3
$\begingroup$

My task is this;

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:)

Thanks in advance!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Oh yes ofcourse, i completely missed the y variable in the integrand, that's why i thought it was rather hard. Thanks for answering! $\endgroup$ – Thomas Mar 8 '16 at 7:01
1
$\begingroup$

Your setup is wrong. See the picture below:

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.