# How to setup a double integral when the region is bounded by a circle and a parabola?

Calculate$$\iint\limits_{A}y\:dA.$$

Where $A$ is the region in the $xy-$plane such that $x^2\leq y,\: x^2 + y^2 \leq 2$.

My work so far:

Our region $A$ is in the first and seccond quadrant above the parabola $x^2$ and below the circle centered at the origin with a radius of $\sqrt{2}$. Switching to polar coordinates gives us (remember the jacobian):$$\int\limits_{0}^{\pi}\int\limits_{r^2\cos^2(\theta)}^{\sqrt{2}}r^2\sin(\theta)\: dr\:d\theta.$$

However this setup leads to an answer with a variable $r$ and since the answer is a real number i must have set this one up wrong. Hints are welcome, and don't show calculations that reveal the answer as i would very much like to do that on my own:)

Stick with rectangular coordinates. I get

$$\int_{-1}^1 dx \, \int_{x^2}^{\sqrt{2-x^2}} dy \, y$$

which I imagine you should be able to do easily.

• Oh yes ofcourse, i completely missed the y variable in the integrand, that's why i thought it was rather hard. Thanks for answering! – Thomas Mar 8 '16 at 7:01

Your setup is wrong. See the picture below:

If $0\le \theta \le \frac{\pi}{4}$ or $\frac{3\pi}{4}\le \theta \le \pi$, then we get $r \le \frac{\sin \theta}{\cos^2\theta}$, but if $\frac{\pi}{4} \le \theta\le\frac{3\pi}{4}$, then we get just $r\le \sqrt{2}$.

Thus you have to integrate $$\int_0^{\frac{\pi}{4}}\int_0^{\frac{\sin\theta}{\cos^2\theta}} r^2\sin\theta dr d\theta +\int_\frac{\pi}{4}^\frac{3\pi}{4}\int_0^\sqrt{2} r^2\sin \theta drd\theta+\int_\frac{3\pi}{4}^{\pi}\int_0^{\frac{\sin\theta}{\cos^2\theta}}r^2\sin\theta dr d\theta$$