Isotropic second moment of a ball (${\int }_{|x|\le R}{x}_i{x}_j{d}^{d}x=\frac{V_d{R}^{d+2}}{d+2}{\delta }_{ij}$)

Here $V_d$ is the volume of the unit ball in $d$-dimensional Euclidean space. Rotational symmetry makes the integral an isotropic second-rank tensor $C{\delta }_{ij}$. Taking its trace gives $dC={\int }_{|x|\le R}|x{|}^2{d}^{d}x=d{V}_d{\int }_0^R{r}^{d+1}dr$. This proves the formula. For a constant matrix $M$, contraction gives $\int x^{T}Mx{d}^{d}x=V_d{R}^{d+2}\mathrm{tr}M/(d+2)$. In three dimensions the coefficient is $4\pi R^5/15$.