Chapter 15

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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

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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

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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

dy = x=0

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

y 3

1

√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

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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

π

π/2

x2

x

9.

1 cos(y/x)dy dx = x

0

1

0

x2

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

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 = 1

0

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

√ 1−x2

−1

4

16/x

4

4 8

2

(b)

−1

8

2

4

y

8

x dx dy =

16/y

y

4

=

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 =

1

x dy dx +

x

17. (a)

0

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

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Exercise Set 15.2

659

1

19. (a)

√ − 1−x2

−1

1

√

−1 5

(3x − 2y)dy dx =

√1−y2

(b)

√ 1−x2

−

(3x − 2y) dxdy = 1−y 2

√ 25−x2

(b)

4

√ y

Ï€

x

22. 0

6−y

2

y2

0 √ 1/ 2

24.

Ï€/4

1 cos 2y dy = 1/8 4

x dx dy = 0

sin y 1

0

x

1

(x − 1)dy dx =

25.

(−x4 + x3 + x2 − x)dx = −7/60

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 =

Ï€/4

4

0

23. 0

25 − y 2 − 5 + y dy = 125/6

x sin x dx = π

0 2

Ï€

x cos y dy dx =

x(1 + y 2 )−1/2 dx dy =

0

1 − y 2 dy = 0

0

0

0

y

5−y

21.

5

y dxdy =

−1

−4y

0

√25−y2

0

1

(5x − x2 )dx = 125/6

5−x

5

6x 1 − x2 dx = 0

5

y dy dx =

−1

20. (a) 0

1

√ 1/ 2

0

2x 2

x dy dx +

26. 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

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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 − y/2) dy dx =

33. 0

[(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12

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 dy dx =

(4 − x2 ) dx = 16/3 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

(9 − x )dx dy =

40. V = y 2 /3

5

41. V = 8 0

0

√ 25−x2

6 −3/2

3

3/2

+ 3)dy dx =

9 − 4x2 dx = 27π/2

3

(18 − 3y 2 + y 6 /81)dy = 216/7

2

0

(6 9 − x2 − 2x 9 − x2 )dx = 27π

1

(x2 + 3y 2 )dy dx =

36. V =

3

0

25 − x2 dy dx = 8

5

(25 − x2 )dx = 2000/3 0

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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

e−y dx dy = 2

0 1

4

0

x2

x3

2

3

54.

1 2

x dx dy = ey

0

2

ln 3

(9 − e2y )dy = 0

e

56.

2

1

(ex − xex )dx = e/2 − 1

x dy dx = ex

0

1 (9 ln 3 − 4) 2

0

0 1

f (x, y)dy dx x2

0

y 2 sin(y 3 )dy = (1 − cos 8)/3

sin(y 3 )dx dy = 0

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 y 2 sin πy 3 dy = − cos πy 3 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)

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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

1

1

61. Area of triangle is 1/2, so f¯ = 2 0

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

x

1 dy dx = 2 1 + x2

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 dA = st0 x dydx = (4 − 2x)x dx = = C Tave = x − x 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)

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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

e−r r dr dθ = 2

0

64 √ 2π 3

0

r sin θ drdθ = 9

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

16 32 8 dθ = + π 3 9 3

dθ = (1 − e−1 )π

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664

Chapter 15

Ď€/2

3

24.

r 0

9−

r2

Ď€/4

0

dθ = 9Ď€/2 0

2

0 π/2

Ď€/2

dr dθ = 9

0

25.

1 1 r dr dθ = ln 5 1 + r2 2

2 cos θ

Ď€/4

dθ = 0

16 2r sin θ dr dθ = 3

Ď€/2

2

26. π/4

0

Ď€/2

1

r3 dr dθ =

27. 0

0 2Ď€

2

0 π/2

Ď€/2

dθ = Ď€/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

Ď€/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)

= 2Ď€

a

hr dr dθ =

35. V = 0

0 π/2

36. (a) V = 8 0

(b) V ≈

0

√ sec3 θ tan3 θ dθ = 2( 2 + 1)/45

a

Ď€/2

tan−1 (3/4)

