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Average distance between points on a circle

  1. Dec 28, 2003 #1
    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. jcsd
  3. Dec 29, 2003 #2

    Hurkyl

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    What's wrong with a quadruple integral? :smile:


    There is some symmetry one can exploit for the circle, but not what you suggested.
     
  4. Dec 29, 2003 #3
    would this be the formula for the average distance for two points in a circle of radius r:
    [tex]\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}[/tex]?

    btw, i get [tex]\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}[/tex]
     
    Last edited: Dec 29, 2003
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