Conversion of an integral in cartesian to polar coordinates

This question is more specifically about converting the limits of the integral.

I have an integral $\int_0^1 dx \int_0^{(1-x^2)^{1/2}}(1-x^2-y^2)^{1/2} dy$, and I need to convert this into polar coordinates to solve it.

So far I've substituted $x=rcos\theta$ and $y=rsin\theta$ into $(1-x^2-y^2)^{1/2}$ and got $(1-r^2)^{1/2}$ as the thing I'm integrating.

I have also converted $dydx$ to $rdrd\theta$.

This leaves me with $\int_0^1 d\theta \int_0^{(1-x^2)^{1/2}}r(1-r^2)^{1/2} dr$ so far.

I don't know how to convert the limits though. Substituting $x=rcos\theta$ into $(1-x^2)^{1/2}$ just gives $(1-r^2cos^2\theta)^{1/2}$, which isn't what I need.

I know from the answers that I need limits of 0 to $\pi/2$ for the $d\theta$ integral, and 0 to 1 for the $dr$ integral.

So, what am I missing? I assume it's relatively simple, but I can't see it.

I don't need help solving the integral, it's not hard once I'm past this bit. Just changing the limits, how and why you do what you do.

In the starting iterated integral, $x$ goes $0$ to $1$ while $y$ goes $0$ to $\sqrt{1-x^2}.$ This describes the quarter of the unit circle in quadrant I, which in polar coordinates means $\theta \in [0,\pi/2],\ r \in [0,1].$
• So, would I substitute in x at 0 and 1 to the y limits to find that y varies between 0 and 1 as well, and then pick r and $\theta$ values that would correspond to these, or is this just a coincidence? As it is, it feels like I am just noticing the coincidence that the values correspond with a circle, and if I encounter problems like this in the future (likely) I will have to rely on recognising the limits needed instead of following a set of instructions/using a rule to determine the correct limits, which would be my preference. – user218124 Feb 21 '15 at 13:36
• @Spectre1235 I think here it's more a matter of just "recognizing" that the upper $y$ limit $\sqrt{1-x^2}$ is part of the graph of the unit circle, and then since that circle is simple in polar coordinates the transform is easy. I've come across more complicated coordinate changes where the new region was not really that simple in the new variables, but still made the resulting integral easier than the original one. – coffeemath Feb 21 '15 at 17:08