Ď€/2

0

a sin θ

Ď€/4

0

25 25 Ď€ − tan−1 (3/4) − 6 = tan−1 (4/3) − 6 2 2 2

2Ď€

a2 dθ = Ď€a2 h 2

c 2 4c (a − r2 )1/2 r dr dθ = − Ď€(a2 − r2 )3/2 a 3a

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

a = 0

4 2 πa c 3

4 Ď€(6378.1370)2 6356.5231 ≈ 1,083,168,200,000 km3 3

37. V = 2

0

Ď€ sin 1 4

0

h 0

dθ =

r Ď€ √ dr dθ = ( 5 − 1) 2 4 1+r

Ď€/2

Ď€/2

r Ď€ 1 − 1/ 1 + a2 dr dθ = 2 3/2 2 (1 + r )

a

31. 0

dθ = (1 − e−4 )Ď€

0

Ď€/2

1 sin 1 2

0 π/2

2Ď€

0

cos(r2 )r dr dθ = 0

cos3 θ dθ = 16/9

1

30.

cos3 θ sin θ dθ = 1/3 Ď€/4

0

e−r r dr dθ =

28. 0

1 4

Ď€ ln 5 8

0

Ď€/2

(1 − cos3 θ)dθ = (3Ď€ − 4)a2 c/9 0

Ď€/4

cos 2θ dθ = 2a2 0

January 27, 2005 11:55

L24-CH15

Sheet number 11 Page number 665

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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 π/4

0

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

e−x e−y dx dy = 2

2

0

π/2

+∞

(b) I 2 = 0

e−r r dr dθ = 2

0

1 2

0

+∞

0

0

+∞

e−(x

2

+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 0 (1 + r ) 0 0 (1 + x + y ) 0 πA2 and the integral on the right equals 4(1 + A2 ) 2 2πA π . 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

re−r dr dθ = π 4

0

1

re−r dr ≈ 1.173108605 4

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

January 27, 2005 11:55

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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

√ 1 3 1 ρ 15. x = ρ cos θ, y = ρ sin θ, z = 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

January 27, 2005 11:55

L24-CH15

Sheet number 13 Page number 667

black

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 =

4

36. z = 8 − 2x − 2y, zx2 + zy2 + 1 = 4 + 4 + 1 = 9, S =

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 √ = (5 5 − 1) dθ = (5 5 − 1)π/6 12 0

January 27, 2005 11:55

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Sheet number 14 Page number 668

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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 a + h2 1 2 2 S= dθ = πa a2 + h2 r dr dθ = a a + h a 2 0 0 0

49. zx =

January 27, 2005 11:55

L24-CH15

Sheet number 15 Page number 669

black

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 ≤ 2Ď€. 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 ≤ 2Ď€. Set a0 = a + b cos v and b0 = a + b sin v and use the ďŹ rst 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 Ă— ∂r/∂v = b(a + b cos v); 2Ď€ 2Ď€ S= b(a + b cos v)du dv = 4Ď€ 2 ab 0

0

52. ru Ă— rv =

√

u2

4Ď€

+ 1; S = 0

5

0

u2

+ 1 du dv = 4Ď€

5

u2 + 1 du = 174.7199011

0

53. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, ru Ă— rv = | cos v|; 2Ď€ 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 ≤ Ď€, 0 ≤ θ < 2Ď€. 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 Ď€/2 Ď€/2 S=8 rθ Ă— rφ dφ dθ 0

0

Ď€/2

= 72 0

Ď€/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 Ă—rθ = 2 6 sin2 φ cos θ, 4 sin2 φ sin θ, 3 cos φ sin φ , rφ Ă— rθ = 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ, 2Ď€ Ď€ S= 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ dφ dθ ≈ 111.5457699 0

0

January 27, 2005 11:55

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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

Ï€

−1

1

1/2

Ï€

2

0

y2

0

1/3

z

2

−1

−1

0

Ï€/4

1

0

0

x2

Ï€/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

Ï€/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

January 27, 2005 11:55

L24-CH15

Sheet number 17 Page number 671

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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

y

2

√

1

13. 0

−2

1

1

√ 1−x2

π/6

√1−x2 −y2

0

e−x

2

−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

2

0

2 (1 − x)3/2 dx = 4/15 3

4

x2

1

y

dz dx dy = 0

1

0

0

18. V =

y dy dx =

2

(4 − y)dy dx = 2

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

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

0

0

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

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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

x+3

dz dy dx

23. V = 2 −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

1−x−y

1

1−x

dz dy dx = 1/6, fave = 6

27. V = 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

January 27, 2005 11:55

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Sheet number 19 Page number 673

black

Exercise Set 15.5

673

3Ď€ 29. The volume V = √ , and thus 2 √ 2 x2 + y 2 + z 2 dV rave = 3Ď€ G

√ 1/√2 √1−2x2 6−7x2 −y2 2 x2 + y 2 + z 2 dz dy dx ≈ 3.291 = √ √ 3Ď€ −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 ďŹ rst 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

y

0

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 suďŹƒces to include all points inside the sphere; hence the maximum value is taken on the region G = {x2 + y 2 + z 2 ≤ 1}. 2Ď€ Ď€ 1 Ď€2 (b) 4.934802202 (c) (1 − Ď 2 )Ď dĎ 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

