Respostas Calc - Ie Ii 8ed - Sm Ch15

Transcript

CHAPTER 15 Multiple Integrals EXERCISE SET 15.1
1

2 1.
1 (x + 3)dy dx = 0
(2x + 6)dx = 7 0 4
0
1 4 2
0 2
ln 3
ln 2 0
2

1
0 2 0

5
1 1
0
π

1
2 0 4
2 ln 2
1 1

2

1 −2
1
0 1

−3 10dx = 20 4  dx = 1 − ln 2 3 1 1 − x+1 x+2  dx = ln(25/24) 0 dx = 0 −1 2 π/2 
 2 x 1 − x dy dx = 3 15. 0 4 xy  dy dx = x2 + y 2 + 1 14. 0 6 dy dx = 1 4xy 3 dy dx = 13. −1 1 1− x+1
7 1 x (e − 1)dx = (1 − ln 2)/2 2 0 1 dy dx = (x + y)2 12. 3
π xy ey x dy dx =
(3 + 3y 2 )dy = 14 −2 π/2 11. 0 0 (sin 2x − sin x)dx = −2 1 ln 2  1 x cos xy dy dx =
6 8. 0
2 10. π/2 −1 4 x dy dx = (xy + 1)2 9. 0
3 dy = 3 −1 2

2 (x2 + y 2 )dx dy = −2 0 dx dy = −1
4. 1 sin x dx = (1 − cos 2)/2 2 0 7. 0 4x dx = 16 1 0 y sin x dy dx = 0 −1 3 (2x − 4y)dy dx = ex dx = 2 0 6.

1 ln 3 ex+y dy dx = 5.
1 1 y dy = 2 3 x2 y dx dy = 3. 3 2.
1 √ √ [x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3 0 1 x(1 − x2 )1/2 dx = 1/3 0

π/3 (x sin y − y sin x)dy dx = 16. 0 0 0 π/2   x π2 − sin x dx = π 2 /144 2 18 17. (a) x∗k = k/2 − 1/4, k = 1, 2, 3, 4; yl∗ = l/2 − 1/4, l = 1, 2, 3, 4,

4  4  4 4   f (x, y) dxdy ≈ f (x∗k , yl∗ )∆Akl = [(k/2−1/4)2 +(l/2−1/4)](1/2)2 = 37/4
R 2
k=1 l=1 k=1 l=1 2 (x2 + y) dxdy = 28/3; the error is |37/4 − 28/3| = 1/12 (b) 0 0 655 656 Chapter 15 18. (a) x∗k = k/2 − 1/4, k = 1, 2, 3, 4; yl∗ = l/2 − 1/4, l = 1, 2, 3, 4,

4  4  4 4   f (x, y) dxdy ≈ f (x∗k , yl∗ )∆Akl = [(k/2 − 1/4) − 2(l/2 − 1/4)](1/2)2 = −4 R
2 k=1 l=1
k=1 l=1 2 (x − 2y) dxdy = −4; the error is zero (b) 0 0 z 19. (a) z (b) (0, 0, 5) (1, 0, 4) (0, 4, 3) y y (2, 5, 0) (3, 4, 0) x x z 20. (a) z (b) (2, 2, 8) (0, 0, 2) y y (2, 2, 0) (1, 1, 0) x x
5
(2x + y)dy dx = 3
1 3
5 (2x + 3/2)dx = 19 3
2 3 (3x3 + 3x2 y)dy dx = 22. V = 1
2
1
3 2 0 0 3
0
4 3 5(1 − x/3)dy dx = 24. V = 0 1/2 2 3x2 dx = 8 x dy dx =
(6x3 + 6x2 )dx = 172 0 23. V =

2 21. V =
0 5(4 − 4x/3)dx = 30 0
π 1/2 2 25. x cos(xy) cos πx dy dx = 0 0 0 0
π cos2 πx sin(xy) dx 1/2 cos2 πx sin πx dx = − = 0 1/2 1 1 cos3 πx = 3π 3π 0 Exercise Set 15.2 657
z 26. (a) (b) 5
5
3 y dy dx + 0 (0, 2, 2)
2 V = 0 (−2y + 6) dy dx 0 2
π/2 = 10 + 5 = 15 y 3 5 (5, 3, 0) x 2 π 27. fave =
π/2 1 2 30. fave 1 y sin xy dx dy = 0 0 1 28. average = 3 29. Tave =

1 0
3
 0 1 = A(R)
b 3 dx dy = 0 d x=0 dy = 2 π (1 − cos y) dy = 1 − 0 2 π √ 1 [(1 + y)3/2 − y 3/2 ]dy = 2(31 − 9 3)/45 9
1  0 44 − 16x2 3   dx = 14 3 ◦ C 1 (b − a)(d − c)k = k A(R) k dy dx = c 31. 1.381737122 32. 2.230985141


33. x=1  0  1 10 − 8x2 − 2y 2 dy dx = 2
a  − cos xy 0 2 π/2
1/2 x(x + y)

1 2 0 2 π b
f (x, y)dA =
c

b = g(x)dx d
b g(x)h(y)dy dx = a R d g(x) a d h(y)dy dx c h(y)dy a c 34. The integral of tan x (an odd function) over the interval [−1, 1] is zero. 35. The first integral equals 1/2, the second equals −1/2. No, because the integrand is not continuous. EXERCISE SET 15.2
1

x 1 1 4 (x − x7 )dx = 1/40 3 2 1. xy dy dx = x2 0
3/2 0

3−y 2. 3/2 (3y − 2y 2 )dy = 7/24 y dx dy = 1
1 y 3
√9−y2 3.
3 y dx dy = 0

x 4. 1/4  9 − y 2 dy = 9 0 0 1 y x2 
1
x x/y dy dx = 1/4 x2 x1/2 y −1/2 dy dx =
1 2(x − x3/2 )dx = 13/80 1/4 658 Chapter 15
5. √ 2π
sin(y/x)dy dx = √ π
1

x2 −1
1 −x2
x 8. (x2 − y)dy dx =
x2
0 2
2x4 dx = 4/5 −1 1 y2 x/y 2 e x2

1 x 9. 0
1 cos(y/x)dy dx = x 0 π/2 x2
sin x dx = 1 π/2
 2 2 y x − y dy dx = 0 π 1 0 1 3 x dx = 1/12 3 2 (e − 1)y 2 dy = 7(e − 1)/3 dx dy = 0
π 7. 0 10. 1
1 xe dx = (e − 1)/2 e dy dx = 0 [−x cos(x2 ) + x]dx = π/2 √ π 0 6. √ 2π
x3 1
2

x2 11. (a) f (x, y) dydx 0


√ x 12. (a) f (x, y) dydx
2

3 4
f (x, y) dydx +
f (x, y) dxdy

3 5
3 f (x, y) dydx + −2x+5 3 √ y y2 0 13. (a) 1
1 f (x, y) dxdy (b) x2 0 2 √ y 0 0 1
4 (b) 2 f (x, y) dydx 1 4 2x−7 (y+7)/2 f (x, y) dxdy (b) 1
(5−y)/2 1 14. (a) √ − 1−x2 −1
2 √ 1−x2

f (x, y) dydx
√1−y2 1 (b) √ −1
x2 15. (a)
− f (x, y) dxdy 1−y 2 2 1 5 16 x dx = 3 0 2
3 xy dx dy = (3y 2 + 3y)dy = 38 xy dy dx = 0
0 3
(y+7)/2 (b) −(y−5)/2 1
1
1
√ x 16. (a) 1 (x3/2 + x/2 − x3 − x4 /2)dx = 3/10 (x + y)dy dx = x2 0
1
(b) √ − 1−x2 −1
8
0 √ 1−x2
x dy dx + −1
x 4
16/x 4
4 8
2 (b) −1
8 2
4
y 8 x dx dy = 16/y 4 y =
1 y dy dx = 2 x dxdy + 2 √ − 1−x2
 2x 1 − x2 dx + 0 = 0 (x3 − 16x)dx = 576
8 √ 1−x2
8 x2 dy dx = 17. (a) 1 
8 512 4096 512 − y 3 dy + − dy 3 3y 3 3 4 640 1088 + = 576 3 3 2 1 4 y dy = 31/10 1 0 1 2
1
2
2
2
(b) xy 2 dydx + xy 2 dydx = xy 2 dx dy = 18. (a) 0 1 1 x 0
1 7x/3 dx + 1 2 8x − x4 dx = 7/6 + 29/15 = 31/10 3 Exercise Set 15.2
659
1 19. (a) √ − 1−x2 −1
1 √ −1 5
(3x − 2y)dy dx =
√1−y2 (b)
√ 1−x2 −
1−y 2 (3x − 2y) dxdy = √ 25−x2

√25−y2
4
√ y π
x 0

6−y 2 y2 0 √ 1/ 2
24.
π/4 1 cos 2y dy = 1/8 4 x dx dy = 0
sin y 1

0
x 25. x3 0 √ 1 y(1 + y 2 )−1/2 dy = ( 17 − 1)/2 2 1 (36y − 12y 2 + y 3 − y 5 )dy = 50/3 2 xy dx dy = π/4 4 0 23. 0  25 − y 2 − 5 + y dy = 125/6 x sin x dx = π 0 2  π x cos y dy dx =

x(1 + y 2 )−1/2 dx dy = 0
 1 − y 2 dy = 0 0 0 22.
y 5−y 21. 0 5 y dxdy =
−1 −4y (5x − x2 )dx = 125/6
(b) 0 1 0 5−x 5  6x 1 − x2 dx = 0 5 y dy dx =
−1
20. (a) 0 1 √ 1/ 2 1 (−x4 + x3 + x2 − x)dx = −7/60 (x − 1)dy dx = 0

2x 2 26. x dy dx + 0 x
1
1/x √ 1/ 2 2 x dy dx = √ 1/ 2 x dx + 0 x
3 1 √ (x 1/ 2 − x3 )dx = 1/8 y 27. (a) 4 3 2 1 x –2 –1 0.5 1.5 (b) x = (−1.8414, 0.1586), (1.1462, 3.1462)


x dA ≈ (c) R

1.1462
R 1.1462 x dydx = −1.8414
3.1462 −1.8414 ex

ln y x dA ≈ (d)
x+2 3.1462 x dxdy = 0.1586 y−2 0.1586  x(x + 2 − ex ) dx ≈ −0.4044 ln2 y (y − 2)2 − 2 2 dy ≈ −0.4044 660 Chapter 15 y 28. (a) (b) (1, 3), (3, 27) 25 R 15 5 x 1
3
2
4x3 −x4 (c) 1 3−4x+4x2 π/4

x[(4x3 − x4 ) − (3 − 4x + 4x2 )] dx = 1
cos x π/4 (cos x − sin x)dx = dy dx = 0
sin x
1

−4 3y−4
3 (−y 2 − 3y + 4)dy = 125/6
9−y 2 31. A = 3 8(1 − y 2 /9)dy = 32 dx dy = −3
1 −3 1−y 2 /9

cosh x 32. A = 1 dy dx = 0
sinh x (cosh x − sinh x)dx = 1 − e−1 0
6−3x/2 4 [(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12 (3 − 3x/4 − y/2) dy dx = 33. 0
0 2
0 √ 4−x2  34. 0 2−1 1 dx dy = −4 √ 224 15 0 −y 2 30. A = 4 3 x dy dx = 29. A =
3
2 (4 − x2 ) dx = 16/3 4 − x2 dy dx = 0 0

3 35. V = √ − 9−x2 −3
1 √ 9−x2

(3 − x)dy dx = −3
x
(2x3 − x4 − x6 )dx = 11/70 x2 0 3
0
2 2 37. V = 3 2 (18x2 + 8/3)dx = 170 (9x + y )dy dx = 0
0
1 0
(1 − x)dx dy = y2
3/2 39. V = 3
1 (1/2 − y 2 + y 4 /2)dy = 8/15 −1 √ 9−4x2
(y √ − 9−4x2 −3/2

