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. Z 1 0 Z 2 0 ( x +3) dy dx = Z 1 0 (2 x +6) dx =7 2. Z 3 1 Z 1 1 (2 x 4 y ) dy dx = Z 3 1 4 xdx =16 3. Z 4 2 Z 1 0 x 2 ydxdy = Z 4 2 1 3 ydy =2 4. Z 0 2 Z 2 1 ( x 2 + y 2 ) dx dy = Z 0 2 (3+3 y 2 ) dy =14 5. Z ln 3 0 Z ln 2 0 e x + y dy dx = Z ln 3 0 e x dx 6. Z 2 0 Z 1 0 y sin xdydx = Z 2 0 1 2 sin =(1 cos 2) / 2 7. Z 0 1 Z 5 2 dx dy = Z 0 1 3 dy =3 8. Z 6 4 Z 7 3 dy dx = Z 6 4 10 dx =20 9. Z 1 0 Z 1 0 x ( xy +1) 2 dy dx = Z 1 0 µ 1 1 x +1 dx =1 ln 2 10. Z π π/ 2 Z 2 1 x cos xy dy dx = Z π 2 (sin 2 x sin x ) dx = 2 11. Z ln 2 0 Z 1 0 xy e y 2 x dy dx = Z ln 2 0 1 2 ( e x 1) dx ln 2) / 2 12. Z 4 3 Z 2 1 1 ( x + y ) 2 dy dx = Z 4 3 µ 1 x 1 x +2 dx = ln(25 / 24) 13. Z 1 1 Z 2 2 4 xy 3 dy dx = Z 1 1 0 dx =0 14. Z 1 0 Z 1 0 xy p x 2 + y 2 dy dx = Z 1 0 [ x ( x 2 +2) 1 / 2 x ( x 2 1 / 2 ] dx =(3 3 4 2+1) / 3 15. Z 1 0 Z 3 2 x p 1 x 2 dy dx = Z 1 0 x (1 x 2 ) 1 / 2 dx / 3 16. Z 2 0 Z 3 0 ( x sin y y sin x ) dy dx = Z 2 0 µ x 2 π 2 18 sin x dx = π 2 / 144 17. (a) x k = k/ 2 1 / 4 ,k , 2 , 3 , 4; y l = l/ 2 1 / 4 ,l , 2 , 3 , 4 , ZZ R f ( x, y ) dxdy 4 X k =1 4 X l =1 f ( x k ,y l )∆ A kl = 4 X k =1 4 X l =1 [( 2 1 / 4) 2 +( l/ 2 1 / 4)](1 / 2) 2 =37 / 4 (b) Z 2 0 Z 2 0 ( x 2 + y ) dxdy =28 / 3; the error is | 37 / 4 28 / 3 | / 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 , 2 , 3 , 4 , ZZ R f ( x, y ) dxdy 4 X k =1 4 X l =1 f ( x k ,y l )∆ A kl = 4 X k =1 4 X l =1 [( 2 1 / 4) 2( l/ 2 1 / 4)](1 / 2) 2 = 4 (b) Z 2 0 Z 2 0 ( x 2 y ) dxdy = 4; the error is zero 19. V = Z 5 3 Z 2 1 (2 x + y ) dy dx = Z 5 3 (2 x +3 / 2) dx =19 20. V = Z 3 1 Z 2 0 (3 x 3 x 2 y ) dy dx = Z 3 1 (6 x 3 +6 x 2 ) dx = 172 21. V = Z 2 0 Z 3 0 x 2 dy dx = Z 2 0 3 x 2 dx =8 22. V = Z 3 0 Z 4 0 5(1 x/ 3) dy dx = Z 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. Z 1 / 2 0 Z π 0 x cos( xy ) cos 2 πx dy dx = Z 1 / 2 0 cos 2 πx sin( xy ) i π 0 dx = Z 1 / 2 0 cos 2 sin πxdx = 1 3 π cos 3 i 1 / 2 0 = 1 3 π
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622 Chapter 15 26. (a) y x z 5 3 (0, 2, 2) (5, 3, 0) (b) V = Z 5 0 Z 2 0 ydydx + Z 5 0 Z 3 2 ( 2 y +6) dy dx =10+5=15 27. f ave = 2 π Z π/ 2 0 Z 1 0 y sin xy dx dy = 2 π Z 2 0 µ cos xy i x =1 x =0 dy = 2 π Z 2 0 (1 cos y ) dy =1 2 π 28. average = 1 3 Z 3 0 Z 1 0 x ( x 2 + y ) 1 / 2 dx dy = Z 3 0 1 9 [(1 + y ) 3 / 2 y 3 / 2 ] dy = 2(31 9 3) / 45 29. T ave = 1 2 Z 1 0 Z 2 0 ( 10 8 x 2 2 y 2 ) dy dx = 1 2 Z 1 0 µ 44 3 16 x 2 dx = µ 14 3 30. f ave = 1 A ( R ) Z b a Z d c kdydx = 1 A ( R ) ( b a )( d c ) k = k 31. 1 . 381737122 32. 2 . 230985141 33. ZZ R f ( x, y ) dA = Z b a " Z d c g ( x ) h ( y ) dy ± dx = Z b a g ( x ) " Z d c h ( y ) dy ± dx = " Z b a g ( x ) dx ±" Z d c h ( y ) dy ± 34. The integral of tan x (an odd function) over the interval [ 1 , 1] is zero. 35. The Frst integral equals 1/2, the second equals 1 / 2. No, because the integrand is not continuous. EXERCISE SET 15.2 1. Z 1 0 Z x x 2 xy 2 dy dx = Z 1 0 1 3 ( x 4 x 7 ) dx / 40 2. Z 3 / 2 1 Z 3 y y ydxdy = Z 3 / 2 1 (3 y 2 y 2 ) dy =7 / 24 3. Z 3 0 Z 9 y 2 0 = Z 3 0 y p 9 y 2 dy =9 4. Z 1 1 / 4 Z x x 2 p x/y dy dx = Z 1 1 / 4 Z x x 2 x 1 / 2 y 1 / 2 dy dx = Z 1 1 / 4 2( x x 3 / 2 ) dx =13 / 80
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Exercise Set 15.2 623 5.
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Calculus: Early Transcendentals, by Anton, 7th Edition,ch15...

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