January 27, 2005 11:55

L24-CH15

Sheet number 20 Page number 674

black

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

January 27, 2005 11:55

L24-CH15

Sheet number 21 Page number 675

black

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

11. M =

1

0

R

1

1

|x + y − 1| dx dy

δ(x, y)dA = 0

1−x

(1 − x − y) dy +

= 0

0

1

(x + y − 1) dy dx =

1

1

1−x 1

1−x

0

0

1 x(x + y − 1) dy dx = 2 1−x

x(1 − x − y) dy +

xδ(x, y) dy dx = 3

x=3

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 y dy dx = 4/9, y = Mx /M = ; mass π/4, center of gravity 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

January 27, 2005 11:55

L24-CH15

Sheet number 22 Page number 676

black

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 V = dx dz dy = , x = x dx dz dy = , y = 0 by symmetry, 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 √a2 −x2 −y2 1 2 1 a 1 a (a − x2 − y 2 )dy dx z dz dy dx = z= V −a −√a2 −x2 2 V −a −√a2 −x2 0 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)

January 27, 2005 11:55

L24-CH15

Sheet number 23 Page number 677

black

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

January 27, 2005 11:55

L24-CH15

Sheet number 24 Page number 678

black

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

January 27, 2005 11:55

L24-CH15

Sheet number 25 Page number 679

black

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 ab (2πy), y = b/3. A cone of volume πa2 b is generated when the region is ab so πab = 2 3 2 3 1 2 1 region is revolved about the y-axis so πa b = ab (2πx), x = a/3. The centroid is (a/3, b/3). 2 3 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

January 27, 2005 11:55

L24-CH15

Sheet number 26 Page number 680

black

680

Chapter 15

7. f (Ď , φ, θ) = Ď cos φ

8. f (Ď , φ, θ) = 1 z

z

a

y x x

2Ď€

3

9

2Ď€

0

r2

0

2Ď€

0

√ 9−r 2

10. V = 2

0

2Ď€

2

r dz dr dθ = 2 0

0

2Ď€

r(9 − r2 )dr dθ = 0

2

3

r dz dr dθ =

9. V =

y

r 0

0

=

0

9 − r2 dr dθ

√ 2 (27 − 5 5) 3

81 dθ = 81Ď€/2 4

2Ď€

√ dθ = 4(27 − 5 5)Ď€/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: 2Ď€ 2 r2 2Ď€ √20 √20−r2 V= r dz dr dθ + r dz dr dθ √ √ 0

− 20−r 2

0

2Ď€

2

r r2 +

= 0

=

0

2

2Ď€

20 − r2 dr dθ +

0

− 20−r 2 √ 20

2r

0

20 − r2 dr dθ

2

2Ď€ √ 80 √ 4 128 2Ď€ 152 (10 5 − 13) Ď€+ Ď€ 5 dθ + dθ = 3 3 0 3 3 0

12. z = hr/a intersects z = h in a circle of radius a, 2Ď€ a h 2Ď€ a 2Ď€ h 1 2 (ar − r2 )dr dθ = a h dθ = Ď€a2 h/3 V = r dz dr dθ = a 6 0 0 hr/a 0 0 0

2Ď€

Ď€/3

4

2Ď€

Ď€/3

Ď 2 sin φ dĎ dφ dθ =

13. V = 0

0 2Ď€

0 π/4

0

2

0 2Ď€

Ď€/4

Ď 2 sin φ dĎ dφ dθ =

14. V = 0

0

1

0

0

64 32 sin φ dφ dθ = 3 3

2Ď€

dθ = 64Ď€/3 0

√ 7 7 sin φ dφ dθ = (2 − 2) 3 6

2Ď€

√ dθ = 7(2 − 2)Ď€/3

0

15. In spherical coordinates the sphere and the plane z = a are Ď = 2a and Ď = a sec φ, respectively. They intersect at φ = Ď€/3, 2Ď€ Ď€/3 a sec φ 2Ď€ Ď€/2 2a V= Ď 2 sin φ dĎ dφ dθ + Ď 2 sin φ dĎ dφ dθ 0

