calculus 3 math problem I really need help:(>.<):?

Note that in polar coordinates:
x^2 + y^2 = 2y
==> r^2 = 2r sin θ
==> r = 2 sin θ, which is completely traced out with θ in [0, π].

So, the volume equals
∫∫ (3x + 4y + 30) dA
= ∫(θ = 0 to π) ∫(r = 0 to 2 sin θ) (3r cos θ + 4r sin θ + 30) * (r dr dθ), via polar coordinates
= ∫(θ = 0 to π) ∫(r = 0 to 2 sin θ) (3r^2 cos θ + 4r^2 sin θ + 30r) dr dθ
= ∫(θ = 0 to π) (r^3 cos θ + (4/3)r^3 sin θ + 15r^2) {for r = 0 to 2 sin θ} dθ
= ∫(θ = 0 to π) [8 sin^3(θ) cos θ + (32/3) sin^4(θ) + 60 sin^2(θ)] dθ
= ∫(θ = 0 to π) [8 sin^3(θ) cos θ + (32/3) * ((1/2)(1 - cos(2θ)))^2 + 60(1/2)(1 - cos(2θ))] dθ
= ∫(θ = 0 to π) [8 sin^3(θ) cos θ + (8/3)(1 - 2 cos(2θ) + cos^2(2θ) + 30(1 - cos(2θ))] dθ
= ∫(θ = 0 to π) [8 sin^3(θ) cos θ + (8/3)(1 - 2 cos(2θ) + (1/2)(1 + cos(4θ)) + 30(1 - cos(2θ))] dθ
= ∫(θ = 0 to π) [8 sin^3(θ) cos θ + (4/3)(3 - 4 cos(2θ) + cos(4θ)) + 30(1 - cos(2θ))] dθ
= [2 sin^4(θ) + (4/3)(3θ - 2 sin(2θ) + (1/4)sin(4θ)) + 30(θ - (1/2) sin(2θ))] {for θ = 0 to π}
= 34π.

I hope this helps!