1 38. V = −1 40. V = y 2 /3
5
41. V = 8 0 0 9 − 4x2 dx = 27π/2 3 (18 − 3y 2 + y 6 /81)dy = 216/7 (9 − x )dx dy = √ 25−x2 6 −3/2
3  3/2 + 3)dy dx = 2 0   (6 9 − x2 − 2x 9 − x2 )dx = 27π 1 (x2 + 3y 2 )dy dx = 36. V = 3 0 
5 (25 − x2 )dx = 2000/3 25 − x2 dy dx = 8 0 Exercise Set 15.2 661
√1−(y−1)2  1 [1 − (y − 1)2 ]3/2 + y 2 [1 − (y − 1)2 ]1/2 dy, 3 0 0 0
π/2  1 3 2 let y − 1 = sin θ to get V = 2 cos θ + (1 + sin θ) cos θ cos θ dθ which eventually yields −π/2 3 V = 3π/2
2
√ 1−x2
1 0
0 √ 4−x2
2
2 44. V =
0 0
2 45. f (x, y)dx dy 1

f (x, y)dx dy

0
2 e−y dx dy =
0 1
4 0
x2
x3 2
3 54. 1 2 x dx dy = ey 0
2
ln 3 (9 − e2y )dy = 0
sin(y 3 )dx dy = 0
e 56. 1 (9 ln 3 − 4) 2 2 y 2 sin(y 3 )dy = (1 − cos 8)/3 1 (ex − xex )dx = e/2 − 1 x dy dx = ex 0 f (x, y)dy dx x2 0 0 0 1
50. 0 3
y2 55.
√ x 0 0 ln 3
x2 ex dx = (e8 − 1)/3 e dy dx =
f (x, y)dy dx 1 0 53. 0
sin x 2x cos(x2 )dx = sin 1 0 2 f (x, y)dy dx ln x 1
2 1 cos(x2 )dy dx =

1 −y2 ye dy = (1 − e−16 )/8 4
2x 52. 0 π/2 e2 47. 0 0 y/4
x/2 49. ey 4
8 0 51. (1 − x2 )3/2 dx = π/2 0 f (x, y)dy dx e 48. 0 1 46. y2 0

   1 2 2 3/2 2 x 4 − x + (4 − x ) dx = 2π 3 2 2 (x + y )dy dx = 0 √ 2 2 8 3 (1 − x2 − y 2 )dy dx = 43. V = 4

(x2 + y 2 )dx dy = 2 42. V = 2 0
4
57. (a) 2 √ x 0
2 0
y2 sin πy 3 dy dx; the inner integral is non-elementary.   sin πy 3 dx dy = 0 0
1

2     1 cos πy 3 y 2 sin πy 3 dy = − 3π 2 =0 0 π/2 sec2 (cos x)dx dy ; the inner integral is non-elementary. (b) sin−1 0
π/2
y
sin x π/2 2 sec2 (cos x) sin x dx = tan 1 sec (cos x)dy dx = 0 0
2
0 √ 4−x2
2 (x2 + y 2 ) dy dx = 4 58. V = 4
= 0 0 π/2 0   x2  0 64 64 128 + sin2 θ − sin4 θ 3 3 3  dθ = 1 4 − x2 + (4 − x2 )3/2 3  dx 64 π 64 π 128 π 1 · 3 + − = 8π 3 2 3 4 3 2 2·4 (x = 2 sin θ) 662 Chapter 15 59. The region is symmetric with respect to the y-axis, and the integrand is an odd function of x, hence the answer is zero. 60. This is the volume in the first octant under the surface z = π the sphere of radius 1, thus . 6 61. Area of triangle is 1/2, so f¯ = 2
1
0
x 1 dy dx = 2 1 + x2
0 1  1 − x2 − y 2 , so 1/8 of the volume of 1 π x dx = − ln 2 − 1 + x2 1 + x2 2 2 (3x − x2 − x) dx = 4/3, so 62. Area = 3 f¯ = 4 1  0
2
3x−x2 (x2 − xy)dy dx = 0 x 1 A(R) 63. Tave = 3 4
2 (−2x3 + 2x4 − x5 /2)dx = − 0 2 3 8 =− 4 15 5

(5xy + x2 ) dA. The diamond has corners (±2, 0), (0, ±4) and thus has area R 1 A(R) = 4 2(4) = 16m2 . Since 5xy is an odd function of x (as well as y), 2

5xy dA = 0. Since R x2 is an even function of both x and y,   2



4 1 2 4−2x 2 1 2 1 4 3 1 4 2◦ 2 2 x − x x dA = st0 x dydx = (4 − 2x)x dx = = C Tave = 16 4 4 0 4 3 2 3 0 0 R x,y>0 64. The area of the lens is πR2 = 4π and the average thickness Tave is
2
√4−x2
  4 1 21 (4 − x2 )3/2 dx 1 − (x2 + y 2 )/4 dydx = Tave = 4π 0 0 π 0 6
π 1 8 8 1·3π = in = sin4 θ dθ = 3π 0 3π 2 · 4 2 2 65. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
a
sin x  V = 1 + x + y dy dx = 0.676089 0 x/2 EXERCISE SET 15.3
π/2
1.
sin θ π/2 r cos θdr dθ = 0
0 π
0
1+cos θ 2. π r dr dθ = 0
0 π/2 0

a sin θ 1 (1 + cos θ)2 dθ = 3π/4 2 π/2 2 3. r dr dθ = 0
0 π/6
4. 0
cos 3θ r dr dθ = 0 0 0 1 sin2 θ cos θ dθ = 1/6 2 π/6 a3 2 sin3 θ dθ = a3 3 9 1 cos2 3θ dθ = π/24 2 (x = 2 cos θ) Exercise Set 15.3
π
663
1−sin θ π 1 (1 − sin θ)3 cos θ dθ = 0 3 r2 cos θ dr dθ = 5. 0
0 π/2 0

cos θ π/2 1 cos4 θ dθ = 3π/64 4 3 6. r dr dθ = 0 0
0 2π

1−cos θ 7. A = 0 0

π/2
sin 2θ
sin2 2θ dθ = π/2 0 π/2
0
1 9. A = π/2 r dr dθ = π/4 sin 2θ

π/3 π/4
2 10. A = 2


4 sin θ f (r, θ) r dr dθ 2

π/2 3 r


9 − r2 dr dθ
π/2 r 0

π/2 2 sin θ r2 dr dθ = 0
128 √ 2 3 9 − r2 dr dθ = 1 18. V = 2 0
π/2 cos θ 16 3

0
0
π/2
dr dθ = 8
1 π/2

π/2 (2 cos2 θ − cos4 θ)dθ = 5π/32 π/2 dθ = 4π
3 sin θ π/2 sin4 θ dθ = 0 π/2 0
2 22. V = 2
0 π/2 = 0
23. 0 1 
π
4 − r2 r drdθ + 2 2 cos θ 16 (1 − cos2 θ)3/2 θ dθ + 3 2 0 64 √ 2π 3 0 r sin θ drdθ = 9
dθ = e−r r dr dθ = 1 (1 − e−1 ) 2 1 sin3 θ dθ = 32/9 2 0 π/2 3 dr dθ 0 21. V =
0 1 (1 − r )r dr dθ = 2 3 20. V = 4 0 π/2 π/2 2 19. V = 2 2 sin θ 0 0 3
16. V = 4 0 17. V = 8 1+cos θ r2 dr dθ
cos θ (1 − r )r dr dθ
π/2 0 15. V = 2 1 14. V = 2 2 0
f (r, θ)r dr dθ 1 π/2 3π/2 12. A = π/2 13. V = 8 0 3 0 11. A = π/6 √ (4 − sec2 θ)dθ = 4π/3 − sec θ 5π/6 1 (1 − sin2 2θ)dθ = π/16 2 π/3 r dr dθ = 0 0 π/2 r dr dθ = 2 0 2π 1 (1 − cos θ)2 dθ = 3π/2 2 0 8. A = 4
2π r dr dθ =
π 0 2π  4 − r2 r drdθ π/2 π/2
2 27 π 16 0 32 8 16 dθ = + π 3 9 3 dθ = (1 − e−1 )π 664 Chapter 15
π/2
3 24. r 0

9− r2 π/4
0 dθ = 9π/2 0 2 0 π/2
π/2 dr dθ = 9 0 25.
 1 1 r dr dθ = ln 5 1 + r2 2 2 cos θ
π/4 dθ = 0 16 2r sin θ dr dθ = 3
π/2 2 26. 0 π/4
π/2
1 r3 dr dθ = 27. 0
0 2π
2

0 π/2
π/2 dθ = π/8 2 1 (1 − e−4 ) 2 2 cos θ 0
0 π/2

8 3 r2 dr dθ = 29.


0 π/4

sec θ tan θ 1 r dr dθ = 3
π/4 2 32. 0
0 π/4
2 √ 33. 0
0 5 1 r dr dθ = 2 3 csc θ 34. tan−1 (3/4) 2π
a 0 0 π/2
36. (a) V = 8 0 (b) V ≈ √ sec3 θ tan3 θ dθ = 2( 2 + 1)/45 0 0 a
π/2 tan−1 (3/4)
2π h a2 dθ = πa2 h 2 c 2 4c (a − r2 )1/2 r dr dθ = − π(a2 − r2 )3/2 a 3a a = 0 4 2 πa c 3 4 π(6378.1370)2 6356.5231 ≈ 1,083,168,200,000 km3 3 π/2
a sin θ 37. V = 2 0 π/4 0
c 2 2 (a − r2 )1/2 r dr dθ = a2 c a 3 √ a 2 cos 2θ
r dr dθ = 4a2 38. A = 4 0 (25 − 9 csc2 θ)dθ  25 25  π − tan−1 (3/4) − 6 = tan−1 (4/3) − 6 2 2 2 0

0 π sin 1 4 = hr dr dθ = 35. V =
dθ = r π √ dr dθ = ( 5 − 1) 2 4 1+r
π/2
π/2   r π 1 − 1/ 1 + a2 dr dθ = 2 3/2 2 (1 + r ) a 31. 0 dθ = (1 − e−4 )π 0 π/2 1 sin 1 2 0 π/2 2π 0 cos(r2 )r dr dθ = 0
cos3 θ dθ = 16/9 1 30. cos3 θ sin θ dθ = 1/3 π/4 0 e−r r dr dθ = 28. 0 1 4 π ln 5 8 0
π/2 (1 − cos3 θ)dθ = (3π − 4)a2 c/9 0 π/4 cos 2θ dθ = 2a2 0 Exercise Set 15.4
665
π/4 39. A = √ 8 cos 2θ π/6

4 sin θ
π/2 4 sin θ r dr dθ + r dr dθ 0 π/4
π/4 π/2 (8 sin2 θ − 4 cos 2θ)dθ + = π/6
φ
√ 8 sin2 θ dθ = 4π/3 + 2 3 − 2 π/4
2a sin θ φ 1 sin2 θ dθ = a2 φ − a2 sin 2φ 2 0 0 0 
+∞ 
+∞
+∞ 
+∞ 2 2 2 2 41. (a) I 2 = e−x dx e−y dy = e−x dx e−y dy r dr dθ = 2a2 40. A = 0 +∞
0
+∞ = 0
2 2 e−x e−y dx dy = 0 π/2
+∞ (b) I 2 = 0 2 e−r r dr dθ = 0 1 2 0

