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?
 A: 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
$$ \int_{\frac{1}{2}}^1 \frac{8}{3}r^3 - \frac{2}{3}r \: dr = \frac{2}{3}r^4 - \frac{1}{3}r^2\Bigr|_{\frac{1}{2}}^1 = \frac{3}{8}$$
A: Let
$$ R_1=\{(x,y): x^2+y^2\le1\},R_2=\{(x,y): x^2+4y^2\le1\}. $$
Then $R=R_1\setminus R_2$. So
$$ \iint_R|xy|dxdy=\iint_{R_1}|xy|dxdy-\iint_{R_2}|xy|dxdy. $$
But for $R_1$, we can use $x=r\cos t,y=r\sin t$, $0\le r\le 1$ and $0\le t \le 2\pi$, and so
$$ \iint_{R_1}|xy|dxdy=\int_0^1\int_0^{2\pi}r^3|\cos t\sin t|drdt=\frac12. $$
Similarly, for $R_2$, we can use $x=r\cos t,y=\frac12r\sin t$, $0\le r\le 1$ and $0\le t \le 2\pi$, and so
$$ \iint_{R_1}|xy|dxdy=\int_0^1\int_0^{2\pi}\frac14r^3|\cos t\sin t|dt=\frac18. $$
Thus
$$ \iint_R|xy|dxdy=\frac12-\frac18=\frac38. $$
