What is $\int_{|\vec x| = 1, z \geq 0}(x^2+y^2)^pz^q$ for $p,q \geq 0$? What is $\int_{|\vec x = (x,y,z)| = 1, z \geq  0}(x^2+y^2)^pz^q$ for $p,q \geq 0$?
A hint is to use the Gamma function. I plugged in spherical coordinates and got:
$I = 2\pi \int \int r^{2p+q+2}sin^{2p+1}(\theta)cos^{q}(\theta)$. 
I don't see how to use the Gamma function here, and would appreciate any help.
 A: Note that on the sphere, $r=|\vec x|=1$.  Therefore, using spherical coordinates $(r=1,\theta,\phi)$, we have 
$$\begin{align}
I&=\int_0^{2\pi} \int_0^{\pi/2} \left(\sin(\theta)\right)^{2p+1}\left(\cos(\theta)\right)^q\,d\theta \,d\phi\\\\
&2\pi \int_0^{\pi/2} \left(\sin(\theta)\right)^{2p+1}\left(\cos(\theta)\right)^q\,d\theta\\\\
&=\pi B\left(p+1,\frac{q+1}{2}\right)\\\\
&=\pi \frac{\Gamma\left(p+1\right)\Gamma\left(\frac{q+1}{2}\right)}{\Gamma\left(\frac{2p+q+3}{2}\right)} 
\end{align}$$

Alternatively, using cylindrical coordinates $(\rho=\sqrt{1-z^2},\phi,z)$, we can write
$$I=2\pi \int_0^1 (1-z^2)^p z^q \,dz \tag 1$$
Enforcing the substitution $z\to \sqrt{z}$ into $(1)$ yields
$$\begin{align}
I&=\pi \int_0^1 (1-z)^{p}z^{(q-1)/2}\,dz\\\\
&=\pi B\left(p+1,\frac{q+1}{2}\right)\\\\
&=\pi \frac{\Gamma\left(p+1\right)\Gamma\left(\frac{q+1}{2}\right)}{\Gamma\left(\frac{2p+q+3}{2}\right)}
\end{align}$$
as expected!

And finally, using cylindrical coordinates $(\rho,\phi,z=\sqrt{1-\rho^2})$, we can write
$$\begin{align}
I&=2\pi \int_0^1 (\rho^2)^p (1-\rho^2)^{q/2} \frac{\rho}{\sqrt{1-\rho^2}}\,d\rho \\\\
&=2\pi\int_0^1 \rho^{2p+1}(1-\rho^2)^{(q-1)/2}\,d\rho\\\\
&=\pi\int_0^1 t^p(1-t)^{(q-1)/2}\,dt\\\\
&=\pi B\left(p+1,\frac{q+1}{2}\right)\\\\
&=\pi \frac{\Gamma\left(p+1\right)\Gamma\left(\frac{q+1}{2}\right)}{\Gamma\left(\frac{2p+q+3}{2}\right)}
\end{align}$$
as expected!
