y 2 + z 2 dS = 2(x2 + y 2 ) + 2 + 1 dA
x2 +y 2 x + y2
σ R

= lim 2 x2 + y 2 dA
r0 →0+
R

where R is the annular region enclosed by x2 +y 2 = 1 and x2 +y 2 = r0 with r0 slightly larger
2

x2 y2
than 0 because + 2 + 1 is not defined for x2 + y 2 = 0, so
x2 + y 2 x + y2
2π 1
4π 4π
x2 + y 2 + z 2 dS = lim 2 r2 dr dθ = lim (1 − r0 ) =
3
.
r0 →0+ 0 r0 r0 →0+ 3 3
σ

(c) The cone is contained in the locus of points satisfying φ = π/4, so it can be parametrized
with spherical coordinates ρ, θ:
1 1 1 √
r(ρ, θ) = √ ρ cos θi + √ ρ sin θj + √ ρk, 0 ≤ θ ≤ 2π, r < ρ ≤ 2. Then
2 2 2
1 1 1 1 1
rρ = √ cos θi + √ sin θj + √ k, rθ = − √ ρ sin θi + √ ρ cos θ j
2 2 2 2 2
ρ 1
rρ × rθ = (− cos θi − sin θj + k) and rρ × rθ = √ ρ, and thus
2 2 √
√
2π 2 2
1 1 1
f (x, y, z) dS = lim ρ √ ρ dρ dθ = lim 2π √ ρ3
r→0 0 r 2 r→0 23 r
σr
√
2 √ 4π
= lim 2 2 − r3 π = .
r→0 3 3

11. (a) Subdivide the right hemisphere σ ∩ {x > 0} into patches, each patch being as small as desired
(i). For each patch there is a corresponding patch on the left hemisphere σ ∩ {x < 0} which
is the reflection through the yz-plane. Condition (ii) now follows.

Chapter 16

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Chapter 16