# How to integrate $y^2$ over a circular annulus?

The area to be integrated over is the quarter annulus that lies in the first quadrant contained between circles $x^2+y^2=r_i^2$ and $x^2+y^2=r_o^2$.

$\int{x^2}\,dA$=$\int{x^2}\,dx\,dy=\int r^2\cos^2\theta\,dx\,dy$.

How to simplify $dx$ and $dy$ in terms of $r$ and $\theta$ so that integrating easily takes place. I am getting $dx=cos\theta\,dr-r\,sin\theta\,d\theta$, similarly for $dy$ which leads to very complicated integrand. Are there any tricks out there to integrate it?

• @ GoCodes you need to use Jacobian of transformation. – incognito Mar 9 '15 at 3:32

The integral in the polar coordinates will be: $$I = \int\limits_{0}^{\pi/2}\int\limits_{r_i}^{r_o} r^2 \cos^2{\theta} r dr d\theta$$ $$\implies I = \frac{1}{16} \pi \left(r_{o}^{4}-r_{i}^{4}\right)$$
• The limits are obvious as the integration has to be performed in the first quadrant so $\theta$ is between $0$ and $\pi/2$, and the region of integration is the annulus so $r_i\le r\le r_o$. – incognito Mar 9 '15 at 3:43