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ansch12review

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

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ANSWERS TO EVEN ASSIGNED REVIEW PROBLEMS FROM CH. 12 2. As in Exercise 1. , we have m = n = 3 and Δ A = 1. Using the contour map to estimate the value of f at the center of each subsquare, we have 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 4 and the x -axis. Setting up the integral as a type II region, gives R f ( x, y ) dA = 4 0 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 0 1 x 1 y 1 0 x 2 0 ye x 2 x 3 dydx = 1 0 e x 2 x 3 y 2 2 x 2 0 dx = 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:
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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 ³ 1 x 2 + 1 − x x 2 + 1 ´ dx = ± arctan x − 1 2 ln( x 2 + 1) ² 1 = arctan 1 − 1 2 ln 2 − (arctan 0 − 1 2 ln 1) = π 4 − 1 2 ln 2 22. The limits for θ are 0 ≤ θ ≤ π 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 π/ 2 Z √ 2 1 r 2 cos θdrdθ = Z π/ 2 cos θ ± r 3 3 ² √ 2 1 dθ = (2 √ 2 − 1) 3 Z π/ 2 cos θdθ = (2 √ 2 − 1) 3 [sin θ ] π/ 2 = (2 √ 2 − 1) 3 1...
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