How to convert this cartesian double integral to polar

So most of the converting cartesian to polar integrals I've seen (both on this website and in class) involve some sort of circular integral region.

I was doing practice problems for my exam and I ran across one that does not and I'm stuck on how to solve it.

The integral is the following and the instructions say to solve it by converting to polar coordinates (which is annoying because this integral would be so easy to do normally):

$$\int _0 ^1 \int_x^1 x^2 dydx$$

How should I approach this? I tried setting 1 = y and x = yand got 1 $=rcos(\theta)$ and $1 = tan\theta$ but these bounds don't really seem to help me integrate.

• If I see $x^2$ in an integral I want to convert to polar, I look for a way to make it $x^2+y^2=r^2$ Since $x$ is a dummy variable, you could add this integral to one with $x$ and $y$ interchanged. Now the problem is that the limits of the inner integral don't match. Can you fix that? – Ross Millikan Oct 25 '16 at 3:48

Don't try to do this sort of thing by "pure algebra" - always draw the region of integration. If you do this you will see easily that $\theta$ varies from $\pi/4$ to $\pi/2$. So we have $$I=\int_{\pi/4}^{\pi/2}\int_?^? x^2\,J\,dr\,d\theta$$ where $J$ is the Jacobian. To find the limits for $r$, draw a line on your diagram starting at the origin and heading in the $\theta$ direction (where $\theta$ is between $\pi/4$ and $\pi/2$). You can see that the values of $r$ which are in your region and on this line go from a minimum of $0$ to a maximum on the horizontal line $y=1$. To find the $r$ value on this line we have $$r\sin\theta=y=1$$ and so $r_{\rm max}=1/\sin\theta$. Hence $$I=\int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} x^2\,J\,dr\,d\theta\ .$$ You should also know that the Jacobian for polar coordinates is $r$ and that $x=r\cos\theta$. Hence $$I=\int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} (r\cos\theta)^2\,r\,dr\,d\theta\ .$$