Evaluate $\iint |xy|\,dx\,dy$ using polar coordinates If $R$ is the region bounded by $x^2+4y^2 \ge 1$ and $x^2+y^2 \le 1$. Then find the integral $$I=\iint_R |xy|\,dx\,dy.$$
I tried using Cartesian system and got the answer as $\frac{3}{8}$ using symmetry. Can we do this in Polar coordinates?

First, use symmetry to only do the integral in the first quadrant (this is allowed because both the integrand and the region of integration share the same symmetry):
$$\iint_R |xy|\: dA = 4 \iint_{R_1} |xy|\:dA$$
Next, we'll have to figure out the bounds. The upper bounds aren't difficult, but the lower ones are given by
$$r^2(\cos^2\theta+4\sin^2\theta) = 1 \implies \begin{cases} r = \frac{1}{\sqrt{1+3\sin^2\theta}} \\ \theta = \sin^{-1}\left(\frac{\sqrt{1-r^2}}{r\sqrt{3}}\right) \\ \end{cases} $$
which gives us two choices of integrals, either $dr$ first:
$$4\int_0^{\frac{\pi}{2}} \int_{\frac{1}{\sqrt{1+3\sin^2\theta}}}^1 r^3\sin\theta\cos\theta \: dr \: d\theta = \int_0^{\frac{\pi}{2}} \sin\theta \cos\theta \left(1- \frac{1}{(1+3\sin^2\theta)^2}\right)\: d\theta $$
or doing $d\theta$ first:
$$4\int_{\frac{1}{2}}^1 \int_{\sin^{-1}\left(\frac{\sqrt{1-r^2}}{r\sqrt{3}}\right)}^{\frac{\pi}{2}} r^3 \sin\theta \cos\theta \: d\theta \: dr = \int_{\frac{1}{2}}^1 2r^3\left(1 - \frac{1-r^2}{3r^2} \right)\: dr$$
This is one of the few times where doing the angular integral first is easier than the other way around. Finishing, we get
$$ \int_{\frac{1}{2}}^1 \frac{8}{3}r^3 - \frac{2}{3}r \: dr = \frac{2}{3}r^4 - \frac{1}{3}r^2\Bigr|_{\frac{1}{2}}^1 = \frac{3}{8}$$
