Formula for integral, $\int\limits_{B(0,1)} x^a y^b z^c dx\ dy\ dz$, over unit ball

Question

Is there any general formula for the integral $$\int\limits_{B(0,1)} x^a y^b z^c dx\ dy\ dz$$ where $B(0,1)$ is the unit ball centered at $x=y=z=0$ and $a,b,c$ are positive integers.

Answer 1

Change to polar coordinates:

$\int\limits_{r=0}^1 \int\limits_{\theta = 0}^{2 \pi} \int\limits_{\phi = 0}^{\pi} (r \cos (\theta) \sin (\phi))^a (r \sin (\theta) \sin (\phi))^b (r \cos (\phi))^c r^2 \sin (\theta)\ dr\ d\theta\ d\phi$

which equals:

$$-\frac{\sqrt{\pi } \left((-1)^a+1\right) \left((-1)^b-1\right) \Gamma \left(\frac{a+1}{2}\right) \Gamma \left(\frac{b}{2}+1\right) \Gamma \left(\frac{1}{2} (a+b+c+1)\right)}{2 (a+b+c+3) \Gamma \left(\frac{1}{2} (a+b+3)\right) \Gamma \left(\frac{1}{2} (a+b+c+2)\right)}$$

In rectilinear coordinates (with assumptions about $a,b,c$):

$8 \int\limits_{x=0}^1 \int\limits_{y=0}^{\sqrt{1-x^2}} \int\limits_{z=0}^{\sqrt{1 - x^2 - y^2}} x^a y^b z^c dx\ dy\ dz =$

$$\frac{\Gamma \left(\frac{a+1}{2}\right) \Gamma \left(\frac{b+1}{2}\right) \Gamma \left(\frac{c+1}{2}\right)}{\Gamma \left(\frac{1}{2} (a+b+c+5)\right)}$$

I'm not sure if the assumptions underlying the $a$, $b$, and $c$ make these two results equal.