# 4 13 g is the conical solid 2 div f dv 2 g 1 1 x

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Unformatted text preview: 1 1 3π = 2(10)(4) + 2(20)(2) − 2 10(2) + 20(4) − 10(20) 2 2 2 60 10π = 60 + √ 3 π dx √ = 3 + , the length of BC is increasing. dt 6 (b) d dt ∂z ∂x = ∂ ∂x ∂z ∂x ∂ dx + dt ∂y ∂z ∂x ∂ 2 z dx ∂ 2 z dy dy = + by the Chain Rule, and 2 dt dt ∂x ∂y∂x dt d dt 38. (a) ∂z ∂y = ∂ ∂x ∂z ∂y ∂ dx + dt ∂y ∂z ∂y ∂ 2 z dx ∂ 2 z dy dy = +2 dt ∂x∂y dt ∂y dt ∂z dx ∂z dy dz = + , dt ∂x dt ∂y dt dx d2 z = 2 dt dt ∂ 2 z dy ∂ 2 z dx + ∂x2 dt ∂y∂x dt + ∂z d2 x dy + ∂x dt2 dt ∂ 2 z dx ∂ 2 z dy +2 ∂x∂y dt ∂y dt + ∂z d2 y ∂y dt2 CHAPTER 16 Multiple Integrals EXERCISE SET 16.1 1 2 1. 1 (x + 3)dy dx = 0 0 4 1 4 x2 y dx dy = 2 1 2 0 0 0 5 dx dy = −1 1 0 3 dy = 3 0 π 1 1− 0 ln 2 1 0 2 1x (e − 1)dx = (1 − ln 2)/2 2 4 1 1 − x+1 x+2 0 1 dy dx = (x + y )2 12. 1 1 2 −1 −2 1 1 xy x2 + y 2 + 1 0 0 dx = 0 1 dy dx = √ √ [x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3 0 3 1 x 1 − x2 dy dx = 15. 0 2 x(1 − x2 )1/2 dx = 1/3 0 π /2 π /3 π /2 (x sin y − y sin x)dy dx = 16. 0 dx = ln(25/24) −1 14. 1 3 1 4xy 3 dy dx = 13. 0 10dx = 20 4 dx = 1 − ln 2 ln 2 2 xy ey x dy dx = 11. 3 −3 π/2 1 4 1 x+1 6 dy dx = (sin 2x − sin x)dx = −2 x cos xy dy dx = 0 (3 + 3y 2 )dy = 14 −2 π 2 10. π/2 7 8. 4 x dy dx = (xy + 1)2 9. −1 1 sin x dx = (1 − cos 2)/2 2 −1 2 1 −2 0 7. 0 (x2 + y 2 )dx dy = 0 y sin x dy dx = 0 2 4. ex dx = 2 0 2 4x dx = 16 1 ln 3 ex+y dy dx = 6. 0 1 y dy = 2 3 ln 2 5. 0 −1 1 3 (2x − 4y )dy dx = 6 0 ln 3 1 2. 0 3. 2 3 (2x + 6)dx = 7 0 0 5 2 5 (2x + y )dy dx = 19. V = 3 1 3 (2x + 3/2)dx = 19 3 2 3 (3x3 + 3x2 y )dy dx = 20. V = 1 0 x π2 − sin x dx = π 2 /144 2 18 (6x3 + 6x2 )dx = 172 1 573 Exercise Set 16.1 574 2 3 2 x2 dy dx = 21. V = 0 0 3 3x2 dx = 8 0 4 3 5(1 − x/3)dy dx = 22. V = 0 0 5(4 − 4x/3)dx = 30 0 z 23. (a) z (b) (0, 0, 5) (1, 0, 4) y y (2, 5, 0) (3, 4, 0) x x z 24. (a) z (b) (2, 2, 8) (0, 0, 2) y y (2, 2, 0) (1, 1, 0) x x 1/2 π 1/2 x cos(xy ) cos2 πx dy dx = 25. 0 0 cos2 πx sin(xy ) π dx 0 0 1/2 = 0 1 cos2 πx sin πx dx = − cos3 πx 3 1/2 = 0 1 3π z 26. (a) y 3 5 x (b) The projection onto the xy -plane consists of R1 : [0, 5] × [0, 1], over which lie each of the two skew planes, and R2 : [0, 5] × [1, 3], over which is only the plane z = −2y + 6, so 5 1 0 27. fave = 2 π 5 3 ((−2y + 6) − y ) dy dx + V= 0 0 π /2 1 y sin xy dy dx = 0 (−2y + 6) dy dx = 0 2 π 1 π /2 − cos xy 0 1 dx = 0 2 π 85 45 + 20 = 2 2 π /2 (1 − cos x) dx = 1 − 0 2 π 575 Chapter 16 28. average = 0 √ 1 [(1 + y )3/2 − y 3/2 ]dy = 2(31 − 9 3)/45 9 3 0 0 2 10 − 8x2 − 2y 2 dy dx = 0 0 1 A(R) 30. fave = 1 x(x2 + y )1/2 dx dy = 1 1 2 29. Tave = 3 1 3 b d 0 c 31. 