Distance between two uniform points in a three-dimensional ball ($f_R(r)=3r^2/R^3-9r^3/(4R^4)+3r^5/(16R^6)$)

For two independent random variables taking values in a three-dimensional ball with constant volume probability density, the density of their distance is the displayed expression for $0\le r\le 2R$, and zero otherwise. Both independence and the common radius $R>0$ are essential. To derive it, put $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$ and integrate the product volume differential form over orientations. The law of cosines gives the angular Jacobian determinant $v/(ur)$, leaving $f_R(r)=9r\int uvdudv/(2R^6)$ over $0\le u,v\le R$, $|u-v|\le r\le u+v$. With $x=u+v$, $y=u-v$, the integrand becomes $(x^2-y^2)dxdy/8$. Direct integration gives $\int uvdudv=2R^{3}r/3-R^2{r}^2/2+r^4/24$. An independent geometric derivation uses the overlap volume of two radius-$R$ balls displaced by $r$, namely $\pi (4R+r)(2R-r{)}^2/12$, multiplied by $4\pi r^2$ and divided by the square of $4\pi R^3/3$. The density integrates to one and has mean $36R/35$.