0

2Ď€

0 π/3

0 2Ď€

1 3 a sec3 φ sin φ dφ dθ + 3 0 0 0 2Ď€ 2Ď€ 1 4 = a3 dθ + a3 dθ = 11Ď€a3 /3 2 3 0 0

π/3 π/2

=

Ď€/3

0

8 3 a sin φ dφ dθ 3

January 27, 2005 11:55

L24-CH15

Sheet number 27 Page number 681

black

Exercise Set 15.7

2Ď€

681

Ď€/2

Ď€/2

Ď€/4

a

0

0

0

0

a2 −r 2

Ď€

Ď€/2

1

eâˆ’Ď Ď 2 sin φ dĎ dφ dθ = 3

18. 0

0

Ď€/2

√ 8

Ď€/4

0

0 2Ď€

Ď€/4

√ √ 9 2 2Ď€ 9 sin φ dφ dθ = dθ = 9 2Ď€ 2 0

a

(a2 r3 − r5 ) cos2 θ dr dθ 1 6 a 12

0 π/2

cos2 θ dθ = Ď€a6 /48 0

1 (1 − e−1 ) 3

Ď€

0

Ď€/2

sin φ dφ dθ = (1 − e−1 )Ď€/3

0

√ Ď 4 cos2 φ sin φ dĎ dφ dθ = 32(2 2 − 1)Ď€/15

19.

Ď€/2

0

0

=

0

Ď€/2

r3 cos2 θ dz dr dθ =

17.

Ď sin φ dĎ dφ dθ = 0

2Ď€

2

16. V =

3

0

Ď€

3

Ď 3 sin φ dĎ dφ dθ = 81Ď€

20. 0

0

0

2

4

4 Ď€/3 1 3 √ dz r dr tan θ dθ 1 + z2 −2 1 Ď€/6 15 4 1 √ − ln 3 ≈ 16.97774196 = −2 ln( 5 − 2) 2 3 2

Ď€/3

r tan3 θ √ dθ dr dz = 1 + z2

21. (a) −2

1

Ď€/6

2

The region is a cylindrical wedge. (b) To convert to rectangular √ √ coordinates observe that the rays θ = Ď€/6, θ = Ď€/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 x/ 3 −2 1

Ď€/2

√ Ď€/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

Ď€/4

22. 0

0

2Ď€

a

√ a2 −r 2

23. (a) V = 2 0

2Ď€

0

Ď€

a

Ď 2 sin φ dĎ dφ dθ = 4Ď€a3 /3

(b) V = 0

2

r dz dr dθ = 4Ď€a3 /3

0

0

√ 4−x2

0

√4−x2 −y2 xyz dz dy dx

24. (a) 0

0

0

2

√ 4−x2

=

Ď€/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

Ď€/2

=

Ď€/2

Ď€/2

0

0

2

1 3 8 (4r − r5 ) sin θ cos θ dr dθ = 2 3

Ď€/2

sin θ cos θ dθ = 4/3 0

2

Ď 5 sin3 φ cos φ sin θ cos θ dĎ dφ dθ

(c) 0

0

0

Ď€/2

= 0

0

Ď€/2

32 8 sin3 φ cos φ sin θ cos θ dφ dθ = 3 3

Ď€/2

sin θ cos θ dθ = 4/3 0

January 27, 2005 11:55

L24-CH15

Sheet number 28 Page number 682

black

682

Chapter 15

2π

3

3

2π

3

(3 − z)r dz dr dθ =

25. M = 0

0

0

0

a

h

26. M =

2π

dθ = 27π/4 0

2π 1 2 1 dθ = πka2 h2 /2 kh r dr dθ = ka2 h2 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

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¯ = zr dz dr dθ = (4π) = 3/2; centroid (0, 0, 3/2) 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

January 27, 2005 11:55

L24-CH15

Sheet number 29 Page number 683

black

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 M= (r2 + z 2 )r dz dr dθ = π/4, 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 1 π 65π 3 = sin θ dθ = (65π/4) = 195/152; centroid (0, 195/152, 0) V 0 8 38π

