Use polar coordinates d r 0 2 0 r 3 and fx

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Unformatted text preview: (16 + 24 cos θ + 9(1 + cos 2θ)/2) dθ 20 41 =π 2 5. Use polar coordinates, D = {(r, θ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3}, and fx = 2x, fy = 2y. So 3 2π (2x)2 + (2y )2 + 1 dA = A= D 0 3 0 √ π√ ( 4r2 + 1)r dr dθ = (37 37 − 1) 6 6. Solution 1: Use a double integral, D is the intersection of z = 3x2 + 3y 2 z = 4 − x2 − y 2 x2 + y 2 = 1. ⇒ So D = {(r, θ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1}, and V= D 2π (top surface − bottom surface) dA 1 = 0 0 ((4 − r2 ) − 3r2 )r dr dθ = 2π. Solution 1: Use triple integral in cylindrical coordinates,...
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