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One way to calculate it would be doing:

$\dfrac{\int_0{}^{r}\int_0{}^{r}\sqrt{(x_1-x_2)^2+(\sqrt{r^2-x_1^2}+\sqrt{r^2-x_2^2})^2} \ \ \ dx_1 \ dx_2}{r^2}$

This seems impossible to integrate it, but knowing that $x_1=rcos\alpha \ and \ x_2=rcos\beta$ we get

$r\sqrt{2}\int_0{}^{\pi/2}sin\beta\int_0{}^{\pi/2}sin\alpha\sqrt{1-cos(\alpha-\beta)} \ \ \ d\alpha \ d\beta $

However this integral gives as a result $0$, but it should give around $r·0.41975$, this last result has been calculated just iterating with a program.

can anyone figure out why its not giving the right result? $\ \ $ Thank you.

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  • $\begingroup$ How do you define the average length of a chord in the first quadrant of a circle? $\endgroup$ – Christian Blatter Jun 28 '17 at 16:23
  • $\begingroup$ As the average distance between two points with positive x and y coordinates of a circumference $\endgroup$ – maxbp Jun 29 '17 at 10:46
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If you are talking about average you have to specify the measure used to compute that average. For the average of $n$ numbers we usually use ${1\over n}\cdot$ counting measure, but here we have a geometric situation. I shall use the following setup:

Consider two points chosen independently and uniformly with respect to arc length on the first quarter of the unit circle. What is their expected distance $E(D)$?

If the two points have polar angles $\phi$ and $\psi$ then $$D=2\sin{|\psi-\phi|\over2}\ .$$ It follows that $$E(D)={4\over\pi^2}\int_0^{\pi/2}\int_0^{\pi/2}2\sin{|\psi-\phi|\over2}\>{\rm d}(\psi,\phi)\ .$$ Elementary computation gives $$E(D)={32\over\pi^2}\left({\pi\over2}-\sqrt{2}\right)\doteq0.507685\ .$$

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