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.$$ I am not sure how to proceed from here.
Let me set $$k^2=\frac{4ar}{(r+a)^2} <1. $$ If we replace $\theta$ with $2\theta$ we reduce this to an integral; $$ 2 \underbrace{\int_0^\pi \sqrt{1k^2\cos^2\theta} d\theta}_{=I}. $$ Now set $$ x=\cos\theta $$ so that $$ dx=\sqrt{1x^2} d\theta $$ and $$ I=\int_{1}^1\frac{\sqrt{1k^2x^2}}{\sqrt{1x^2}} dx= 2\underbrace{\int_0^1 \frac{\sqrt{1k^2x^2}}{\sqrt{1x^2}} dx}_{=:E_2(k)}. $$ The integral $E_2(k)$ is called Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997. 

