double integral convert from cartesian to polar

Question

I have this question but I'm stuck on the integral - usually the r gets cancelled out but this time they're both r^2? I can't tell what I'm not seeing or doing wrong.

Answer 1

You've done the transformation correctly: the resulting integral is $$I = \int_{\theta=-\pi/4}^{\pi/3} \cos \theta \, d\theta \int_{r=1}^2 r^2 \sin r^2 \, dr.$$

Although the integral with respect to $\theta$ is simply $(\sqrt{2} + \sqrt{3})/2$, the integral with respect to $r$ does not have an elementary closed form. I suspect the author of the problem failed to realize that the Jacobian of the transformation introduces another factor of $r$. The antiderivative can be expressed in terms of Fresnel integrals but this is not useful. The approximate numerical value is $$I \approx 1.1047531349442256551\ldots.$$