+∞ 0
0 +∞ 2 e−(x +y 2 ) dx dy 0 √ π/2 dθ = π/4 (c) I= π/2 0 √ 42. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outside of the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold:
π/2
A
A
A
π/2
√2A 1 1 1 rdr dθ ≤ dx dy ≤ rdr dθ 2 2 2 2 2 (1 + r2 )2 0 0 (1 + r ) 0 0 (1 + x + y ) 0 0 πA2 and the integral on the right equals 4(1 + A2 ) 2 π 2πA . Since both of these quantities tend to as A → +∞, it follows by sandwiching that 2 4(1 + 2A ) 4
+∞
+∞ 1 π dx dy = . 2 + y 2 )2 (1 + x 4 0 0 The integral on the left can be evaluated as
43. (a) 1.173108605 0 2π

R D(r)r dr dθ = 0
2π

R 44. V = tan−1 (2) 0 0
tan−1 (1/3)

2 4 re−r dr dθ = π
0 1 4 re−r dr ≈ 1.173108605 0 ke−r r dr dθ = −2πk(1 + r)e−r R = 2πk[1 − (R + 1)e−R ] 0
tan−1 (2) tan−1 (2) cos2 θ dθ = 2 tan−1 (1/3) 0 1 0 r3 cos2 θ dr dθ = 4 45. π (b) (1 + cos(2θ)) dθ tan−1 (1/3) √ √ = 2(tan−1 2 − tan−1 (1/3)) + 2/ 5 − 1/ 10 EXERCISE SET 15.4 z 1. (a) z (b) z (c) y x x y x y 666 Chapter 15 z 2. (a) z (b) x y y x z (c) y x 3. (a) x = u, y = v, z = 5 3 + u − 2v 2 2 (b) x = u, y = v, z = u2 4. (a) x = u, y = v, z = v 1 + u2 (b) x = u, y = v, z = 1 2 5 v − 3 3 5. (a) x = 5 cos u, y = 5 sin u, z = v; 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1 (b) x = 2 cos u, y = v, z = 2 sin u; 0 ≤ u ≤ 2π, 1 ≤ v ≤ 3 6. (a) x = u, y = 1 − u, z = v; −1 ≤ v ≤ 1 (b) x = u, y = 5 + 2v, z = v; 0 ≤ u ≤ 3 8. x = u, y = eu cos v, z = eu sin v 7. x = u, y = sin u cos v, z = sin u sin v 9. x = r cos θ, y = r sin θ, z = 1 1 + r2 10. x = r cos θ, y = r sin θ, z = e−r 2 11. x = r cos θ, y = r sin θ, z = 2r2 cos θ sin θ 12. x = r cos θ, y = r sin θ, z = r2 (cos2 θ − sin2 θ) 13. x = r cos θ, y = r sin θ, z = √ 9 − r2 ; r ≤ √ 5 14. x = r cos θ, y = r sin θ, z = r; r ≤ 3 √ 3 1 1 15. x = ρ cos θ, y = ρ sin θ, z = ρ 2 2 2 16. x = 3 cos θ, y = 3 sin θ, z = 3 cot φ 17. z = x − 2y; a plane 18. y = x2 + z 2 , 0 ≤ y ≤ 4; part of a circular paraboloid 19. (x/3)2 + (y/2)2 = 1; 2 ≤ z ≤ 4; part of an elliptic cylinder Exercise Set 15.4 667 20. z = x2 + y 2 ; 0 ≤ z ≤ 4; part of a circular paraboloid 21. (x/3)2 + (y/4)2 = z 2 ; 0 ≤ z ≤ 1; part of an elliptic cone 22. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid 23. (a) x = r cos θ, y = r sin θ, z = r, 0 ≤ r ≤ 2; x = u, y = v, z = 24. (a) I: x = r cos θ, y = r sin θ, z = r2 , 0 ≤ r ≤ √ √ u2 + v 2 ; 0 ≤ u2 + v 2 ≤ 4 2; II: x = u, y = v, z = u2 + v 2 ; u2 + v 2 ≤ 2 25. (a) 0 ≤ u ≤ 3, 0 ≤ v ≤ π (b) 0 ≤ u ≤ 4, −π/2 ≤ v ≤ π/2 26. (a) 0 ≤ u ≤ 6, −π ≤ v ≤ 0 (b) 0 ≤ u ≤ 5, π/2 ≤ v ≤ 3π/2 27. (a) 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π (b) 0 ≤ φ ≤ π, 0 ≤ θ ≤ π 28. (a) π/2 ≤ φ ≤ π, 0 ≤ θ ≤ 2π (b) 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2 29. u = 1, v = 2, ru × rv = −2i − 4j + k; 2x + 4y − z = 5 30. u = 1, v = 2, ru × rv = −4i − 2j + 8k; 2x + y − 4z = −6 31. u = 0, v = 1, ru × rv = 6k; z = 0 32. ru × rv = 2i − j − 3k; 2x − y − 3z = −4 √ √ 33. ru × rv = ( 2/2)i − ( 2/2)j + (1/2)k; x − y + √ π 2 2 z= 2 8 √ 34. ru × rv = 2i − ln 2k; 2x − (ln 2)z = 0   9 − y 2 , zx = 0, zy = −y/ 9 − y 2 , zx2 + zy2 + 1 = 9/(9 − y 2 ),
2
3
2 3  S= dy dx = 3π dx = 6π 9 − y2 0 −3 0 35. z =
36. z = 8 − 2x − 2y, zx2 + zy2 + 1 = 4 + 4 + 1 = 9, S = 4

4−x 0 0 4 3(4 − x)dx = 24 3 dy dx = 0 37. z 2 = 4x2 + 4y 2 , 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
1
x√ √
1 √ zx2 + zy2 + 1 = (16x2 + 16y 2 )/z 2 + 1 = 5, S = 5 dy dx = 5 (x − x2 )dx = 5/6 0 x2 0 38. z 2 = x2 + y 2 , zx = x/z, zy = y/z, zx2 + zy2 + 1 = (z 2 + y 2 )/z 2 + 1 = 2,

√
π/2
2 cos θ √ √
π/2 √ S= 2 dA = 2 2 r dr dθ = 4 2 cos2 θ dθ = 2π R 0 0 0 39. zx = −2x, zy = −2y, zx2 + zy2 + 1 = 4x2 + 4y 2 + 1,


2π
1  S= 4x2 + 4y 2 + 1 dA = r 4r2 + 1 dr dθ R 0 0
2π √ 1 √ = dθ = (5 5 − 1)π/6 (5 5 − 1) 12 0 668 Chapter 15 40. zx = 2, zy = 2y, zx2 + zy2 + 1 = 5 + 4y 2 ,
1
y
1  √ S= 5 + 4y 2 dx dy = y 5 + 4y 2 dy = (27 − 5 5)/12 0 0 0 41. ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj,
2π
2  √ √ √ ∂r/∂u × ∂r/∂v = u 4u2 + 1; S = u 4u2 + 1 du dv = (17 17 − 5 5)π/6 0 1 42. ∂r/∂u = cos vi + sin vj + k, ∂r/∂v = −u sin vi + u cos vj, √
π/2
2v √ √ 2 3 ∂r/∂u × ∂r/∂v = 2u; S = 2 u du dv = π 12 0 0 43. zx = y, zy = x, zx2 + zy2 + 1 = x2 + y 2 + 1,


π/6
3 
π/6 √ √ 1 S= x2 + y 2 + 1 dA = r r2 + 1 dr dθ = (10 10 − 1) dθ = (10 10 − 1)π/18 3 0 0 0 R 44. zx = x, zy = y, zx2 + zy2 + 1 = x2 + y 2 + 1,

2π
√8 

 26 2π 2 2 x + y + 1 dA = r r2 + 1 dr dθ = dθ = 52π/3 S= 3 0 0 0 R 45. On the sphere, zx = −x/z and zy = −y/z so zx2 + zy2 + 1 = (x2 + y 2 + z 2 )/z 2 = 16/(16 − x2 − y 2 ); the planes z = 1 and z = 2 intersect the sphere along the circles x2 + y 2 = 15 and x2 + y 2 = 12;


2π
√15
2π 4 4r √  S= dA = dr dθ = 4 dθ = 8π √ 16 − r2 16 − x2 − y 2 0 0 12 R 46. On the sphere, zx = −x/z and zy = −y/z so zx2 + zy2 + 1 = (x2 + y 2 + z 2 )/z 2 = 8/(8 − x2 − y 2 ); the cone cuts the sphere in the circle x2 + y 2 = 4;
2π
2 √ √
2π √ 2 2r √ dr dθ = (8 − 4 2) dθ = 8(2 − 2)π S= 2 8−r 0 0 0 47. r(u, v) = a cos u sin vi + a sin u sin vj + a cos vk, ru × rv  = a2 sin v,
π
π
2π 2 2 S= a sin v du dv = 2πa sin v dv = 4πa2 0 0 0
h
48. r = r cos ui + r sin uj + vk, ru × rv  = r; S = 2π r du dv = 2πrh 0 0 x y h h h2 x2 + h2 y 2  , zy =  , zx2 + zy2 + 1 = 2 2 + 1 = (a2 + h2 )/a2 , a x2 + y 2 a x2 + y 2 a (x + y 2 )
2π
a √ 2
2π  1  2 a + h2 2 S= r dr dθ = a a + h dθ = πa a2 + h2 a 2 0 0 0 49. zx = Exercise Set 15.4 669 50. (a) Revolving a point (a0 , 0, b0 ) of the xz-plane around the z-axis generates a circle, an equation of which is r = a0 cos ui + a0 sin uj + b0 k, 0 ≤ u ≤ 2π. A point on the circle (x − a)2 + z 2 = b2 which generates the torus can be written r = (a + b cos v)i + b sin vk, 0 ≤ v ≤ 2π. Set a0 = a + b cos v and b0 = a + b sin v and use the first result: any point on the torus can thus be written in the form r = (a + b cos v) cos ui + (a + b cos v) sin uj + b sin vk, which yields the result. 51. ∂r/∂u = −(a + b cos v) sin ui + (a + b cos v) cos uj, ∂r/∂v = −b sin v cos ui − b sin v sin uj + b cos vk, ∂r/∂u × ∂r/∂v = b(a + b cos v);
2π
2π S= b(a + b cos v)du dv = 4π 2 ab 0 0 52. ru × rv  = √
u2 4π
5
 u2 + 1; S = 0 0 + 1 du dv = 4π 5  u2 + 1 du = 174.7199011 0 53. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, ru × rv  = | cos v|;
2π
2.86204 S= | cos v| dv du ≈ 9.099 0 0.27955 54. (a) Let S1 be the set of points (x, y, z) which satisfy the equation x2/3 + y 2/3 + z 2/3 = a2/3 , and let S2 be the set of points (x, y, z) where x = a(sin φ cos θ)3 , y = a(sin φ sin θ)3 , z = a cos3 φ, 0 ≤ φ ≤ π, 0 ≤ θ < 2π. If (x, y, z) is a point of S2 then x2/3 + y 2/3 + z 2/3 = a2/3 [(sin φ cos θ)3 + (sin φ sin θ)3 + cos3 φ] = a2/3 so (x, y, z) belongs to S1 . If (x, y, z) is a point of S1 then x2/3 + y 2/3 + z 2/3 = a2/3 . Let x1 = x1/3 , y1 = y 1/3 , z1 = z 1/3 , a1 = a1/3 . Then x21 + y12 + z12 = a21 , so in spherical coordinates x1 = a1 sin φ cos θ, y1 = a1 sin φ sin θ, z1 = a1 cos φ, with   y 1/3 z 1/3 y1 z1 −1 = tan−1 , φ = cos−1 = cos−1 . Then θ = tan x1 x a1 a x = x31 = a31 (sin φ cos θ)3 = a(sin φ cos θ)3 , similarly y = a(sin φ sin θ)3 , z = a cos φ so (x, y, z) belongs to S2 . Thus S1 = S2 (b) Let a = 1 and r = (cos θ sin φ)3 i + (sin θ sin φ)3 j + cos3 φk, then
π/2
π/2 S=8 rθ × rφ dφ dθ 0
0 π/2
= 72 0 π/2  sin θ cos θ sin4 φ cos φ cos2 φ + sin2 φ sin2 θ cos2 θ dθ dφ ≈ 4.4506 0 55. (a) (x/a)2 +(y/b)2 +(z/c)2 = sin2 φ cos2 θ+sin2 φ sin2 θ+cos2 φ = sin2 φ+cos2 φ = 1, an ellipsoid (b) r(φ, θ) = 2 sin φ cos θ, 3 sin φ sin θ, 4 cos φ ; rφ ×rθ = 2 6 sin2 φ cos θ, 4 sin2 φ sin θ, 3 cos φ sin φ ,  rφ × rθ  = 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ,
2π
π  S= 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ dφ dθ ≈ 111.5457699 0 0 670 Chapter 15 56. (a) x = v cos u, y = v sin u, z = f (v), for example z (c) x 57. 58. x 2 y + a x 2 + a 59. − x = v cos u, y = v sin u, z = 1/v 2 (b) x 2 y 2 b y 2 − b − a + y 2 b z 2 c z 2 c + = 1, ellipsoid = 1, hyperboloid of one sheet z 2 c = 1, hyperboloid of two sheets EXERCISE SET 15.5

