$\int_K |x|^m |y|^n dx dy$ Let $K=\{(x,y)\in \mathbb{R}^2 \,|\, x^2+y^2 \leq 1\}$.
I want to find $\int_K |x|^m |y|^n \,\,dx\,\, dy$.

My approach is:
$...\,\,=4\cdot \int_0^1 \int_0^{\sqrt{1-y^2}} |x|^m |y|^n\,dx \, dy = 4\cdot \int_0^1 |y|^n \cdot \int_0^{\sqrt{1-y^2}} |x|^m \,dx \, dy$
But I have no idea how to solve $\int_0^{\sqrt{1-y^2}} |x|^m \,dx$. I tried a transformation $x \mapsto \sqrt[m]{x}$, but this doesn't make it any better.
I looked $\int_0^{\sqrt{1-y^2}} |x|^m \,dx$ up on Wolfram-Alpha, it seems quite complex. And then I would have to put in the $\sqrt{1-y^2}$ from the integral bounds. So, is this really a good path to go? I'm lacking ideas...
 A: Since $x$ belongs to the the interval $[0,\sqrt{1-y^2}],$ $|x|^m=x^m.$ However, I think this method of solving the integral is not very convenient, since it means you will need to evaluate $$\int_0^1\frac{y^n\sqrt{1-y^2}^m}{m+1}\,\mathrm{d}y,$$ which is nightmarish. I think you should consider using polar coordinates. Let $x(r,\theta)=r\cos(\theta)$ and $y(r,\theta)=r\sin(\theta).$ Hence $$\int_K|x|^m|y|^n\,\mathrm{d}x\,\mathrm{d}y=\int_0^{2\pi}\int_0^1r^{m+n+1}|\cos(\theta)|^m|\sin(\theta)|^n\,\mathrm{d}r\,\mathrm{d}\theta.$$ By Fubini's theorem, you can say $$\int_0^{2\pi}\int_0^1r^{m+n+1}|\cos(\theta)|^m|\sin(\theta)|^n\,\mathrm{d}r\,\mathrm{d}\theta=\left(\int_0^1r^{m+n+1}\,\mathrm{d}r\right)\left(\int_0^{2\pi}|\cos(\theta)|^m|\sin(\theta)|^n\,\mathrm{d}\theta\right)$$ $$=\frac1{m+n+1}\int_0^{2\pi}|\cos(\theta)|^m|\sin(\theta)|^n\,\mathrm{d}\theta.$$
This is now a much simpler calculus problem that I believe you can solve on your own. To simplify the integral further, think about the symmetries of the trigonometric functions on the interval $[0,2\pi].$
A: Rewrite as $$\int_0^{2\pi} \int_0^1 r^{m+n+1} |\cos \theta|^m |\sin \theta|^n \, drd\theta = \frac{4}{m+n+2} \int_0^{\pi/2} \cos^m\theta \sin^n \theta \, d\theta$$ and focus on the trigonometric integral.

Addendum: the resulting integral can be evaluated using the Beta function:
$$\int_0^{\pi/2} \cos^m\theta \sin^n \theta \, d\theta = \frac 12 B\left( \frac {m+1}2,\frac{n+1}2 \right).$$
