Calculus: Early Transcendentals, by Anton, 7th Edition,ch15

# Calculus - Early Transcendentals

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620 CHAPTER 15 Multiple Integrals EXERCISE SET 15.1 1. 1 0 2 0 ( x + 3) dy dx = 1 0 (2 x + 6) dx = 7 2. 3 1 1 1 (2 x 4 y ) dy dx = 3 1 4 x dx = 16 3. 4 2 1 0 x 2 y dx dy = 4 2 1 3 y dy = 2 4. 0 2 2 1 ( x 2 + y 2 ) dx dy = 0 2 (3 + 3 y 2 ) dy = 14 5. ln 3 0 ln 2 0 e x + y dy dx = ln 3 0 e x dx = 2 6. 2 0 1 0 y sin x dy dx = 2 0 1 2 sin x dx = (1 cos 2) / 2 7. 0 1 5 2 dx dy = 0 1 3 dy = 3 8. 6 4 7 3 dy dx = 6 4 10 dx = 20 9. 1 0 1 0 x ( xy + 1) 2 dy dx = 1 0 1 1 x + 1 dx = 1 ln 2 10. π π/ 2 2 1 x cos xy dy dx = π π/ 2 (sin 2 x sin x ) dx = 2 11. ln 2 0 1 0 xy e y 2 x dy dx = ln 2 0 1 2 ( e x 1) dx = (1 ln 2) / 2 12. 4 3 2 1 1 ( x + y ) 2 dy dx = 4 3 1 x + 1 1 x + 2 dx = ln(25 / 24) 13. 1 1 2 2 4 xy 3 dy dx = 1 1 0 dx = 0 14. 1 0 1 0 xy x 2 + y 2 + 1 dy dx = 1 0 [ x ( x 2 + 2) 1 / 2 x ( x 2 + 1) 1 / 2 ] dx = (3 3 4 2 + 1) / 3 15. 1 0 3 2 x 1 x 2 dy dx = 1 0 x (1 x 2 ) 1 / 2 dx = 1 / 3 16. π/ 2 0 π/ 3 0 ( x sin y y sin x ) dy dx = π/ 2 0 x 2 π 2 18 sin x dx = π 2 / 144 17. (a) x k = k/ 2 1 / 4 , k = 1 , 2 , 3 , 4; y l = l/ 2 1 / 4 , l = 1 , 2 , 3 , 4 , R f ( x, y ) dxdy 4 k =1 4 l =1 f ( x k , y l )∆ A kl = 4 k =1 4 l =1 [( k/ 2 1 / 4) 2 +( l/ 2 1 / 4)](1 / 2) 2 = 37 / 4 (b) 2 0 2 0 ( x 2 + y ) dxdy = 28 / 3; the error is | 37 / 4 28 / 3 | = 1 / 12

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Exercise Set 15.1 621 18. (a) x k = k/ 2 1 / 4 , k = 1 , 2 , 3 , 4; y l = l/ 2 1 / 4 , l = 1 , 2 , 3 , 4 , R f ( x, y ) dxdy 4 k =1 4 l =1 f ( x k , y l )∆ A kl = 4 k =1 4 l =1 [( k/ 2 1 / 4) 2( l/ 2 1 / 4)](1 / 2) 2 = 4 (b) 2 0 2 0 ( x 2 y ) dxdy = 4; the error is zero 19. V = 5 3 2 1 (2 x + y ) dy dx = 5 3 (2 x + 3 / 2) dx = 19 20. V = 3 1 2 0 (3 x 3 + 3 x 2 y ) dy dx = 3 1 (6 x 3 + 6 x 2 ) dx = 172 21. V = 2 0 3 0 x 2 dy dx = 2 0 3 x 2 dx = 8 22. V = 3 0 4 0 5(1 x/ 3) dy dx = 3 0 5(4 4 x/ 3) dx = 30 23. (a) (1, 0, 4) (2, 5, 0) x y z (b) (0, 0, 5) (3, 4, 0) (0, 4, 3) z x y 24. (a) (1, 1, 0) (0, 0, 2) x y z (b) (2, 2, 0) (2, 2, 8) x y z 25. 1 / 2 0 π 0 x cos( xy ) cos 2 πx dy dx = 1 / 2 0 cos 2 πx sin( xy ) π 0 dx = 1 / 2 0 cos 2 πx sin πx dx = 1 3 π cos 3 πx 1 / 2 0 = 1 3 π
622 Chapter 15 26. (a) y x z 5 3 (0, 2, 2) (5, 3, 0) (b) V = 5 0 2 0 y dy dx + 5 0 3 2 ( 2 y + 6) dy dx = 10 + 5 = 15 27. f ave = 2 π π/ 2 0 1 0 y sin xy dx dy = 2 π π/ 2 0 cos xy x =1 x =0 dy = 2 π π/ 2 0 (1 cos y ) dy = 1 2 π 28. average = 1 3 3 0 1 0 x ( x 2 + y ) 1 / 2 dx dy = 3 0 1 9 [(1 + y ) 3 / 2 y 3 / 2 ] dy = 2(31 9 3) / 45 29. T ave = 1 2 1 0 2 0 ( 10 8 x 2 2 y 2 ) dy dx = 1 2 1 0 44 3 16 x 2 dx = 14 3 30. f ave = 1 A ( R ) b a d c k dy dx = 1 A ( R ) ( b a )( d c ) k = k 31. 1 . 381737122 32. 2 . 230985141 33. R f ( x, y ) dA = b a d c g ( x ) h ( y ) dy dx = b a g ( x ) d c h ( y ) dy dx = b a g ( x ) dx d c h ( y ) dy 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. 1 0 x x 2 xy 2 dy dx = 1 0 1 3 ( x 4 x 7 ) dx = 1 / 40 2. 3 / 2 1 3 y y y dx dy = 3 / 2 1 (3 y 2 y 2 ) dy = 7 / 24 3. 3 0 9 y 2 0 y dx dy = 3 0 y 9 y 2 dy = 9 4. 1 1 / 4 x x 2 x/y dy dx = 1 1 / 4 x x 2 x 1 / 2 y 1 / 2 dy dx = 1 1 / 4 2( x x 3 / 2 ) dx = 13 / 80

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Exercise Set 15.2 623 5. 2 π π x 3 0 sin( y/x ) dy dx = 2 π π [ x cos( x 2 ) + x ] dx = π/ 2 6. 1 1 x 2 x 2 ( x 2 y ) dy dx = 1 1 2 x 4 dx = 4 / 5 7.
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