1
2
1
1 −1
0 1/2
π −1

1
1/2 π
2 0
y2 0
1/3
z 2

−1 −1 0 π/4
1


0 0
x2
π/4 3


x 3 0 0 √ 9−z 2
xy dy dx dz = 0
0 0 3
x2
0
ln z 3 0
1
0 x 2
√ 4−x2
1 x
3−x2 −y 2 2
0 0 0
0
2
2
8. 1 z 0 y dx dy dz = x2 + y 2
1 √ 1 cos y dy = 2/8 4 1 (81 − 18z 2 + z 4 )dz = 81/5 8   1 5 3 3 x − x + x2 dx = 118/3 2 2 √ 4−x2 0 2 = √ 3y 3 1/3 1 1 (1 − cos πx)dx = + 2 12 [2x(4 − x2 ) − 2xy 2 ]dy dx x dz dy dx = −5+x2 +y 2 3 0 (xz − x)dz dx = 1 7.

x2 xey dy dz dx = 6. 1 3 x dx dz = 2 1/2 π/4 x cos y dx dy = 0 5.
1
 47 1 7 1 5 1 y + y − y dy = 3 2 6 3 3 0 √ 9−z 2  (yz + yz)dz dy = −1 x cos y dz dx dy = 0 2 2 0 4. 1 x sin xy dy dx = 2 y2 yz dx dz dy = 3. 0 (10/3 + 2z 2 )dz = 8 −1 0 zx sin xy dz dy dx = 1/3 1 (1/3 + y 2 + z 2 )dy dz = 0 2.
2 (x2 + y 2 + z 2 )dx dy dz = 1. 2
z 2 4 x(4 − x2 )3/2 dx = 128/15 3 π dy dz = 3
1 2 π (2 − z)dz = π/6 3 √ 3−2 4π Exercise Set 15.5

π 671
1
π/6 9.
π
0 0
1 0
1−x2
y 10.
1 0 0
√ 2


x −1 −1 0 √ 2
2−x2 11.
x
π/2 0 0
π/2 0
0
xy 12. π/2

3
2
√ 1 13. 0 −2 1
1 √ 1−x2
π/6
√1−x2 −y2 0
2 e−x −y 2 −z 2
4 y
0 (4−x)/2
0 4 dz dy dx ≈ 2.381
(12−3x−6y)/4 4
(4−x)/2 0 1 0 0

1−x
√ y
1 1−x dz dy dx = 16. V = 0 0

2 4
0 0 4−y
17. V = 2
y 0 18. V = 2

1 0
0 0 1 2 (1 − x)3/2 dx = 4/15 3
4 x2 y dz dx dy = 0
y dy dx = 0 0
√1−y2 √ 0 dz dy dx = 2 x2 0 1 0 1 (12 − 3x − 6y)dy dx 4 3 (4 − x)2 dx = 4 16 =
√ y cos y dy = (5π − 6 3)/12 π/6 dz dy dx =
π/2 0 15. V =
0 1 3 x (2 − x2 )2 dx = 1/6 4 x + z2 dz dy dx ≈ 9.425 y 14. 8 0 √ 2 y sin x dx dy = 0 y
π/2 cos(z/y)dz dx dy = π/6 1 (1 − x2 )3 dx = 32/105 3 1 xy(2 − x2 )2 dy dx = 2 xyz dz dy dx = 0 1 y 2 dy dx = 0 (1 − 3/π)x dx = π(π − 3)/2 0 1−x2 y dz dy dx = −1 π x[1 − cos(πy/6)]dy dx = xy sin yz dz dy dx = 0
1 2  (4 − y)dy dx = 2 0 
1 − y 2 dx dy = 0  1 4 8 − 4x + x dx = 256/15 2 2 1 y  1 − y 2 dy = 1/3 0 19. The projection of the curve of intersection onto the xy-plane is x2 + y 2 = 1,
1
√1−x2
4−3y2 (a) V = f (x, y, z)dz dy dx √ −1
1 − 1−x2 4x2 +y 2
√1−y2
(b) V = √ −1 − 4−3y 2 f (x, y, z)dz dx dy 1−y 2 4x2 +y 2 20. The projection of the curve of intersection onto the xy-plane is 2x2 + y 2 = 4,
√2
√4−2x2
8−x2 −y2 f (x, y, z)dz dy dx (a) V = √ √ − 2
2 (b) V = −2 − 4−2x2 3x2 +y 2
√(4−y2 )/2
√ − 8−x2 −y 2 f (x, y, z)dz dx dy (4−y 2 )/2 3x2 +y 2 672 Chapter 15 21. The projection of the curve of intersection onto the xy-plane is x2 + y 2 = 1,
1
√1−x2
4−3y2 V =4 dz dy dx 0 4x2 +y 2 0 22. The projection of the curve of intersection onto the xy-plane is 2x2 + y 2 = 4,
√2
√4−2x2
8−x2 −y2 V =4 dz dy dx 0
3x2 +y 2 0
3 √ 9−x2 /3
23. V = 2
x+3 dz dy dx −3 0 √ 1−x2
√ 1−x2 dz dy dx 0 z
24. V = 8 0 25. (a) 1 0 0 z (b) z (c) (0, 9, 9) (0, 0, 1) (0, 0, 1) y y (0, –1, 0) x (1, 0, 0) (3, 9, 0) (1, 2, 0) y x x z 26. (a) z (b) (0, 0, 2) (0, 0, 2) (0, 2, 0) y y (3, 9, 0) (2, 0, 0) x x z (c) (0, 0, 4) y (2, 2, 0) x
1
1−x
27. V =
1−x−y 1
1−x
dz dy dx = 1/6, fave = 6 0 0 0 1−x−y (x + y + z) dz dy dx = 0 0 0 28. The integrand is an odd function of each of x, y, and z, so the answer is zero. 3 4 Exercise Set 15.5 673 3π 29. The volume V = √ , and thus 2 √


 2 x2 + y 2 + z 2 dV rave = 3π G √
1/√2
√1−2x2
6−7x2 −y2  2 x2 + y 2 + z 2 dz dy dx ≈ 3.291 = √ √ 3π −1/ 2 − 1−2x2 5x2 +5y2 30. V = 1, dave =
a

1 V
1 0 1
0 b(1−x/a)
1  (x − z)2 + (y − z)2 + z 2 dx dy dz ≈ 0.771 0
c(1−x/a−y/b)
b
a(1−y/b) c(1−x/a−y/b) dz dy dx, 31. (a) 0
c 0

a(1−z/c) 0 dz dx dy,
b(1−x/a−z/c) 0
a 0
c(1−x/a) 0 b(1−x/a−z/c) dy dx dz, 0
0 c

b(1−z/c) 0 dy dz dx,
a(1−y/b−z/c) 0
b 0 c(1−y/b) 0
a(1−y/b−z/c) dx dy dz, 0 0 0 dx dz dy 0 0 0 (b) Use the first integral in Part (a) to get
a
b(1−x/a)
a x 2 1 x y 1 dy dx = bc 1 − c 1− − dx = abc a b 2 a 6 0 0 0 √ √

2 2
2 2 2 2 a b 1−x /a c 1−x /a −y /b 32. V = 8 dz dy dx 0
2 0 0 √ 4−x2

5 33. (a) f (x, y, z) dz dy dx
0 9
0 √ 3− x 0

√ 3− x (b) f (x, y, z) dz dy dx 0
0 3
8−y f (x, y, z) dz dy dx 0
√9−x2 −y2 34. (a) 4−x2 (c) y √ 9−x2

2 0 y f (x, y, z)dz dy dx
0 4
0 x/2
(b) 0
2 f (x, y, z)dz dy dx 0 0
2 4−x2
4−y (c) f (x, y, z)dz dy dx 0 0 x2 0 35. (a) At any point outside the closed sphere {x2 + y 2 + z 2 ≤ 1} the integrand is negative, so to maximize the integral it suffices to include all points inside the sphere; hence the maximum value is taken on the region G = {x2 + y 2 + z 2 ≤ 1}.
2π
π
1 π2 (b) 4.934802202 (c) (1 − ρ2 )ρ dρ dφ dθ = 2 0 0 0





b d  36. b d f (x)g(y)h(z)dz dy dx = a c k  f (x)g(y) a h(z)dz dy dx c

b = k f (x) = k
b  h(z)dz c f (x)dx a
g(y)dy dx a
d
d g(y)dy c  h(z)dz k 674 Chapter 15 

x dx 1 37. (a) −1 

1 y dy 0

ln 3 2x (b) sin z dz = (0)(1/3)(1) = 0 0 1 π/2 2 e dx e dy 0 ln 2 y −z e 0 dz = [(e2 − 1)/2](2)(1/2) = (e2 − 1)/2 0 EXERCISE SET 15.6 1. (a) m1 and m3 are equidistant from x = 5, but m3 has a greater mass, so the sum is positive. (b) Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is 5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35a = 0, a = 50/7. The fulcrum should be placed 50/7 units to the right of m1 . 2. (a) The sum must be negative, since m1 , m2 and m3 are all to the left of the fulcrum, and the magnitude of the moment of m1 about x = 4 is by itself greater than the moment of m about x = 4 (i.e. 40 > 28), so even if we replace the masses of m2 and m3 with 0, the sum is negative. (b) At equilibrium, 10(0 − 4) + 3(2 − 4) + 4(3 − 4) + m(6 − 4) = 0, m = 25
1
1 1 x dy dx = , y = 2 3. A = 1, x = 0 1
1 y dy dx = 0 0 1 2

1 2 4. A = 2, x = 0
x dy dx, and the region of integration is symmetric with respect to the x-axes G and the integrand is an odd function of x, so x = 0. Likewise, y = 0.


