Hw15-4 - Math 1920 Solutions for 15.4 4 We have x=1 y = 3x...

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Math 1920 Solutions for 15.4 4. We have: x = 1 r = sec θ y = 3 x θ = π 3 Thus the region can be described as: 0 θ π 3 0 r sec θ 5. We have x 2 + y 2 = 1 r = 1 x = 2 3 r = 2 3 sec θ y = 2 r = 2csc θ We have to split the region in two parts, because of the boundary of the rect- angle. The corner is given at 2 3 sec θ = 2csc θ θ = π 6 Thus the region can be given as: 0 θ π 6 , 1 r 2 3 sec θ π 6 θ π 2 , 1 r 2 csc θ 1
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16. We have: Z 2 2 Z y 4 - y 2 dydx = Z π 2 π 4 Z 2 csc θ 2 rdrdθ = Z π 2 π 4 ( 2 csc 2 θ - 2 ) = [ - 2 cot θ - 2 θ ] π 2 π 4 = 2 - π 2 20. We have: Z 1 - 1 Z 1 - y 2 - 1 - y 2 ln( x 2 + y 2 + 1) dxdy = 4 Z π 2 0 Z 1 0 ln( r 2 + 1) rdrdθ = 2 Z π 2 0 ln 4 - 1 = π (ln 4 - 1) 30. The area is given by:
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Hw15-4 - Math 1920 Solutions for 15.4 4 We have x=1 y = 3x...

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