Average distance between points on a circle

1. Dec 28, 2003

phoenixthoth

of radius r or a square of side length a? do you need some kind of quadruple or double integral or is there a trick?

sorry, i meant two points either in the interior of said shape or on the boundary.

for the circle, by symmetry, is that the same as the average distance between a point and the origin? that is, i think, 2r/3. note that the max distance is 2r and the min distance is 0.

Last edited: Dec 29, 2003
2. Dec 29, 2003

Hurkyl

Staff Emeritus
What's wrong with a quadruple integral?

There is some symmetry one can exploit for the circle, but not what you suggested.

3. Dec 29, 2003

phoenixthoth

would this be the formula for the average distance for two points in a circle of radius r:
$$\frac{\int_{-r}^{r}\int_{-\sqrt{r^{2}-y2^{2}}}^{\sqrt{r^{2}-y2^{2}}}\int_{-r}^{r}\int_{-\sqrt{r^{2}-y1^{2}}}^{\sqrt{r^{2}-y1^{2}}}\sqrt{\left( x2-x1\right) ^{2}+\left( y2-y1\right) ^{2}}dx1dy1dx2dy2}{\int_{-r}^{r}\int_{-\sqrt{r^{2}-y2^{2}}}^{\sqrt{r^{2}-y2^{2}}}\int_{-r}^{r}\int_{-\sqrt{r^{2}-y1^{2}}}^{\sqrt{r^{2}-y1^{2}}}1dx1dy1dx2dy2}$$?

btw, i get $$\frac{\int_{-r}^{r}\int_{-\sqrt{r^{2}-y1^{2}}}^{\sqrt{r^{2}-y1^{2}}}\sqrt{x1^{2}+y1^{2}}dx1dy1}{\int_{-r}^{r}\int_{-\sqrt{r^{2}-y1^{2}}}^{\sqrt{r^{2}-y1^{2}}}1dx1dy1}=\frac{2r}{3}$$

Last edited: Dec 29, 2003