So assume we have: $x^2+y^2\le2y$ , $0 \leqslant z \leqslant y$ evaluating over $\iiint\sqrt{x^2+y^2}\,dV$
Given the graph which lies in the 1st and 2nd quadrants, as well as the radius and given z intervals the integration intervals are as follows: $$0 \leqslant \theta \leqslant \pi$$ $$0 \leqslant r \leqslant 2y$$ $$ 0 \leqslant z \leqslant y$$
If we convert using cylindrical coordinates ($x=r\cos\theta, y=r\sin\theta, dV=r\,dz\,dr\,d\theta)$ they become: $$0 \leqslant \theta \leqslant \pi$$ $$0 \leqslant r \leqslant 2\sin\theta$$ $$0 \leqslant z \leqslant r\sin\theta$$
And thus we get (using $r^2=x^2+y^2)$: $$\int_0^\pi \int_0^{2\sin\theta}\int_0^{r\sin\theta}r^2 \,dz\,dr\,d\theta$$
My question is, is this correct so far? If you do all the messy work after I think the answer is 64/15, but I really just need to know if I set it up correctly.