I need to find the centroid of the following body but first I need to find the volume which I'm currently stuck with but I think I'm close
$z=2y$, $z=x^2+y^2$
The integral I found is
$$\int _0^4\:\int _0^{2\pi }\:\int _0^{\sqrt{z-1}}\:rdrd\theta \:dz$$
I think I did something wrong for the bounds of r
I also tried it using cartesian coordinates which gave me
$$\int _0^4\:\int _0^{\frac{z}{2}}\:\int _{-\sqrt{z-y^2}}^{\sqrt{z-y^2}}\:dxdydz$$
this also gave me the wrong answer
The volume should be $\frac{\pi}{2}$
$z = 2y$ and $z = x^2 + y^2$ should give you $x^2+y^2 = 2y \implies x^{2} + \left(y - 1\right)^{2} = 1$. This is the constraint. Looks like a circle centered at $\left(0,1\right)$. Cylindrical coordinated about that point? $(x,y,z) \mapsto (r\cos(\theta),1 + r\sin\theta,z)$, where $0\leq r \leq 1$, $0 \leq \theta \leq 2\pi$, and $r^{2} + 2 r \sin{\left(\theta \right)} + 1\leq z \leq 2r\sin\theta + 2$. So we set up the integral: $$\int_0^1 \int_0^{2\pi}\int_{r^{2} + 2 r \sin{\left(\theta \right)} + 1}^{2r\sin\theta + 2} r\, dz\, d\theta \, dr \\= \int\limits_{0}^{1}\int\limits_{0}^{2 \pi}\int\limits_{r^{2}}^{1} r\, dz\, d\theta\, dr = \frac{\pi}{2}$$