0.6211310829 dx = 14 3 ◦ 32. 2.230985141 b d b f (x, y )dA = 33. 44 − 16x2 3 1 (b − a)(d − c)k = k A(R) k dy dx = a 1 1 2 g (x)h(y )dy dx = a R c h(y )dy dx a b c d g (x)dx = d g (x) 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. 16 4 4 ∗ f (x∗ , yk )∆A∗ = k k 36. (a) i=1 j =1 k=1 2 2 0 −2 1 j − 24 1 2 2 = −4 2 (x − 2y ) dy dx = (b) 1 i − 24 (2x − 4) dx = −4 0 0 EXERCISE SET 16.2 x 1 1 xy 2 dy dx = 1. x2 0 0 3−y 3/2 3/2 (3y − 2y 2 )dy = 7/24 y dx dy = 2. y 1 √ 3 1 9−y 2 3 y dx dy = 3. 0 0 x 1 9 − y 2 dy = 9 y 0 x 1 4. x/y dy dx = x2 1/4 √ 2π 5. 14 (x − x7 )dx = 1/40 3 √ x2 1/4 1 √ 0 x2 −1 −x2 2(x − x3/2 )dx = 13/80 1/4 [−x cos(x2 ) + x]dx = π/2 π (x − y )dy dx = 4 2x dx = 4/5 −1 x2 π 1 2 6. 1 √ 2π x3 sin(y/x)dy dx = π x1/2 y −1/2 dy dx = 7. π/2 0 1 cos(y/x)dy dx = x π sin x dx = 1 π/2 Exercise Set 16.2 x 1 576 1 2 0 0 (e − 1)y 2 dy = 7(e − 1)/3 x2 2 0 0 0 (y +7)/2 3 3 (3y 2 + 3y )dy = 38 xy dx dy = (b) −(y −5)/2 1 1 √ x 1 12. (a) 1 (x3/2 + x/2 − x3 − x4 /2)dx = 3/10 (x + y )dy dx = x2 √ 1−x2 0 1 (b) √ − 1−x2 −1 0 x dy dx + √ y 15. 4 x(1 + y 2 )−1/2 dx dy = √ 1−x2 √ − 1−x2 5 1 (3x − 2y )dy dx = 0 6−y 2 19. xy dx dy = y2 0 1 (36y − 12y 2 + y 3 − y 5 )dy = 50/3 2 √ 1/ 2 π /4 x dx dy = 20. sin y 0 0 x3 1 1 cos 2y dy = 1/8 4 x (x − 1)dy dx + 21. 6x 1 − x2 dx = 0 (5x − x2 )dx = 125/6 5−x π /4 −1 5 y dy dx = 18. −1 √ 1 y (1 + y 2 )−1/2 dy = ( 17 − 1)/2 2 x sin x dx = π √ 25−x2 0 x (x − 1)dy dx 0 x3 0 = −1 √ 1/ 2 1 (x4 − x3 − x2 + x)dx + 2x 22. x dy dx + x (−x4 + x3 + x2 − x)dx = −1/2 0 1 2 0 1 0 −1 0 0 14 y dy = 31/10 2 π 0 2 2 xy 2 dx dy = 1 x cos y dy dx = 0 y 14. 4 x 17. 2x 1 − x2 dx + 0 = 0 2 0 π −1 (x3 − 16x)dx = 576 0 16. 1 y dy dx = 8 16/x 1 √ − 1−x2 −1 x2 dy dx = 0 √ 1−x2 1 x 13. 0 16 15 x dx = 2 3 xy dy dx = 11. (a) 4 0 1 2 4 0 13 x dx = 1/12 3 2 2 0 8 1 x2 − y 2 dy dx = y 0 ex/y dx dy = 10. 1 9. 0 y2 2 x 1 2 xex dx = (e − 1)/2 ex dy dx = 8. √ 1/ 2 √ 1/ 2 1/x 2 x 1 x3 dx + x dy dx = 0 √ 1/ 2 (x − x3 )dx = 1/8 577 Ch...
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