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# vbnbvn - | (1 / 2)(-1 / 2)-(1 / 2)(1 / 2) | = | -1 / 2 | =...

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MATH 241H - EXAM 3 - April 16, 2007 No calculators. Show your work. 1. Z 2 - 2 Z 4 - x 2 0 p x 2 + y 2 dydx = Z π 0 Z 2 0 r 2 drdθ = 8 π/ 3 2. r u × r v = (2 u i + 2 u j ) × ( i - j + k ) = 2 u i - 2 u j - 4 u k so k r u × r v k = 24 u 2 = 2 u 6. Thus surface area = Z 1 0 Z u - u 2 u 6 dvdu = 4 6 / 3 . 3. The top z = 4 - r 2 intersects the bottom z = 1 when 4 - r 2 = 1 so that r = 3. Thus Z Z Z D z dV = Z 2 π 0 Z 3 0 Z 4 - r 2 1 zr dzdrdθ = 9 π/ 4 4. Z 2 - 2 Z 4 - x 2 - 4 - x 2 Z x 2 + y 2 0 ( x 2 + y 2 + z ) dzdydx = Z 2 π 0 Z 2 0 Z r 0 ( r 2 + z ) r dzdrdθ = Z 2 π 0 Z π/ 2 π/ 4 Z 2 / sin φ 0 ( ρ 2 sin 2 φ + ρ cos φ ) ρ 2 sin φ dρdφdθ [20] 5. Absolute value of Jacobian is
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Unformatted text preview: | (1 / 2)(-1 / 2)-(1 / 2)(1 / 2) | = | -1 / 2 | = 1 / 2. x + y = u, x-y = v , and x = y becomes v = 0, y =-x becomes u = 0, y =-x + / 2 becomes u = / 2, and y = x-5 becomes v = 5. Thus the integral is Z / 2 Z 5 v cos u (1 / 2) dvdu = 25 / 4...
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## This note was uploaded on 04/04/2010 for the course MATH 113 taught by Professor Staff during the Spring '08 term at Maryland.

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