2π

π

R

−(ρ/R)3 2

δ0 e

41. M = 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

January 27, 2005 11:55

L24-CH15

Sheet number 30 Page number 684

black

684

Chapter 15

(c)

d cos φ = − sin φ so from the Mean-Value Theorem cos φ2 −cos φ1 = −(φ2 âˆ’Ď†1 ) sin Ď†âˆ— where dφ d 3 Ď†âˆ— is between φ1 and φ2 . Similarly Ď = 3Ď 2 so Ď 32 âˆ’Ď 31 = 3Ď âˆ—2 (Ď 2 âˆ’Ď 1 ) where Ď âˆ— is between dĎ Ď 1 and Ď 2 . Thus cos φ1 −cos φ2 = sin Ď†âˆ— âˆ†Ď† and Ď 32 âˆ’Ď 31 = 3Ď âˆ—2 âˆ†Ď so ∆V = Ď âˆ—2 sin Ď†âˆ— âˆ†Ď âˆ†Ď†âˆ†θ.

2Ď€

a

h

2Ď€

a

h

2

43. Iz =

r3 dz dr dθ =

r δ r dz dr dθ = δ 0

0

2Ď€

0

a

0

0

0

h

2Ď€

0

0

0

2Ď€

=δ 0

2Ď€

0

a2

h

0

a1 2Ď€

Ď€

2Ď€

a2

0

h

r3 dz dr dθ = 0

a1

0

a

2Ď€

Ď€

0

0

0

1 δĎ€h(a42 − a41 ) 2

a

(Ď 2 sin2 φ)δ Ď 2 sin φ dĎ dφ dθ = δ

46. Iz =

Ď 4 sin3 φ dĎ dφ dθ = 0

EXERCISE SET 15.8

4.

1 (hr3 cos2 θ + h3 r)dr dθ 3

Ď€ 1 4 1 Ď€ a h cos2 θ + a2 h3 dθ = δ a4 h + a2 h3 4 6 4 3 r2 δ r dz dr dθ = δ

0

3.

a

45. Iz =

1.

(r2 cos2 θ + z 2 )δr dz dr dθ = δ

44. Iy =

1 δĎ€a4 h 2

0

0

∂(x, y) 1 4 = ∂(u, v) 3 −5

= −17

∂(x, y) cos u = ∂(u, v) sin u

− sin v = cos u cos v + sin u sin v = cos(u − v) cos v

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

5 1 2 ∂(x, y) 2/9 2 = 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 δĎ€a5 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

January 27, 2005 11:55

L24-CH15

Sheet number 31 Page number 685

black

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

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

u 1 dAuv = v 5

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

veuv dAuv = S

3

1

S

18. x =

2

1 2

4

4

1

u 3 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 v = 0, v = u, and u = 1 so 2 sin u cos vdAuv = 2 sin u cos v dv du = 1 − sin 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

January 27, 2005 11:55

L24-CH15

Sheet number 32 Page number 686

black

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

2 1 3 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)

January 27, 2005 11:55

L24-CH15

Sheet number 33 Page number 687

black

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 √ sin u 1 π/4 v sin u 1 1 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 ∂(x, y) 1 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 eu/v dAuv = e du dv = (e − e−1 ) 2 1 −v 4 2 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, ∂(x, y, z) 1 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

January 27, 2005 11:55

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Sheet number 34 Page number 688

black

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 ∂x ∂u ∂y ∂v ∂y ∂y ∂u ∂x ∂v ∂x ∂y ∂y ∂u ∂y ∂v ∂y ∂y ∂u ∂y ∂v + = = 0, + = =1 ∂x ∂u ∂y ∂v ∂y ∂y ∂u ∂x ∂v ∂x 41. (a)

∂(x, y) 1 − v −u y = = u; u = x + y, v = , v u ∂(u, v) x+y 1 x y 1 ∂(u, v) 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 1 ∂(u, v) ∂(x, y) ; =1 =− =− √ √ 2 2 2uv ∂(x, y) ∂(u, v) 1/(2 x − y) −1/(2 x − y) 2 x −y u u

1 1 ∂(u, v) ∂(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) 1 1 ∂(x, y) 1 = xy = so = 8xy so = ; xy ∂(x, y) ∂(u, v) 8xy ∂(u, v) 8xy 8 16 4 1 1 dAuv = du dv = 21/8 8 8 9 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

January 27, 2005 11:55

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Sheet number 35 Page number 689

black

Review Exercises, Chapter 15

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 1 u ∂(x, y) dv du = f (x + y)dA = f (u) uf (u) 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 =1 0

−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 EXERCISES, CHAPTER 15

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) dA = 3du dv = 3 du dv = 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

January 27, 2005 11:55

L24-CH15

Sheet number 36 Page number 690

black

690

Chapter 15

1

2

ex ey dx dy

13. 0

2y

0

y

15.

