# HW09 - HW 9 Solutions Math 1920 Spring 2009 April 1 2009...

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Unformatted text preview: HW 9 Solutions Math 1920 Spring 2009 April 1, 2009 Problem 15.Rev.40 Due to symmetry we can integrate just over a quarter of the region. V = 4 R 2 R √ 4- x 2 R 4- x 2 dz dy dx V = 4 R 2 R √ 4- x 2 4- x 2 dy dx V = 4 R 2 ( 4- x 2 ) √ 4- x 2 dx Use the following trig sub: x = 2 sinθ dx = 2 cosθ V = 64 R π 2 ( 1- sin 2 θ ) √ 1- sin 2 θ cosθdθ V = 64 R π 2 cos 4 θdθ V = 64 R π 2 1+ cos (2 θ ) 2 1+ cos (2 θ ) 2 dθ V = 16 R π 2 1 + 2 cos (2 θ ) + cos 2 (2 θ ) dθ V = 16 R π 2 3 2 + 2 cos (2 θ ) + cos (4 θ ) 2 dθ V = 16 h 3 2 θ + sin (2 θ ) + sin (4 θ ) 4 i π 2 = 12 π Problem 15.Rev.52 - Apple I z = R R R V x 2 + y 2 δ dV I z = R 2 π R π R 1- cos ( φ ) ( ρ 2 sin 2 φ ) ρ 2 sinφ δ dρdφdθ I z = 2 π δ R π R 1- cos ( φ ) ρ 4 sin 3 φdρdφ I z = 2 π δ 5 R π (1- cosφ ) 5 sin 3 φdφ I z = 2 π δ 5 R π (1- cosφ ) 6 (1 + cosφ ) sinφdφ Use following subsitution: u = 1- cosφ du = sinφ 1 + cosφ = 2- u 1 I z = 2 π δ 5 R 2 2 u 6- ( u ) 7 du...
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HW09 - HW 9 Solutions Math 1920 Spring 2009 April 1 2009...

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