Question

how to find the average squared distance between the points of the unit disk and the point (1,1)

Answer 1

The unit disk can be parameterized by the function

$\mathbf p(r,\theta)=(r\cos\theta,r\sin\theta)$

where $0\le r\le 1$ and $0\le\theta\le2\pi$. The squared distance between any point in this region $(x,y)=(r\cos\theta,r\sin\theta)$ and the point (1, 1) is

$(x-1)^2+(y-1)^2=(r\cos\theta-1)^2+(r\sin\theta-1)^2$
$=(r^2\cos^2\theta-2r\cos\theta+1)+(r^2\sin^2\theta-2r\sin\theta+1)$
$=r^2(\cos^2\theta+\sin^2\theta)-2r(\cos\theta-\sin\theta)+2$
$=r^2-2r(\cos\theta-\sin\theta)+2$

The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.

$\dfrac{\displaystyle\iint_{x^2+y^2<1}((x-1)^2-(y-1)^2)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_{x^2+y^2<1}\mathrm dx\,\mathrm dy}$
$=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}$
$=\dfrac{\frac{5\pi}2}\pi=\dfrac52$