5. A = 1/2, 1


x x dA = x dy dx = 1/3, 0 R
1
y dA = 0 y dy dx = 1/6; 0 R x 0 centroid (2/3, 1/3)
1


x2 6. A = dy dx = 1/3, 0 0


1
1
R

2−x2 dy dx = 7/6, x


1
π 8. A = , 4 0
1
2−x2 x dA = x dy dx = 5/12, 0 R x 2−x2 y dA = y dy dx = 19/15; centroid (5/14, 38/35) 0 R x dy dx = 1/4, 0 0 7. A = 0 x2 y dy dx = 1/10; centroid (3/4, 3/10) 0

1 x2 y dA = R
x dA = x


1
x dA = x dy dx = 0 R √ 1−x2 0 1 4 4 ,x= , y= by symmetry 3 3π 3π 9. x = 0 from the symmetry of the region,


π
b 2 4(b3 − a3 ) 1 y dA = r2 sin θ dr dθ = (b3 − a3 ); centroid x = 0, y = A = π(b2 − a2 ), . 2 3 3π(b2 − a2 ) 0 a R Exercise Set 15.6 675 10. y = 0 from the symmetry of the region, A = πa2 /2,  


π/2
a 4a ,0 x dA = r2 cos θ dr dθ = 2a3 /3; centroid 3π −π/2 0 R


1
0 R

1 1 |x + y − 1| dx dy δ(x, y)dA = 11. M = 0
1−x (1 − x − y) dy + = 0
0
1 x=3 (x + y − 1) dy dx =
1 1 1−x 1 
1−x 0 1 x(x + y − 1) dy dx = 2 1−x
x(1 − x − y) dy + xδ(x, y) dy dx = 3 0 1 3 0 0 1 1 By symmetry, y = as well; center of gravity (1/2, 1/2) 2

1 12. x = xδ(x, y) dA, and the integrand is an odd function of x while the region is symmetric M G with respect to the y-axis, thus x = 0; likewise y = 0.
1

√ x 13. M = 1
(x + y)dy dx = 13/20, Mx = 0 0
1
(x + y)y dy dx = 3/10, 0 √ x My = √ x 0 (x + y)x dy dx = 19/42, x = My /M = 190/273, y = Mx /M = 6/13; 0 0 the mass is 13/20 and the center of gravity is at (190/273, 6/13).
π
sin x 14. M = y dy dx = π/4, x = π/2 from the symmetry of the density and the region, 0 0
π
sin x 16 ; mass π/4, center of gravity y dy dx = 4/9, y = Mx /M = 9π  2 Mx = 0 0
π/2
 π 16 , . 2 9π a r3 sin θ cos θ dr dθ = a4 /8, x = y from the symmetry of the density and the 0
π/2
a = r4 sin θ cos2 θ dr dθ = a5 /15, x = 8a/15; mass a4 /8, center of gravity 15. M = 0 region, My 0 0 (8a/15, 8a/15).
π
1 r3 dr dθ = π/4, x = 0 from the symmetry of density and region, 16. M = 0
0 π
1 0 8 ; mass π/4, center of gravity r sin θ dr dθ = 2/5, y = 5π
1  4 Mx = 0
1
17. V = 1, x = 0 0 0 1 1 1 x dz dy dx = , similarly y = z = ; centroid 2 2


2 18. V = πr h = 2π, x = y = 0 by symmetry, 2
2π 1 1 1 , , 2 2 2
z dz dy dx = G = (0, 0, 1)
  8 0, . 5π  1 rz dr dθ dz = 2π, centroid 0 0 0 676 Chapter 15 19. x = y = z from the symmetry of the region, V = 1/6,


1 1 1−x 1−x−y x= x dz dy dx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4) V 0 0 0 20. The solid is described by −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , 0 ≤ x ≤ 1 − z; 2



1
1−y2
1−z 1−z 4 1 1 1−y 5 , y = 0 by symmetry, dx dz dy = , x = x dx dz dy = V = 5 V 14 −1 0 0 −1 0 0 2
 

1−z 5 2 1 1 1−y 2 , 0, . z= z dx dz dy = ; the centroid is V −1 0 7 14 7 0 21. x = 1/2 and y = 0 from the symmetry of the region,



1
1
1 1 V = dz dy dx = 4/3, z = z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5) V 0 −1 y 2 G 22. x = y from the symmetry of the region,



2
2
xy 1 dz dy dx = 4, x = x dV = (1/4)(16/3) = 4/3, V = V 0 0 0 G


1 z= z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9) V G 23. x = y = z from the symmetry of the region, V = πa3 /6,

√a2 −x2
√a2 −x2 −y2

√a2 −x2  1 a 1 a x= x dz dy dx = x a2 − x2 − y 2 dy dx V 0 0 V 0 0 0

1 π/2 a 2  2 6 = r a − r2 cos θ dr dθ = (πa4 /16) = 3a/8; centroid (3a/8, 3a/8, 3a/8) V 0 πa3 0 24. x = y = 0 from the symmetry of the region, V = 2πa3 /3

√a2 −x2
√a2 −x2 −y2

√a2 −x2 1 2 1 a 1 a z= z dz dy dx = (a − x2 − y 2 )dy dx V −a −√a2 −x2 0 V −a −√a2 −x2 2

1 2π a 1 2 3 (a − r2 )r dr dθ = = (πa4 /4) = 3a/8; centroid (0, 0, 3a/8) V 0 2πa3 0 2
a
a
a (a − x)dz dy dx = a4 /2, y = z = a/2 from the symmetry of density and 25. M = 0 0 region, x = 0 1 M
a
a
a x(a − x)dz dy dx = (2/a4 )(a5 /6) = a/3; 0 0 0 mass a4 /2, center of gravity (a/3, a/2, a/2) Exercise Set 15.6
a 677 √ a2 −x2
26. M =
√ − a2 −x2 −a 1 M and region, z = h (h − z)dz dy dx = 0


z(h − z)dV = 1 2 2 πa h , x = y = 0 from the symmetry of density 2 2 (πa2 h3 /6) = h/3; πa2 h2 G 2 2 mass πa h /2, center of gravity (0, 0, h/3)
1
1
1−y 2 27. M = yz dz dy dx = 1/6, x = 0 by the symmetry of density and region, −1 y= 0 0


1 M y 2 z dV = (6)(8/105) = 16/35, z =


1 M yz 2 dV = (6)(1/12) = 1/2; G G mass 1/6, center of gravity (0, 16/35, 1/2)
3

9−x2 1 28. M = xz dz dy dx = 81/8, x = 0 1 y= M 0 0


1 M


x2 z dV = (8/81)(81/5) = 8/5,


G 1 xyz dV = (8/81)(243/8) = 3, z = M G xz 2 dV = (8/81)(27/4) = 2/3; G mass 81/8, center of gravity (8/5, 3, 2/3)
1
1 k(x2 + y 2 )dy dx = 2k/3, x = y from the symmetry of density and region, 29. (a) M = 0 x= 1 M 0

kx(x2 + y 2 )dA = 3 (5k/12) = 5/8; center of gravity (5/8, 5/8) 2k R (b) y = 1/2 from the symmetry of density and region,


1
1 1 M= kx dy dx = k/2, x = kx2 dA = (2/k)(k/3) = 2/3, M 0 0 center of gravity (2/3, 1/2) R 30. (a) x = y = z from the symmetry of density and region,
1
1
1 M= k(x2 + y 2 + z 2 )dz dy dx = k, 0 x= 1 M 0


0 kx(x2 + y 2 + z 2 )dV = (1/k)(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12) G (b) x = y = z from the symmetry of density and region,
1
1
1 M= k(x + y + z)dz dy dx = 3k/2, 0 1 x= M 0


0 kx(x + y + z)dV = G 2 (5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9) 3k 678 Chapter 15



31. V = π
sin x
1/(1+x2 +y 2 ) dV = G 1 x= V dz dy dx = 0.666633, 0 0 0


xdV = 1.177406, y = 1 V


G ydV = 0.353554, z = 1 V


G zdV = 0.231557 G 32. (b) Use polar coordinates for x and y to get



2π
a
1/(1+r2 ) V = dV = r dz dr dθ = π ln(1 + a2 ), 0 G z= 0 0


1 V zdV = a2 2(1 + a2 ) ln(1 + a2 ) G Thus lim z = a→0+ lim z = a→0+ 1 ; lim z = 0. 2 a→+∞ 1 ; lim z = 0 2 a→+∞ (c) Solve z = 1/4 for a to obtain a ≈ 1.980291. 33. Let x = r cos θ, y = r sin θ, and dA = r dr dθ in formulas (11) and (12).
2π
a(1+sin θ) r dr dθ = 3πa2 /2, 34. x = 0 from the symmetry of the region, A = y= 1 A
2π
0 a(1+sin θ) r2 sin θ dr dθ = 0 0 0 2 (5πa3 /4) = 5a/6; centroid (0, 5a/6) 3πa2
π/2
sin 2θ 35. x = y from the symmetry of the region, A = x= 1 A
π/2
r dr dθ = π/8, 0 sin 2θ r2 cos θ dr dθ = (8/π)(16/105) = 0 0 0 128 ; centroid 105π  128 128 , 105π 105π  36. x = 3/2 and y = 1 from the symmetry of the region,



x dA = xA = (3/2)(6) = 9, y dA = yA = (1)(6) = 6 R R 37. x = 0 from the symmetry of the region, πa2 /2 is the area of the semicircle, 2πy is the distance traveled by the centroid to generate the sphere so 4πa3 /3 = (πa2 /2)(2πy), y = 4a/(3π)  1 2 38. (a) V = πa 2    1 4a = π(3π + 4)a3 2π a + 3π 3 √   4a 2 a+ so (b) the distance between the centroid and the line is 2 3π   √  1 2 4a 1√ 2 V = a+ = 2π 2π(3π + 4)a3 πa 2 2 3π 6 Exercise Set 15.7 679 39. x = k so V = (πab)(2πk) = 2π 2 abk 40. y = 4 from the symmetry of the region,
2
8−x2 A= dy dx = 64/3 so V = (64/3)[2π(4)] = 512π/3 −2 x2 1 41. The region generates a cone of volume πab2 when it is revolved about the x-axis, the area of the 3   1 1 1 1 2 region is ab so πab = ab (2πy), y = b/3. A cone of volume πa2 b is generated when the 2 3 2 3   1 2 1 ab (2πx), x = a/3. The centroid is (a/3, b/3). region is revolved about the y-axis so πa b = 3 2 42. The centroid of the circle which generates the tube travels a distance
4π  √ √ √ sin2 t + cos2 t + 1/16 dt = 17π, so V = π(1/2)2 17π = 17π 2 /4. s= 0
a
b y 2 δ dy dx = 43. Ix =
0 a
0 b (x2 + y 2 )δ dy dx = Iz = 0
0
2π
1 δab3 , Iy = 3
a b x2 δ dy dx = 0 0 1 δab(a2 + b2 ) 3
a 2 3 44. Ix = 2π
a 4 r3 cos2 θ δ dr dθ = δπa4 /4 = Ix ; r sin θ δ dr dθ = δπa /4; Iy = 0 1 3 δa b, 3 0 0 0 4 Iz = Ix + Iy = δπa /2 EXERCISE SET 15.7
2π
1 √ 1−r 2
1.

