This is a homework problem, so please do not give more than hints. I must convert \begin{align} \int_0^\sqrt{2}\int_x^\sqrt{4-x^2}\sin\left(x^2+y^2\right)\:dy\:dx\tag{1} \end{align} to polar coordinates. This is my attempt: \begin{align} \int_{\pi/4}^{\pi/2}\int_{\color{red}{2\cos\left(\theta\right)}}^{\color{red}{2\sin\left(\theta\right)}}\sin\left(r^2\right)r\:dr\:d\theta,\tag{2} \end{align} but I am unsure about the $\color{red}{\text{red}}$ limits, because while I am solving I end up at \begin{align} \int_{\pi/4}^{\pi/2}\frac{\cos\left(4\cos^2\left(\theta\right)\right)}{2}-\frac{\cos\left(4\sin^2\left(\theta\right)\right)}{2}\:d\theta\tag{3} \end{align} after a single round of $u$-substitution. There's no way it should end up here, unless it's really easy and I'm just not thinking...

I think the upper limit is $\color{red}{2\sin\left(\theta\right)}$ because a substitution of $2\cos\left(\theta\right)$ into $\sqrt{4-x^2}$ results in \begin{align} \sqrt{4-x^2}&=\sqrt{4-4\cos^2\left(\theta\right)}\\ &=2\sin\left(\theta\right), \end{align} and the lower limit is $\color{red}{2\cos\left(\theta\right)}$ by direct substitution as before.

Here is my $u$-substitution:

Let $\xi=r^2$, then $d\xi/2r=dr$, resulting in \begin{align} ;\;\int r\sin\left(r^2\right)\:dr&=\frac{1}{2}\int \sin\left(\xi\right)\:d\xi\\ &=\frac{-\cos\left(\xi\right)}{2}=\frac{-\cos\left(r^2\right)}{2}\bigg| \end{align} Thus, \begin{align} \int_{\pi/4}^{\pi/2}\int_{2\cos\left(\theta\right)}^{2\sin\left(\theta\right)}r\sin\left(r^2\right)\:dr\:d\theta&=\int_{\pi/4}^{\pi/2}\left[\frac{-\cos\left(r^2\right)}{2}\right]_{2\cos\left(\theta\right)}^{2\sin\left(\theta\right)}\;d\theta. \end{align} Where have I gone wrong?

  • 1
    $\begingroup$ The advice is always the same for this sort of question: draw a picture, and then you know what the region actually looks like. $\endgroup$ – Chappers Apr 17 '15 at 0:10

firstly, you must sketch the region enter image description here $$\int_{\pi /4}^{\pi /2}\int_{0}^{2}\sin(r^2)rdrd\theta $$


Below is the region over which you are integrating.

enter image description here

Can you now fix the limits of $r$ and $\theta$?


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.