Example 4 find the volume of the solid that lies under the paraboloid z = 6x2 + 6y2, above the xy-plane, and inside the cylinder x2 + y2 = 2x. solution the solid lies above the disk d whose boundary circle has equation x2 + y2 = 2x or, after completing the square, x − 2 + y2 = . in polar coordinates we have x2 + y2 = r2 and x = r cos(θ), so the boundary circle becomes r2 = 2r cos(θ), or r = 2 cos(θ). thus the disk d is given by d = (r, θ) −π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2 cos(θ) and, by this formula, we have v = d (6x2 + 6y2) da = π/2 −π/2 2 cos(θ) 0 6r2 r dr dθ = π/2 −π/2 6 4r4 2 cos(θ) 0 dθ = π/2 0 cos4(θ) dθ = π/2 0 1 + cos(2θ) 2 2 dθ = π/2 0 1 + 2 cos(2θ) + 1 2 (1 + cos(4θ)) dθ