2π 1 0 1 (1 − r2 )r dr dθ = 2 π/2
zr dz dr dθ = 0
π/2 0 0 0
cos θ

r2 2.
0 π/2
π/2

0
0 2π
π/4
0
π/2 0
π/2
a sec φ 2π 0
2 ρ sin φ dρ dφ dθ = 4. 0 0 0 0 5. f (r, θ, z) = z π/2 0 ρ3 sin φ cos φ dρ dφ dθ = 0 0 r sin θ dr dθ = 1 3. 1 dθ = π/4 8 3 0 0 2π cos θ r sin θ dz dr dθ = 0
0 π/4 1 cos4 θ sin θ dθ = 1/20 4 1 sin φ cos φ dφ dθ = 4 1 3 a sec3 φ sin φ dφ dθ = 3 6. f (r, θ, z) = sin θ z z y x x y
π/2 0
0 2π 1 dθ = π/16 8 1 3 a dθ = πa3 /3 6 680 Chapter 15 7. f (ρ, φ, θ) = ρ cos φ 8. f (ρ, φ, θ) = 1 z z a y x x
2π
3

9 9. V = 2π
r2 0

2π 0 √ 9−r 2 0
2π
2 r dz dr dθ = 2 10. V = 2 0 0 2π r(9 − r2 )dr dθ = 0
2
3 r dz dr dθ = 0 y r 0 0 =  0 9 − r2 dr dθ
√ 2 (27 − 5 5) 3 81 dθ = 81π/2 4 2π √ dθ = 4(27 − 5 5)π/3 0 11. r2 + z 2 = 20 intersects z = r2 in a circle of radius 2; the volume consists of two portions, one inside √ the cylinder r = 20 and one outside that cylinder:
2π
2
r2
2π
√20
√20−r2 V= r dz dr dθ + r dz dr dθ √ √ 0
− 20−r 2 0 2π
2 r r2 + = 0 =  0 2

2π 20 − r2 dr dθ + 0 − 20−r 2 √ 20  2r 0 20 − r2 dr dθ 2

2π √ 4 128 2π 152 80 √ dθ + dθ = (10 5 − 13) π+ π 5 3 3 0 3 3 0 12. z = hr/a intersects z = h in a circle of radius a,
2π
a
h
2π
a
2π h 1 2 (ar − r2 )dr dθ = a h dθ = πa2 h/3 V = r dz dr dθ = a 6 0 0 hr/a 0 0 0
2π
π/3

4 2π
π/3 ρ2 sin φ dρ dφ dθ = 13. V = 0
0 2π
0 π/4
0
2 0 2π
π/4 ρ2 sin φ dρ dφ dθ = 14. V = 0 0 1 0 0 32 64 sin φ dφ dθ = 3 3
2π dθ = 64π/3 0 √ 7 7 sin φ dφ dθ = (2 − 2) 3 6
2π √ dθ = 7(2 − 2)π/3 0 15. In spherical coordinates the sphere and the plane z = a are ρ = 2a and ρ = a sec φ, respectively. They intersect at φ = π/3,
2π
π/3
a sec φ
2π
π/2
2a V= ρ2 sin φ dρ dφ dθ + ρ2 sin φ dρ dφ dθ 0
0 2π
0 π/3
0 2π 1 3 a sec3 φ sin φ dφ dθ + 3 0 0 0
2π
2π 1 4 = a3 dθ + a3 dθ = 11πa3 /3 2 3 0 0
π/3 π/2 = π/3 0 8 3 a sin φ dφ dθ 3 Exercise Set 15.7
2π 681
π/2
π/2 0 π/4
a
0 0 0
a2 −r 2 π
π/2
1 3 e−ρ ρ2 sin φ dρ dφ dθ = 18. 0 0 π/2
√ 8 π/4
0 0 2π
π/4
√
√ 9 2 2π 9 sin φ dφ dθ = dθ = 9 2π 2 0 a (a2 r3 − r5 ) cos2 θ dr dθ 1 6 a 12
0 π/2 cos2 θ dθ = πa6 /48 0 1 (1 − e−1 ) 3
π
0 π/2 sin φ dφ dθ = (1 − e−1 )π/3 0 √ ρ4 cos2 φ sin φ dρ dφ dθ = 32(2 2 − 1)π/15 19.
π/2 0 0 = 0 π/2 r3 cos2 θ dz dr dθ = 17.

ρ sin φ dρ dφ dθ = 0
2π 2 16. V =

3 0 π
3 ρ3 sin φ dρ dφ dθ = 81π 20. 0 0 0
2

4   
4  
π/3 1 3 √ dz r dr tan θ dθ 1 + z2 −2 1 π/6   15  4 1 √ − ln 3 ≈ 16.97774196 = −2 ln( 5 − 2) 2 3 2 π/3 r tan3 θ √ dθ dr dz = 1 + z2 21. (a) −2 1 π/6 
2 The region is a cylindrical wedge. (b) To convert to rectangular √ √ coordinates observe that the rays θ = π/6, θ = π/3 correspond to the lines y = x/ 3, y = 3x. Then dx dy dz = r dr dθ dz and tan θ = y/x, hence
4
√3x
2 (y/x)3 y3 √ √ dz dy dx, so f (x, y, z) = . Integral = √ 1 + z2 x3 1 + z 2 1 x/ 3 −2
π/2
√
π/2 1 2 4,294,967,296 √ cos37 θ cos φ dφ dθ = cos37 θ dθ = 2 ≈ 0.008040 18 36 0 755,505,013,725 π/4 22. 0 0

2π
a √ a2 −r 2 23. (a) V = 2 0
2π 0
π
a ρ2 sin φ dρ dφ dθ = 4πa3 /3 (b) V = 0
2 r dz dr dθ = 4πa3 /3 0 0 √ 4−x2
0
√4−x2 −y2 xyz dz dy dx 24. (a) 0 0 0
2 √ 4−x2
=
π/2
2 √ 4−r 2
0 0 1 1 xy(4 − x2 − y 2 )dy dx = 2 8
2 x(4 − x2 )2 dx = 4/3 0 r3 z sin θ cos θ dz dr dθ (b) 0 0 0
π/2
=
π/2
π/2
0 0 2 1 3 8 (4r − r5 ) sin θ cos θ dr dθ = 2 3
π/2 sin θ cos θ dθ = 4/3 0 2 ρ5 sin3 φ cos φ sin θ cos θ dρ dφ dθ (c) 0 0 0
π/2
= 0 0 π/2 32 8 sin3 φ cos φ sin θ cos θ dφ dθ = 3 3
π/2 sin θ cos θ dθ = 4/3 0 682 Chapter 15
2π

3
3
2π 3 (3 − z)r dz dr dθ = 25. M = 0
0
0 a
h 26. M =

2π dθ = 27π/4 0
2π 1 2 1 kh r dr dθ = ka2 h2 dθ = πka2 h2 /2 2 4 0 0 0 0 0 0
2π
2π
π
2π
π
a 1 4 1 ka sin φ dφ dθ = ka4 kρ3 sin φ dρ dφ dθ = dθ = πka4 27. M = 2 0 0 0 0 0 4 0
2π
2π
π
2
2π
π 3 sin φ dφ dθ = 3 28. M = ρ sin φ dρ dφ dθ = dθ = 6π 0 0 1 0 0 2 0 2π
0 r 1 27 r(3 − r)2 dr dθ = 2 8 2π a k zr dz dr dθ = 29. x ¯ = y¯ = 0 from the symmetry of the region,
2π
1 
2π
1
√2−r2 √ r dz dr dθ = (r 2 − r2 − r3 )dr dθ = (8 2 − 7)π/6, V = 0 z¯ = 1 V r2 0
2π
0 √ 2−r 2
1 zr dz dr dθ = 0  centroid r2 0 7 0, 0, √ 16 2 − 14  0 √ 6 √ (7π/12) = 7/(16 2 − 14); (8 2 − 7)π 30. x ¯ = y¯ = 0 from the symmetry of the region, V = 8π/3,


1 2π 2 2 3 z¯ = (4π) = 3/2; centroid (0, 0, 3/2) zr dz dr dθ = V 0 8π 0 r 31. x ¯ = y¯ = z¯ from the symmetry of the region, V = πa3 /6,


1 π/2 π/2 a 3 6 z¯ = ρ cos φ sin φ dρ dφ dθ = (πa4 /16) = 3a/8; 3 V 0 πa 0 0 centroid (3a/8, 3a/8, 3a/8)
2π
π/3
4 ρ2 sin φ dρ dφ dθ = 64π/3, 32. x ¯ = y¯ = 0 from the symmetry of the region, V = z¯ =
1 V 2π
π/3
0 4 ρ3 cos φ sin φ dρ dφ dθ = 0 0 0 0 0 3 (48π) = 9/4; centroid (0, 0, 9/4) 64π
π/2
2 cos θ
33. y¯ = 0 from the symmetry of the region, V = 2
2 x ¯= V π/2
2 cos θ 2 z¯ = V 0
π/2 0
2 cos θ 0 r2 rz dz dr dθ = 0
0
π/2 0 2 cos θ
16 3
0
π/2
0 0 2 cos θ 0 1 r(4 − r2 )2 dr dθ 2 π/2 (1 − sin6 θ)dθ = (16/3)(11π/32) = 11π/6 0 π/2
π/3

2 π/2
π/3 ρ2 sin φ dρ dφ dθ = 35. V = 0 0 4 (π) = 4/3, 3π zr dz dr dθ = 0 0 4 (5π/6) = 10/9; centroid (4/3, 0, 10/9) 3π
4−r 2 34. M = = r2 r2 cos θ dz dr dθ =
r dz dr dθ = 3π/2, 0
r2 π/6 0 √ = 2( 3 − 1)π/3 0 π/6
π/2 4 √ 8 sin φ dφ dθ = ( 3 − 1) dθ 3 3 0 Exercise Set 15.7
683 2π
π/4

1 2π
√ 1 1 sin φ dφ dθ = (2 − 2) 4 8 π/4 3 36. M = ρ sin φ dρ dφ dθ = 0 0 0 0 0
2π dθ = (2 − √ 2)π/4 0 37. x ¯ = y¯ = 0 from the symmetry of density and region,
2π
1
1−r2 (r2 + z 2 )r dz dr dθ = π/4, M= 0 0
1 z¯ = M 2π 0
1
1−r 2 z(r2 +z 2 )r dz dr dθ = (4/π)(11π/120) = 11/30; center of gravity (0, 0, 11/30) 0 0 0
2π
1
38. x ¯ = y¯ = 0 from the symmetry of density and region, M =
1 z¯ = M 2π
1 zr dz dr dθ = π/4, 0
r 0 0 r z 2 r dz dr dθ = (4/π)(2π/15) = 8/15; center of gravity (0, 0, 8/15) 0 0 0 39. x ¯ = y¯ = 0 from the symmetry of density and region,
2π
π/2
a M= kρ3 sin φ dρ dφ dθ = πka4 /2, 0 z¯ = 0
1 M 2π 0
π/2
a kρ4 sin φ cos φ dρ dφ dθ = 0 0 0 2 (πka5 /5) = 2a/5; center of gravity (0, 0, 2a/5) πka4 40. x ¯ = z¯ = 0 from the symmetry of the region, V = 54π/3 − 16π/3 = 38π/3,




1 π π 3 3 2 1 π π 65 y¯ = sin2 φ sin θ dφ dθ ρ sin φ sin θ dρ dφ dθ = V 0 0 2 V 0 0 4
3 1 π 65π sin θ dθ = (65π/4) = 195/152; centroid (0, 195/152, 0) = V 0 8 38π
2π
π
R 41. M = −(ρ/R)3 2 δ0 e 0 0
2π
ρ sin φ dρ dφ dθ = 0 0 0 π 1 (1 − e−1 )R3 δ0 sin φ dφ dθ 3 4 = π(1 − e−1 )δ0 R3 3 42. (a) The sphere and cone intersect in a circle of radius ρ0 sin φ0 , 
θ2
ρ0 sin φ0
√ρ20 −r2
θ2
ρ0 sin φ0   2 2 2 r ρ0 − r − r cot φ0 dr dθ V= r dz dr dθ = θ1
θ2 = θ1 0 r cot φ0 θ1 0 1 3 1 ρ0 (1 − cos3 φ0 − sin3 φ0 cot φ0 )dθ = ρ30 (1 − cos3 φ0 − sin2 φ0 cos φ0 )(θ2 − θ1 ) 3 3 1 3 ρ (1 − cos φ0 )(θ2 − θ1 ). 3 0 (b) From Part (a), the volume of the solid bounded by θ = θ1 , θ = θ2 , φ = φ1 , φ = φ2 , and 1 1 1 ρ = ρ0 is ρ30 (1 − cos φ2 )(θ2 − θ1 ) − ρ30 (1 − cos φ1 )(θ2 − θ1 ) = ρ30 (cos φ1 − cos φ2 )(θ2 − θ1 ) 3 3 3 so the volume of the spherical wedge between ρ = ρ1 and ρ = ρ2 is 1 1 ∆V = ρ32 (cos φ1 − cos φ2 )(θ2 − θ1 ) − ρ31 (cos φ1 − cos φ2 )(θ2 − θ1 ) 3 3 = = 1 3 (ρ − ρ31 )(cos φ1 − cos φ2 )(θ2 − θ1 ) 3 2 684 Chapter 15 (c) d cos φ = − sin φ so from the Mean-Value Theorem cos φ2 −cos φ1 = −(φ2 −φ1 ) sin φ∗ where dφ d 3 φ∗ is between φ1 and φ2 . Similarly ρ = 3ρ2 so ρ32 −ρ31 = 3ρ∗2 (ρ2 −ρ1 ) where ρ∗ is between dρ ρ1 and ρ2 . Thus cos φ1 −cos φ2 = sin φ∗ ∆φ and ρ32 −ρ31 = 3ρ∗2 ∆ρ so ∆V = ρ∗2 sin φ∗ ∆ρ∆φ∆θ.
2π
a

h 2π
a
h 2 43. Iz = r3 dz dr dθ = r δ r dz dr dθ = δ 0
0 2π
0
a 0 0 0
h 2π
a 1 (hr3 cos2 θ + h3 r)dr dθ 3 (r2 cos2 θ + z 2 )δr dz dr dθ = δ 44. Iy = 0 0
0 2π  =δ 0
2π
0
π  1 4 1 π a h cos2 θ + a2 h3 dθ = δ a4 h + a2 h3 4 6 4 3 a2

