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Pie
3 years ago
10

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

Mathematics
1 answer:
ddd [48]3 years ago
5 0
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
=\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
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