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.
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!