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.


`(\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)`

`=(\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)`

`=\frac{\frac{5\pi}2}\pi=\frac52`