Adjusted polar coordinates

Question

Let $\Omega = \{(x,y)\in (0,\infty)^2 | 4 < x^2+4y^2<16\}$, so the area between two ellipses in the first quadrant. I need to calculate the following integral: $$\int_{\Omega}\frac{xy}{x^2+4y^2}d(x,y)$$ I tried using normal polar coordinates, however the integral gets really messy after the transformation. Does anyone have an idea of better transformation, that would enable me to calculate this integral?

Answer 1

Try to use the substitution $x=rcos\theta, y=\frac{r}{2}sin\theta$ and be careful for the $|\frac{\partial(x,y)}{\partial(r,\theta)}|=\frac{r}{2}$.

Answer 2

First let's describe $\Omega$ in polar coordinates:$$\Omega=\{(x,y)\in (0,\infty)^2 | 4 < x^2+4y^2<16\}\\=\{(r,\theta)\in (0,\infty)\times[0,\dfrac{\pi}{2}] | 4 < r^2(1+3\sin^2\theta)<16\}$$therefore the integral can be changed to:$$I=\int_{\Omega}\frac{xy}{x^2+4y^2}d(x,y)=\int_{\Large0}^{\Large\frac{\Large\pi}{2}}\int_{\Large\frac{2}{\sqrt{1+3sin^2\theta}}}^{\Large\frac{4}{\sqrt{1+3sin^2\theta}}}\frac{\sin\theta\cos\theta}{1+3\sin^2\theta}rdrd\theta=\int_{\Large0}^{\Large\frac{\Large\pi}{2}}\frac{3\sin2\theta}{(1+3\sin^2\theta)^2}d\theta=\int_{\Large0}^{\Large3}\dfrac{du}{(1+u)^2}=\frac{3}{4}$$