Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am interested in finding the average Euclidean distance from a point $(x,y)\in\mathbb{D}_2$, the unit disk $\{(u,v):u^2+v^2\leq 1\}\subseteq\mathbb{R}^2$, to the disk $\mathbb{D}_2$. This amounts to essentially calculating the integral $$\iint_{\mathbb{D}_2} \|(x,y)-\omega\|_2d\omega.$$ By rotational symmetry this gives an integral of the form $$\iint_{\mathbb{D}_2}\|(a,0)-\omega\|_2d\omega$$ where $0\leq a \leq 1$. I believe this can be converted to an elliptic integral of the second kind where the integral over the angle $\theta$ appears in the form $$\int_{0}^{2\pi}\sqrt{1-\frac{4ar}{(r+a)^2}\cos^2(\theta/2)}d\theta.$$ I am not sure how to proceed from here.

share|improve this question
After making a few more substitutions, you should end up with the complete elliptic integral of the second kind. Start by letting $\theta=2u$, then manipulate things so that the only trigonometric function within the square root is $\sin^2$... –  Guess who it is. Aug 1 '12 at 17:54
We kept working and found the angular integral of the form $\int_{0}^{\pi} \sqrt{1+\frac{4ar}{(r-a)^2}\sin^2\theta}d\theta$. Does the sign difference here from the complete elliptic integral of the second kind matter? –  user36119 Aug 1 '12 at 17:56
You can still stand to cut your integration interval in half, since the integrand is symmetric about $\theta=\pi/2$... –  Guess who it is. Aug 1 '12 at 18:00
Yes, that is true, but we have something of the form $\int_{0}^{\pi/2} \sqrt{1+k^2\sin^2\theta}d\theta$, not the desired $\int_0^{\pi/2}\sqrt{1-k^2\sin^2\theta}d\theta$. –  user36119 Aug 1 '12 at 18:08
You might be interested in this, then. –  Guess who it is. Aug 1 '12 at 18:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.