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...

  • 2
    $\begingroup$ Use polar coordinates. $\endgroup$
    – Umberto P.
    Commented Jan 12, 2022 at 20:44

2 Answers 2


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 $[0,2\pi].$

  • $\begingroup$ Where does the $+1$ come from? In $r^{m+n+1}$ $\endgroup$ Commented Jan 12, 2022 at 22:55
  • $\begingroup$ @LuTo peek-a-boo beat me to it, but the +1 comes from the fact that $\mathrm{d}x\,\mathrm{d}y=r\,\mathrm{d}r\,\mathrm{d}\theta,$ which occurs from the change of variables theorem. Essentially, the absolute value of the determinant of the Jacobian for the change of variables is $r,$ so that is why there is an additional factor. $\endgroup$
    – Angel
    Commented Jan 13, 2022 at 13:55

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: $$\int_0^{\pi/2} \cos^m\theta \sin^n \theta \, d\theta = \frac 12 B\left( \frac {m+1}2,\frac{n+1}2 \right).$$

  • $\begingroup$ Where does the $+1$ come from? In $r^{m+n+1}$ $\endgroup$ Commented Jan 12, 2022 at 23:05
  • 3
    $\begingroup$ @LuTo $dx\,dy$ becomes $r\,dr\,d\theta$, the extra factor of $r$ is the Jacobian determinant of the transformation from polar to cartesian coordinates $\left|\det\begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta}\end{pmatrix}\right|$. Look at the change of variables theorem for why this Jacobian determinant factor appears. $\endgroup$
    – peek-a-boo
    Commented Jan 13, 2022 at 2:47

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