Double integral using polar coordinates (rectangular region) I've been learning about polar coordinates recently and the following problem has me stumped:
$\int_{x=0}^{x=1}\int_{y=0}^{y=1}\frac{1}{(1+x^2+y^2)^2}dydx$
I'm required to solve the above problem using polar coordinates. I do know that we could express the integrand in terms of $r$ and $\theta$ as follows:
$\frac{1}{(1+x^2+y^2)^2}dydx=\frac{1}{(1+r^2)^2}rdrd\theta$
But I'm lost as to how to proceed further. Specifically, how to convert the limits of integration in terms of $r$ and $\theta$. I'm used to using polar coordinates with circular regions only, not a rectangular (or in this case square) region.  Thanks for any help.
 A: You are integrating over the square $Q=[0,1]\times[0,1]$. This square lives in the first quadrant. Therefore, fix $\theta \in \left[0,\dfrac{\pi}{2}\right]$ (a picture may help from now on!). 
If $\theta \in \left[0,\dfrac{\pi}{4}\right]$, then $r$ takes values between $0$ and $\dfrac{1}{\cos \theta}$ while, if $\theta \in \left( \dfrac{\pi}{4},\dfrac{\pi}{2}\right]$, $r$ takes values between $0$ and $\dfrac{1}{\sin \theta}$.
Therefore you get
$$\int_0^1 \int_0^1 \frac{1}{(1+x^2+y^2)^2}dxdy =\int_0^\frac{\pi}{4} d\theta \int_0^{\frac{1}{\cos \theta}} dr \frac{r}{(1+r^2)^2}+$$ $$ + \int_\frac{\pi}{4}^\frac{\pi}{2} d\theta \int_0^{\frac{1}{\sin \theta}} dr \frac{r}{(1+r^2)^2} .$$
A: What are the limits of integration?
Your region is a square.  It is entirely in the first quadrant.
It is symmetric across the 45 degree line
Breaking this into two integrals, $\theta \in[0,\frac {\pi}{4}), \theta\in[\frac{\pi}{4},\frac{\pi}{2}]$ would be a good start.
You really only need to look at one of those regions and then you can double it.  But, I might be getting ahead of myself.
Give equations of the curves that bound the region.
$x = 0, y = 0, x = 1, y = 1$
translate into polar.
$r\cos\theta = 0, r\sin\theta = 0, r\cos\theta = 1, r\sin\theta = 1\\
\theta = \frac {\pi}{2}, \theta = 0, r = \sec\theta, r = \csc \theta $
