If $D=\{(x,y):x^2+y^2\geq 1, 0\leq x\leq 1,0\leq y\leq 1\}$, then find $$\iint_{D}\frac{1}{(x^2+y^2)^2}\,dA$$
What i try: I have draw diagram of region $D$
Using polar coordinates $x=r\cos\theta,y=r\sin\theta$
And $x^2+y^2=r^2$ and $dA=rdrd\theta$
So Integral $$I=\iint_{D}\frac{r}{r^4}dr\,d\theta=\iint_{D}\frac{1}{r^3}dr\,d\theta$$
I do not understand how can I set limit for $r$ and $\theta$. Help me please. Thanks
Following up on my comment:
Notice that when $\theta\in[0,\pi/4]$, $r$ varies between the circle ($r=1$) and the line $x=1$, which in polar coordinates is described by $r=1/\cos\theta$.
On the same manner, when $\theta\in[\pi/4,\pi/2],$ $r$ varies between $r=1$ and the line $y=1$, which in polar coordinates is given by $r=1/\sin\theta$. Therefore you are looking for the solution of the following integral:
$$\int_0^{\pi/4}\int_{1}^{1/\cos\theta}r^{-3}drd\theta+\int_{\pi/4}^{\pi/2}\int_{1}^{1/\sin\theta}r^{-3}drd\theta$$