Ď€

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

Ď€/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 = 2Ď€

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

Ď€/2

Ď€/2

2

4r (cos θ sin θ) r dr dθ = −4 cos 2θ 2

20. π/4

2

Ď€/4

2−y/2

21.

2

dx dy = (y/2)1/3

0

0

Ď€/6

2 3 y 4/3 y y 1/3 y2 3 − 2− − dy = 2y − = 2 2 4 2 2 2 0

Ď€/6

cos2 3θ = Ď€/4

r dr dθ = 3 0

2Ď€

cos 3θ

22. A = 6

2

0

0

16

r2 cos2 θ r dz dr dθ =

23. 0

Ď€/2

0

r4

Ď€/2

0

0

2Ď€

2Ď€

0

1

2

r3 (16 − r4 ) dr = 32Ď€

cos2 θ dθ 0

24.

0

1 Ď€ Ď€ Ď€/2 2 Ď sin φ dĎ dφ dθ = 1 − sin φ dφ 1 + Ď 2 4 2 0 Ď€/2 Ď€ Ď€ Ď€ Ď€ (− cos φ) = 1− = 1− 4 2 4 2 0

Ď€/3

a

2Ď€

Ď€/3

(Ď 2 sin2 φ)Ď 2 sin φ dĎ dφ dθ =

25. (a) 0

0 2Ď€

0 √ √ 3a/2 a2 −r 2

(b) 0

(c)

=4

0

√ r/ 3

0 √ 3a/2

√ − 3a/2

√

−

Ď 4 sin3 φ dĎ dφ dθ

2Ď€

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

January 27, 2005 11:55

L24-CH15

Sheet number 37 Page number 691

black

Review Exercises, Chapter 15

4

26. (a)

2Ď€

√ a/ 3

27. V = 0

Ď€/2

4 cos θ

4r cos θ

(b)

r dz dr dθ âˆ’Ď€/2

x2 +y 2 √ a/ 3

a

√ 3r

0

4x

dz dy dx

√ − 4x−x2

0

√ 4x−x2

691

r(a −

r dz dr dθ = 2Ď€

√

3r) dr =

0

0

r2

Ď€a3 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 dx dz dy V =4 0

0

√ 1/ 2

√1−2y2

y 2 +z 2

√ 1/ 2

(1 − 2y − z ) dz dy = 4 2

=4 0

2

0

0

√ 29. ru Ă— rv = 2u2 + 2v 2 + 4, S= 2u2 + 2v 2 + 4 dA =

0

u2 +v 2 ≤4

30. ru Ă— rv =

2Ď€

√

2

3u

1 + u2 , S = 0

2

0

2 (1 − 2y 2 )3/2 dy = 3

√

2Ď€ 4

√ 8Ď€ √ (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 Ă— rv ) u=1 = −2, −4, 1 , tangent plane 2x + 4y − z = 5 v=2

32. u = −3, v = 0, (ru Ă— rv ) u=−3 = −18, 0, −3 , tangent plane 6x + z = −9 v=0

4

2+y 2 /8

4

y2 2− 8

32 ; yÂŻ = 0 by symmetry; 3 −4 −4 4 4 2+y2 /8 3 256 8 8 1 256 3 4 y dy = , x ÂŻ= = ; 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 = Ď€ab/2, x ÂŻ = 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 3Ď€ −a 0 1 35. V = Ď€a2 h, x ÂŻ = yÂŻ = 0 by symmetry, 3 2Ď€ a h−rh/a a r 2 rz dz dr dθ = Ď€ rh2 1 − dr = Ď€a2 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 ÂŻ = 0 by symmetry, centroid 0, , 7 7

36. V =

2

4

4−y

dz dy dx =

2

4

(4 − y)dy dx =

2

January 27, 2005 11:55

L24-CH15

Sheet number 38 Page number 692

black

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 a4 3 2π(2) = a 3 4πa 4 4

1 2 3 2 1 1 3 3 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 8 x − 3y u 1 1 1 1 2 8= dA = du dv = dv u du = 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 ∂(x, y, z) 1 1 1 1 1 1 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

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