Find integral given volume I need to find the integral of the function $F(x,y,z) = z$ in the region $E$ that is top half of a sphere of $r=2$. So $z \geq{0}$ and $x^2+y^2+z^2\leq{4}$. I found the volume to be $\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{\sqrt{4-r^{2}}}rdzdrd\theta = \frac{16\pi}{3}$, but I do not know what to do next. 
$\overline{z}=\frac{1}{V}\int\int\int_{E}zdV$ is what I'm looking for. Any help is appreciated!
 A: Since the function is constant for a given $z$, why don’t we integrate like this:
$$
\begin{aligned}
\frac{1}{V}\int{z}\ dV&=\frac{1}{V}\int_{0}^{2}{z\pi\left(2^{2}-z^{2}\right)}\ dz\\
&=\frac{\pi}{V}\int_{0}^{2}{4z-z^{3}}\ dz
\end{aligned}
$$
Imagine the half sphere as discs with radius $\sqrt{4-z^{2}}$ with infinitesimal thickness $dz$, stacked on top of each other.
A: I presume the integral you want is $I=2\pi\int_0^2\int_0^{\sqrt{4-r^2}}rzdzdr=2\pi\int_0^2 \frac{4-r^2}{2}rdr=4\pi$
A: Take a point $P(x,y,z)$ on sphere. Suppose $\phi$ is the angle which line $OP$ makes with $z-$ axis and $\theta$ is the angle on $xy-$ plane.
Then with polar coordinates  $P(rsin\phi cos\theta, rsin\phi sin\theta, rcos\phi)$; where $0<r<2, 0<\theta <2π, 0<\phi<π/2$ and $\lvert\frac{\partial(x,y,z)}{\partial(r,\theta,\phi)}\rvert=r^2 sin\phi$
$V=\int_0^{π/2}\int_0^{2π}\int_0^2 r^2 sin\phi dr d\theta d\phi$
$\implies V=\int_0^{π/2}sin\phi d\phi\times\int_0^{2π}d\theta\times \int_0^2 r^2 dr$
$=1\times 2π\times \frac{8}{3}=16\frac{π}{3}$
