I've been learning about polar coordinates recently and the following problem has me stumped:


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:


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.


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

$$\int_0^1 \int_0^1 \frac{1}{(1+x^2+y^2)^2}dxdy =\int_0^\frac{\pi}{4} d\theta \int_0^{\frac{1}{\cos \theta}} dr \frac{r}{(1+r^2)^2}+$$ $$ + \int_\frac{\pi}{4}^\frac{\pi}{2} d\theta \int_0^{\frac{1}{\sin \theta}} dr \frac{r}{(1+r^2)^2} .$$

  • 1
    $\begingroup$ Why 1/cosθ instead of √2? The greatest r distance is √2(1, 1 on line y=x) for the region θ=[0, π/4] $\endgroup$ – Aladdin Oct 11 '19 at 16:53

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 $


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.