# tet - 2 sin 2 φ ρ cos φ ρ 2 sin φ dρdφdθ 5 Taking u...

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MATH 241H - EXAM 3 - Solutions - November 1 , 2004 1. R 1 0 R 1 y x 2 cos( xy ) dxdy = R 1 0 R x 0 x 2 cos( xy ) dydx = R 1 0 x sin( x 2 ) dx = (1 - cos 1) / 2. 2. (b) R 2 π 0 R θ 0 r drdθ = R 2 π 0 θ 2 / 2 = 4 π 3 / 3. 3. Z 2 π 0 Z 2 0 Z 5 - r 2 1+ r 2 r r dzdrdθ = Z 2 π 0 Z 2 0 (4 - 2 r 2 ) drdθ = 2 π (4 2 - 4 2 / 3) = 16 π 2 / 3 4. ( a ) Z 2 - 2 Z 4 - x 2 - 4 - x 2 Z 2 x 2 + y 2 ( x 2 + y 2 + z ) dzdydx. ( b ) Z 2 π 0 Z 2 0 Z 2 r ( r 2 + z ) r dzdrdθ ( c ) Z 2 π 0 Z π/ 4 0 Z 2 / cos φ 0 ( ρ
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Unformatted text preview: 2 sin 2 φ + ρ cos φ ) ρ 2 sin φ dρdφdθ 5. Taking u = 2 x + y and v = x-y we get Z 4 1 Z 1-1 ue v (1 / 3) dvdu = 5( e-e-1 ) / 2...
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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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