ansch12review

ansch12review - y = x D 1 x 1 y ZZ D 1 x 2 + 1 dA = Z 1 Z 1...

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ANSWERS TO EVEN ASSIGNED REVIEW PROBLEMS FROM CH. 12 2. As in Exercise 1. , we have m = n =3andΔ A = 1. Using the contour map to estimate the value of f at the center of each subsquare, we have ZZ R f ( x, y ) dA Δ A ( f ( . 5 ,. 5) + f ( . 5 , 1 . 5) + f ( . 5 , 2 . 5) + f (1 . 5 , 1 . 5) + f (1 . 5 , 2 . 5) + f (2 . 5 ,. 5) + f (2 . 5 , 1 . 5) + f (2 . 5 , 2 . 5)) 1 · (1 . 2+2 . 5+5+3 . 2+4 . 5+7 . 1+5 . 2+6 . 5+9 . 0) 44 . 2 10. The region R is enclosed by the lines x =4 y , x = y 4andthe x -axis. Setting up the integral as a type II region, gives ZZ R f ( x, y ) dA = Z 4 0 Z 4 y y 4 f ( x, y ) dxdy. 14. The region D is shown below. The integral is set up as a type II region. Reversing the order of the limits and setting up the integral as a type I gives: x = y D 01 x 1 y Z 1 0 Z x 2 0 ye x 2 x 3 dydx = Z 1 0 e x 2 x 3 ± y 2 2 ² x 2 0 dx = Z 1 0 1 2 xe x 2 dx = 1 2 " e x 2 2 # 1 0 = 1 4 ( e 1) 18. The region D is shown below. Setting up the integral as a Type I region we have: y
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Unformatted text preview: y = x D 1 x 1 y ZZ D 1 x 2 + 1 dA = Z 1 Z 1 x 1 x 2 + 1 dydx = Z 1 1 x 2 + 1 [ y ] 1 x dx = Z 1 1 x 2 + 1 x x 2 + 1 dx = arctan x 1 2 ln( x 2 + 1) 1 = arctan 1 1 2 ln 2 (arctan 0 1 2 ln 1) = 4 1 2 ln 2 22. The limits for are 0 2 (Frst quadrant) and the limits for r are 1 r 2 (between the circles of radius 1 and 2). We then have ZZ D xdA = Z / 2 Z 2 1 r 2 cos drd = Z / 2 cos r 3 3 2 1 d = (2 2 1) 3 Z / 2 cos d = (2 2 1) 3 [sin ] / 2 = (2 2 1) 3 1...
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This note was uploaded on 08/25/2011 for the course CHE 113 taught by Professor Burrows during the Spring '11 term at ASU.

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