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exam1reviewsol - R π 4 R √ 2 f ρ sin φ cos θ ρ sin...

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Practice exam solutions for calc IV, Summer 2007 Exam 1 1. (15 points) a) (1 - 1 2 - 1 5 ) - (( - 3) 3 - 1 2 ( - 3) 4 - 1 5 ( - 3) 5 ) = 96 5 . c) R 1 - 7 R 1 2 ( y +1) - 2 - y x 2 dxdy + R 2 1 R 2 - y 2 - y x 2 dxdy 2. b) u = x - y 2 , v = y , for instance. So x = u + v 2 and y = v d) J ( T ) = ( x, y ) ( u, v ) = ± ± ± ± 1 2 v 0 1 ± ± ± ± = 1 . Areas are preserved under this transformation! e) RR R ( x - y 2 ) 100 dA = R 1 0 R 1 0 u 100 du = 1 101 . 3. y = uv, x = p u v . J ( T ) = 1 2 v - 1 . Answer = R 16 9 R 4 1 ( u 1 / 2 v 1 / 2 ) 1 2 v - 1 dudv = 14 / 3 . 4. a) i) R 1 - 1 R 1 - x 2 - 1 - x 2 R 2 - x 2 - y 2 x 2 + y 2 f ( x, y, z ) dzdydx , ii) R 2 π 0 R 1 0 R 2 - r 2 r f ( r cos θ, r sin θ, z ) rdzdrdθ , iii) R 2 π
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Unformatted text preview: R π/ 4 R √ 2 f ( ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ ) ρ 2 sin φdρdφdθ 5. R 1 R √ 1-z R 1-z x 2 f ( x, y, z ) dydxdz . 6. a) C = 3 25 . b) 3 25 (-10 3 ) ² e-3 5-1 ³ (-5 2 ) ² e-8 5-1 ³ ≈ . 360 c) e-4 5 ² e 4 5-4 e 1 5 + 3 ³ ≈ . 153 7. mR 2 2 1...
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