π
h r2 δ r dz dr dθ = δ 0 a1 2π 0  45. Iz = 0 1 δπa4 h 2
0 2π
a2 a1
h r3 dz dr dθ = 0
a 2π
π
a (ρ2 sin2 φ)δ ρ2 sin φ dρ dφ dθ = δ 46. Iz = 0 0 0 1 δπh(a42 − a41 ) 2 ρ4 sin3 φ dρ dφ dθ = 0 EXERCISE SET 15.8 0 0 1.  ∂(x, y)  1 4 = ∂(u, v)  3 −5    = −17  3.  ∂(x, y)  cos u = ∂(u, v)  sin u  − sin v  = cos u cos v + sin u sin v = cos(u − v) cos v  4.   2(v 2 − u2 )  ∂(x, y)  (u2 + v 2 )2 = ∂(u, v)  4uv   (u2 + v 2 )2 − 2.  4v  = −1 − 16uv −1        = 4/(u2 + v 2 )2 2(v 2 − u2 )  (u2 + v 2 )2  4uv (u2 + v 2 )2  2 5 1 2 ∂(x, y)  2/9 5. x = u + v, y = − u + v; = 9 9 9 9 ∂(u, v)  −1/9 6. x = ln u, y = uv;  ∂(x, y)  1 = ∂(u, v)  4u 8 δπa5 15  ∂(x, y)  1/u = v ∂(u, v)  5/9  1 = 2/9  9  0  =1 u   1  √ √  √ √ √ √ ∂(x, y)  2 2 u + v = 7. x = u + v/ 2, y = v − u/ 2; 1 ∂(u, v)   − √ √ 2 2 v−u   3u1/2  ∂(x, y)  2v 1/2 3/2 1/2 1/2 1/2 = 8. x = u /v , y = v /u ; ∂(u, v)  1/2  −v  2u3/2 u3/2 − 3/2 2v 1 2u1/2 v 1/2 1 √ √ 2 2 u+v 1 √ √ 2 2 v−u     1  =  2v       = √ 1  4 v 2 − u2   Exercise Set 15.8 9. 685   3  ∂(x, y, z) =  1 ∂(u, v, w)  0 1 0 1 0 −2 1    =5     1−v  ∂(x, y, z) =  v − vw ∂(u, v, w)  vw 10.   1/v  ∂(x, y, z) 11. y = v, x = u/y = u/v, z = w − x = w − u/v; = 0 ∂(u, v, w)  −1/v   0  ∂(x, y, z) 12. x = (v + w)/2, y = (u − w)/2, z = (u − v)/2, =  1/2 ∂(u, v, w)  1/2 y 13. −u/v 2 1 u/v 2 1/2 0 −1/2 y 14. 0 0 1 0 −uv uv     = u2 v       = 1/v   1/2 −1/2 0     = −1  4  (3, 4) 4 (0, 2) −u u − uw uw 3 2 1 x x (0, 0) (–1, 0) (0, 0) 2 3 (4, 0) (1, 0) y 15. 1 3 y 16. (0, 3) 2 (2, 0) –3 x 1 3 x –3 17. x = 1 2 2 1 ∂(x, y) 1 1 1 u + v, y = − u + v, = ; 5 5 5 5 ∂(u, v) 5 5

u 1 dAuv = v 5
1 1 1 ∂(x, y) 1 1 1 =− ; u + v, y = u − v, 2 2 2 2 ∂(u, v) 2 2

veuv dAuv = S 3
1 S 18. x = 2 1 2
4 4 1 3 u du dv = ln 3 v 2
1 veuv du dv = 1 0 1 4 (e − e − 3) 2 ∂(x, y) = −2; the boundary curves of the region S in the uv-plane are ∂(u, v)
1
u

1 sin u cos vdAuv = 2 sin u cos v dv du = 1 − sin 2 v = 0, v = u, and u = 1 so 2 2 0 0 19. x = u + v, y = u − v, S  √ 1 ∂(x, y) = − ; the boundary curves of the region S in v/u, y = uv so, from Example 3, ∂(u, v) 2u  



1 1 4 3 2 2 dAuv = the uv-plane are u = 1, u = 3, v = 1, and v = 4 so uv v du dv = 21 2u 2 1 1 20. x = S 686 Chapter 15 ∂(x, y) = 12; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1. ∂(u, v)
2π
1

 12 u2 + v 2 (12) dAuv = 144 r2 dr dθ = 96π Use polar coordinates to obtain 21. x = 3u, y = 4v, 0 S 0 ∂(x, y) = 2; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1. Use ∂(u, v)
2π
1

2 −(4u2 +4v 2 ) e (2) dAuv = 2 re−4r dr dθ = (1 − e−4 )π/2 polar coordinates to obtain 22. x = 2u, y = v, 0 S 0 23. Let S be the region in the uv-plane bounded by u2 + v 2 = 1, so u = 2x, v = 3y,   ∂(x, y)  1/2 0  x = u/2, y = v/3, = = 1/6, use polar coordinates to get 1/3  ∂(u, v)  0 1 6

sin(u2 + v 2 )du dv = 1 6
π/2 1 r sin r2 dr dθ = 0 S 0 1 π π (− cos r2 ) = (1 − cos 1) 24 24 0 ∂(x, y) = ab; A = ab ∂(u, v) 24. u = x/a, v = y/b, x = au, y = bv; 25. x = u/3, y = v/2, z = w,

2π
1 r dr dθ = πab 0 0 ∂(x, y, z) = 1/6; S is the region in uvw-space enclosed by the sphere ∂(u, v, w) u2 + v 2 + w2 = 36 so


u2 1 1 dVuvw = 9 6 54 S = 1 54
2π
π
6 (ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ 0
0 2π
0 π
6 ρ4 sin3 φ cos2 θdρ dφ dθ = 0 0 0 192 π 5 26. Let G1 be the region u2 + v 2 + w2 ≤ 1, with x = au, y = bv, z = cw, spherical coordinates in uvw-space:





Ix = (y 2 + z 2 )dx dy dz = abc (b2 v 2 + c2 w2 ) du dv dw
G 2π
π
G1 1 abc(b2 sin2 φ sin2 θ + c2 cos2 φ)ρ4 sin φ dρ dφ dθ = 0
0 2π = 0 0 abc 2 2 4 (4b sin θ + 2c2 )dθ = πabc(b2 + c2 ) 15 15 27. u = θ = cot−1 (x/y), v = r = 28. u = r = 29. u =   x2 + y 2 x2 + y 2 , v = (θ + π/2)/π = (1/π) tan−1 (y/x) + 1/2 3 2 1 3 x − y, v = − x + y 7 7 7 7 4 30. u = −x + y, v = y 3 ∂(x, y, z) = abc; then use ∂(u, v, w) Exercise Set 15.8 687 31. Let u = y − 4x, v = y + 4x, then x =

1 8 u 1 dAuv = v 8
5
2 S 2 1 5 u du dv = ln v 4 2 0 32. Let u = y + x, v = y − x, then x = − 1 2

uv dAuv = − 1 2 1 ∂(x, y) 1 1 (v − u), y = (v + u) so =− ; 8 2 ∂(u, v) 8
2
1 ∂(x, y) 1 1 (u − v), y = (u + v) so = ; 2 2 ∂(u, v) 2 1 uv du dv = − 0 S 0 33. Let u = x − y, v = x + y, then x = 1 2 1 1 ∂(x, y) 1 (v + u), y = (v − u) so = ; the boundary curves of 2 2 ∂(u, v) 2 the region S in the uv-plane are u = 0, v = u, and v = π/4; thus



√ 1 1 sin u 1 π/4 v sin u dAuv = du dv = [ln( 2 + 1) − π/4] 2 cos v 2 0 2 0 cos v S 34. Let u = y − x, v = y + x, then x = 1 1 ∂(x, y) 1 (v − u), y = (u + v) so = − ; the boundary 2 2 ∂(u, v) 2 curves of the region S in the uv-plane are v = −u, v = u, v = 1, and v = 4; thus



1 1 4 v u/v 15 (e − e−1 ) eu/v dAuv = e du dv = 2 2 1 −v 4 S 35. Let u = y/x, v = x/y 2 , then x = 1/(u2 v), y = 1/(uv) so

1 dAuv = u4 v 3
S 4
1 2 1 1 ∂(x, y) = 4 3; ∂(u, v) u v 1 du dv = 35/256 u4 v 3 ∂(x, y) = 6; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1 ∂(u, v)


2π
1

(9 − x − y)dA = 6(9 − 3u − 2v)dAuv = 6 (9 − 3r cos θ − 2r sin θ)r dr dθ = 54π so 36. Let x = 3u, y = 2v, R 0 S 37. x = u, y = w/u, z = v + w/u,


v2 w dVuvw = u S
2 4
0 1
1 0 ∂(x, y, z) 1 =− ; ∂(u, v, w) u 3 v2 w du dv dw = 2 ln 3 u 38. u = xy, v = yz, w = xz, 1 ≤ u ≤ 2, 1 ≤ v ≤ 3, 1 ≤ w ≤ 4,    1 ∂(x, y, z) = √ x = uw/v, y = uv/w, z = vw/u, ∂(u, v, w) 2 uvw



