# 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