Use cylindrical or spherical coordinates to evaluate triple integral

Use cylindrical or spherical coordinates to evaluate the triple integral ∫ ∫ ∫ zdV where V lies above the paraboloid z=x^2+y^2 and below the plane z=6y...

Really struggling with this one, thanks
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Use cylindrical coordinates.

Clearly, z = x^2 + y^2 to z = 6y
==> z = r^2 to z = 6r sin θ.

The surfaces intersect at x^2 + y^2 = 6y ==> r^2 = 6r sin θ ==> r = 6 sin θ.
(This is traced out completely for θ in [0, π].)

So, ∫∫∫ z dV
= ∫(θ = 0 to π) ∫(r = 0 to 6 sin θ) ∫(z = r^2 to 6r sin θ) z * (r dz dr dθ)
= ∫(θ = 0 to π) ∫(r = 0 to 6 sin θ) (1/2)rz^2 {for z = r^2 to 6r sin θ} dr dθ
= (1/2) ∫(θ = 0 to π) ∫(r = 0 to 6 sin θ) [36r^3 sin^2(θ) - r^5] dr dθ
= (1/2) ∫(θ = 0 to π) [9r^4 sin^2(θ) - r^6/6] {for r = 0 to 6 sin θ} dθ
= 1944 ∫(θ = 0 to π) sin^6(θ) dθ
= 1944 ∫(θ = 0 to π) [(1/2)(1 - cos(2θ))]^3 dθ
= 243 ∫(θ = 0 to π) [1 - 3 cos(2θ) + 3 cos^2(2θ) - cos^3(2θ)] dθ
= 243 ∫(θ = 0 to π) [1 - 3 cos(2θ) + (3/2)(1 + cos(4θ)) - (1 - sin^2(2θ)) cos(2θ)] dθ
= 243 * 5π/2
= 1215π/2.

I hope this helps!