Suppose we have the torus with equation $1-\left(\sqrt{x^2+y^2}-3\right)^2=z^2$. We choose two randomly chosen points inside it with a uniform distribution. I want to find the average distance between two randomly chosen points inside it.
However, I did an attempt to calculate it. I chose an horizontal angle $\theta$, two angles that define a rotation with respect to the cross section of the torus, $\phi_1$ and $\phi_2$. Both points can be placed on two circunferences parallel to the xy plane, with equations:
$$C_1: \left(3+r_1cos(\phi_1)\right)=x^2+y^2,z=r_1sin(\phi_1)$$ $$C_2: \left(3+r_2cos(\phi_2)\right)=x^2+y^2,z=r_2sin(\phi_2)$$ $$ 0\le \phi_1\lt2\pi, 0\le \phi_2\lt2\pi, 0\le \theta\lt2\pi, 0\le r_1\lt 1, 0\le r_1\lt 1$$
And $r_1$ and $r_2$ define a circunference of radius $r_1$ and $r_2$.
Then the weights are $\left(3+r_1cos(\phi_1)\right)$, $\left(3+r_2cos(\phi_2)\right)$, $r_1$ and $r_2$, and:
$$\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^1\int_0^1r_1r_2\left(3+r_1cos(\phi_1)\right)\left(3+r_2cos(\phi_2)\right)dr_1dr_2d\phi_1d\phi_2d\theta=18\pi^3$$
The points are:
$$P_1=\left(3+r_1cos(\phi_1),0,r_1sin(\phi_1)\right)$$ $$P_2=\left(\left(3+r_2cos(\phi_2)\right)cos(\theta),\left(3+r_2cos(\phi_2)\right)sin(\theta),r_2sin(\phi_2)\right)$$
Therefore, the average distance is:
$$A_d=\frac{1}{18\pi^3}\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^1\int_0^1\sqrt{\left(3+r_1cos(\phi_1)-\left(3+r_2cos(\phi_2)\right)cos(\theta)\right)^2+\left(\left(3+r_2cos(\phi_2)\right)sin(\theta)\right)^2+\left(r_1sin(\phi_1)-r_2sin(\phi_2)\right)^2}r_1r_2\left(3+r_1cos(\phi_1)\right)\left(3+r_2cos(\phi_2)\right)dr_1dr_2d\phi_1d\phi_2d\theta$$
However, I don't know wether this integral has closed form, can be evalueted in some way or if there is another method to find the average distance, because I have no idea if a probability density function for the distances can be found. Is this well done?
