Unit Disk Average Distance

It's convenient to do distances from the origin instead, so reflect everything across the line $x=1/2$ to get the following problem: what is the average distance from a point on the disk inside $(x-1)^2+y^2 = 1$ to the origin?

Restrict to the upper half disk, which is ok by symmetry. In polar coordinates, the equation of the boundary circle is $\begin{aligned} x^2-2x+1+y^2 &= 1 \\ x^2+y^2 &=2x \\ r^2 &= 2r\cos \theta \\ r &= 2\cos \theta \end{aligned}$

The distance to the origin is just $r,$ so the appropriate integral is $\begin{aligned} \frac2{\pi} \int_0^{\pi/2} \int_0^{2\cos \theta} r \cdot r \, dr \, d\theta &= \frac2{\pi} \int_0^{\pi/2} \frac{r^3}3 \Big|_0^{2\cos\theta} \, d\theta \\ &= \frac{16}{3\pi} \int_0^{\pi/2} \cos^3\theta \, d\theta \\ &= \frac{16}{3\pi} \left( \sin \theta - \frac13 \sin^3 \theta \right)\Big|_0^{\pi/2} \\ &= \frac{32}{9\pi} \approx \fbox{1.131}. \end{aligned}$