2
3
4 √ √ 1 √ dw dv du = 4( 2 − 1)( 3 − 1) V = dV = 1 1 1 2 uvw G 688 39. Chapter 15   sin3 φ cos3 θ 3ρ sin2 φ cos φ cos3 θ −3ρ sin3 φ cos2 θ sin θ ∂(x, y, z)  = 3ρ sin3 φ sin2 θ cos θ sin3 φ sin3 θ 3ρ sin2 φ cos φ sin3 θ ∂(ρ, φ, θ)  3 2 cos φ −3ρ cos φ sin φ 0       = 9ρ2 cos2 θ sin2 θ cos2 φ sin5 φ,
2π
π
a 4 πa3 V =9 ρ2 cos2 θ sin2 θ cos2 φ sin5 φ dρ dφ dθ = 35 0 0 0 40. (b) If x = x(u, v), y = y(u, v) where u = u(x, y), v = v(x, y), then by the chain rule ∂x ∂u ∂x ∂v ∂x ∂x ∂u ∂x ∂v ∂x + = = 1, + = =0 ∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y ∂y ∂u ∂y ∂v ∂y ∂y ∂u ∂y ∂v ∂y + = = 0, + = =1 ∂u ∂x ∂v ∂x ∂x ∂u ∂y ∂v ∂y ∂y 41. (a)   y ∂(x, y)  1 − v −u  = = u; u = x + y, v = ,  v u ∂(u, v) x+y    ∂(u, v)  x y 1 1 1 1  + = = = ; 2 2 = −y/(x + y) x/(x + y) ∂(x, y) (x + y)2 (x + y)2 x+y u ∂(u, v) ∂(x, y) =1 ∂(x, y) ∂(u, v)  ∂(x, y)  v u (b) = ∂(u, v)  0 2v  √ ∂(u, v)  1/ y = 0 ∂(x, y) (c) 42.  ∂(x, y)  = ∂(u, v)   ∂(u, v)  = ∂(x, y)     = 2v 2 ; u = x/√y, v = √y,   1 1 ∂(u, v) ∂(x, y) −x/(2y 3/2 )  √ = = 2; =1 1/(2 y)  2y 2v ∂(x, y) ∂(u, v)  √ √ v  = −2uv; u = x + y, v = x − y, −v   √ √ 1/(2 x + y) 1/(2 x + y)  1 ∂(u, v) ∂(x, y) 1 ; =1 =− =−  √ √ 2 2 2uv ∂(x, y) ∂(u, v) 1/(2 x − y) −1/(2 x − y)  2 x −y u u ∂(u, v) 1 1 ∂(x, y) = 3xy 4 = 3v so = ; ∂(x, y) ∂(u, v) 3v 3

S 43. sin u 1 dAuv = v 3
1 2
2π π      ∂(x, y)  ∂(u, v) ∂(x, y) 1 1 1   = 8xy so = ; xy  = xy = so  ∂(x, y) ∂(u, v) 8xy ∂(u, v) 8xy 8
16
4

1 1 dAuv = du dv = 21/8 8 9 8 1 S 44. 1 ∂(u, v) ∂(x, y) = −2(x2 + y 2 ) so =− ; 2 ∂(x, y) ∂(u, v) 2(x + y 2 )     x4 − y 4 xy 1 1 4 4 xy  ∂(x, y)  = e = (x2 − y 2 )exy = veu so (x − y )e   2 2 ∂(u, v) 2(x + y ) 2 2


4
3 1 1 7 veu dAuv = veu du dv = (e3 − e) 2 2 3 1 4 S 2 sin u du dv = − ln 2 v 3 Review Exercise Set 689   1 ∂(u, v, w)  = 1 45. Set u = x + y + 2z, v = x − 2y + z, w = 4x + y + z, then ∂(x, y, z)  4



6
2
3 1 ∂(x, y, z) V = du dv dw = 6(4)(12) = 16 dx dy dz = ∂(u, v, w) 18 −6 −2 −3 1 −2 1 2 1 1     = 18, and   R 46. (a) Let u = x + y, v = y, then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes the triangle in the uv-plane with vertices (0, 0), (1, 0), (1, 1), and


1
u
1 ∂(x, y) f (x + y)dA = f (u) uf (u) du dv du = ∂(u, v) 0 0 0 R
1 ueu du = (u − 1)eu (b) 0   cos θ ∂(x, y, z)  =  sin θ ∂(r, θ, z)  0 47. (a) 1 0 =1 −r sin θ r cos θ 0   sin φ cos θ ∂(x, y, z)  (b) =  sin φ sin θ ∂(ρ, φ, θ)  cos φ 0 0 1     = r,      ∂(x, y, z)     ∂(r, θ, z)  = r −ρ sin φ sin θ ρ sin φ cos θ 0 ρ cos φ cos θ ρ cos φ sin θ −ρ sin φ     = ρ2 sin φ;      ∂(x, y, z)  2    ∂(ρ, φ, θ)  = ρ sin φ REVIEW EXERCISE SET




3. (a) dA (b) R

dV  (c) G  1+ 2 ∂z ∂x  + ∂z ∂y 2 dA R 4. (a) x = a sin φ cos θ, y = a sin φ sin θ, z = a cos φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π (b) x = a cos θ, y = a sin θ, z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h
1
1+ 7. √ √ 0 1−
1−y 2 f (x, y) dx dy 2

2x 8. 3
6−x f (x, y) dy dx + 1−y 2 0 x f (x, y) dy dx 2 x 9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2


1
1
1
1 ∂(x, y) (b) du dv = dA = 3du dv = 3 0 0 ∂(u, v) 0 0 R √ 10. If 0 < x, y < π then 0 < sin xy ≤ 1, with equality only on the hyperbola xy = π 2 /4, so
π
π
π
π
π
π √ 0= 0 dy dx < sin xy dy dx < 1 dy dx = π 2 0
0 0 1 2x cos(πx2 ) dx = 11. 1/2
12. 0 2 x2 y3 e 2 x=2y 0 0 0 1 √ 1 = −1/( 2π) sin(πx2 ) π 1/2 3 dy = 2 x=−y
2 2 y3 y e 0 1 3 dy = ey 2 2 = 0  1 8 e −1 2 690 Chapter 15
1

2 ex ey dx dy 13. 0 2y
0 y 15. π x sin x dy dx x 14. 0 16. p /2 1 r = a(1 + cos u) y = sin x y = tan (x/2) r=a x p /6 6

8 y 1/3 2 x sin y dx dy = 3 2 17. 2 0
0 π/2

8 2 1 y sin y dy = − cos y 2 3 8 2 0 = 0 0 1 (1 − cos 64) ≈ 0.20271 3 2 (4 − r2 )r dr dθ = 2π 18. 0 0 2xy , and r = 2a sin θ is the circle x2 + (y − a)2 = a2 , so x2 + y 2
a      2xy 2 − x2 − ln a − 2 − x2 dx = a2 dy dx = x ln a + a a x2 + y 2 0 19. sin 2θ = 2 sin θ cos θ =
a √ a− a2 −x2 0
√ a+ a2 −x2
π/2
π/2 2 2 4r (cos θ sin θ) r dr dθ = −4 cos 2θ 20.
2
π/4
2−y/2 21. 2 dx dy = (y/2)1/3 0
0
π/6    2 y y 1/3 y2 3 3 y 4/3 2− − dy = 2y − = − 2 2 4 2 2 2 0
π/6 cos2 3θ = π/4 r dr dθ = 3 0 2π  cos 3θ 22. A = 6

2 0
0
16 r2 cos2 θ r dz dr dθ = 23. 0
π/2 0 r4
π/2 0
0
2π
2π 0
1 2 cos2 θ dθ r3 (16 − r4 ) dr = 32π 0 24. 0
1 π  π π/2 2 ρ sin φ dρ dφ dθ = 1 − sin φ dφ 1 + ρ2 4 2 0 π/2 π π π π = 1− = 1− (− cos φ) 4 2 4 2 0 π/3

a 2π
π/3
(ρ2 sin2 φ)ρ2 sin φ dρ dφ dθ = 25. (a) 0
0 2π
0 √ √ 3a/2
a2 −r 2 (b) 0
(c) =4 0 π/4 √ r/ 3 0 √ 3a/2 √ − 3a/2
√ − ρ4 sin3 φ dρ dφ dθ 2π
r2 dz rdr dθ = 0
√(3a2 /4)−x2
√a2 −x2 −y2 (3a2 /4)−x2 √ √ x2 +y 2 / 3 a 0 0 0 0 √ √
2 3a/2 a −r 2 √ r/ 3 (x2 + y 2 ) dz dy dx r3 dz dr dθ Review Exercise Set
4
2π √ a/ 3

27. V = 0 π/2
4 cos θ
4r cos θ (b) r dz dr dθ −π/2 x2 +y 2 √ a/ 3
a √ 3r 0
4x dz dy dx √ − 4x−x2 0
√ 4x−x2
26. (a) 691 r(a − r dz dr dθ = 2π √ 3r) dr = 0 0 r2 πa3 9 28. The intersection of the two surfaces projects onto the yz-plane as 2y 2 + z 2 = 1, so
1/√2
√1−2y2
1−y2 V =4 dx dz dy 0
0 √ 1/ 2
√1−2y2 y 2 +z 2
2 √ 1/ 2 2 (1 − 2y − z ) dz dy = 4 =4 0 0 0 √ 29. ru × rv  = 2u2 + 2v 2 + 4,


S= 2u2 + 2v 2 + 4 dA =
0 u2 +v 2 ≤4 30. ru × rv  = 2π √
2
3u 1 + u2 , S = 0 2 0  2 (1 − 2y 2 )3/2 dy = 3 √ 2π 4 √  8π √ (3 3 − 1) 2 r2 + 2 r dr dθ = 3
2 1 + u2 dv du = 0 3u  1 + u2 du = 53/2 − 1 0  31. (ru × rv ) u=1 = −2, −4, 1 , tangent plane 2x + 4y − z = 5 v=2  32. u = −3, v = 0, (ru × rv ) u=−3 = −18, 0, −3 , tangent plane 6x + z = −9 v=0
4

2+y 2 /8 4  y2 2− 8  32 ; y¯ = 0 by symmetry; 3 −4 −4   
4
2+y2 /8
4 3 256 8 8 1 256 3 4 y dy = , x ¯= = ; centroid ,0 2 + y2 − x dx dy = 4 128 15 32 15 5 5 −4 y 2 /4 −4 dx dy = 33. A = y 2 /4 dy = 34. A = πab/2, x ¯ = 0 by symmetry,  
a
b√1−x2 /a2
1 a 2 4b 2 2 2 y dy dx = b (1 − x /a )dx = 2ab /3, centroid 0, 2 −a 3π −a 0 1 35. V = πa2 h, x ¯ = y¯ = 0 by symmetry, 3
2π
a
h−rh/a
a r 2 rz dz dr dθ = π rh2 1 − dr = πa2 h2 /12, centroid (0, 0, h/4) a 0 0 0 0  256 1 , 8 − 4x2 + x4 dx = 2 15 2 2 −2 x 0 −2 x −2 
2
4
4−y
2
4
2 1024 1 6 32 x − 2x4 + dx = y dz dy dx = (4y − y 2 ) dy dx = 3 3 35 −2 x2 0 −2 x2 −2 
2
4
4−y
2
4
2 6 2048 x 1 32 dx = − + 2x4 − 8x2 + z dz dy dx = (4 − y)2 dy dx = 6 3 105 −2 x2 0 −2 x2 2 −2   12 8 x ¯ = 0 by symmetry, centroid 0, , 7 7
36. V = 2
4

4−y dz dy dx = 2

4 (4 − y)dy dx = 2  692 Chapter 15 37. V = 4 3 ¯ 3 πa , d = 3 4πa3


ρdV = 3 4πa3 ρ≤a
π
2π
a ρ3 sin φ dρ dθ dφ = 0 0 0 3 3 a4 2π(2) = a 3 4πa 4 4     1 2  3 2  1 3 3 1 1   38. x = u + v and y = − u + v, hence |J(u, v)| =  , and + =  10 10 10 10 10 10  10


3
4
3
4 x − 3y u 1 1 1 1 2 8 dA = du dv = dv u du = 8= 2 2 2 (3x + y) 10 1 0 v 10 1 v 10 3 15 0 R 1 1 39. (a) Add u and w to get x = ln(u + w) − ln 2; subtract w from u to get y = u − w, substitute 2 2 1 1 1 1 these values into v = y + 2z to get z = − u + v + w. Hence xu = , xv = 0, xw = 4 2 4 u+w 1 1 1 1 1 1 ∂(x, y, z) 1 ; yu = , yv = 0, yz = − ; zu = − , zv = , zw = , and thus = u+w 2 2 4 2 4 ∂(u, v, w) 2(u + w)



3
2
4 1 dw dv du (b) V = = 1 1 0 2(u + w) G = (7 ln 7 − 5 ln 5 − 3 ln 3)/2 = 1 823543 ln ≈ 1.139172308 2 84375