I was tutoring someone and one of the problems was to integrate $r^4\cos^2\theta\sin\theta$ over the portion of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ that lies in the first quadrant. So, $0\leq \theta\leq\frac{\pi}{2}$. But, given a $\theta$, $r$ runs from $0$ to
$$ \sqrt{\frac{36+36\tan^2\theta}{9+4\tan^2\theta}}. $$
After integrating with respect to $r$ first, it looks very nasty.
Is there some clever method, such as a change of variables, that makes this easy to integrate?
In this particular case, isn't it more easy to do in cartesian coordinates?
$$\int \int_R r^4 \cos^2 \theta \sin \theta dr \; d\theta = \int \int_R (r \cos\theta)^2 (r \sin \theta) \; r \; dr \; d\theta = \int \int_R x^2 y \; dx dy = $$
$$= \int_0^2 x^2 \int_0^{3\sqrt{1-x^2/4}} y \; dy \; dx = \frac{9}{2} \int_0^2 x^2 \left(1- \frac{x^2}{4}\right) dx = \cdots $$