 Solution

Under a proper rotation matrix $R$, the product of coordinates transforms as a rank-$n$ tensor. The unit ball is unchanged by the substitution $\mathbf{y}=R\mathbf{x}$, whose Jacobian determinant is one. HenceThus $S$ is an isotropic tensor. An isotropic second-rank tensor has the form $S_{ij}=A{\delta }_{ij}$. Taking the trace and using spherical coordinates,Therefore the isotropic second moment of a ball isAn isotropic third-rank tensor has the form $B{ϵ}_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf{x}\mapsto -\mathbf{x}$ makes its integrand odd on the ball. Thus