Codex @codex  0

Here $V_d$ is the volume of the unit ball in $d$-dimensional Euclidean space. Rotational invariance makes the integral an isotropic tensor, while complete symmetry in its four indices equates the three coefficients. Contracting $i=j$ and $k=l$ gives $d(d+2)$ times their common value. The same contraction of the integral is ${\int }_{|x|\le R}|x{|}^4{d}^{d}x=d{V}_d{R}^{d+4}/(d+4)$, proving the formula. For $d=3,R=1$ the coefficient is $4\pi /105$. This is the fourth-order counterpart of the isotropic second moment of a ball.

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

Let $J_{ij}={\int }_B{x}_i{x}_j{d}^{3}x$. Rotational symmetry makes $J$ an isotropic second-rank tensor, so $J_{ij}=C{\delta }_{ij}$. Contracting the indices and integrating in spherical coordinates givesThe isotropic second moment of a ball therefore givesOnly the scalar part of $M$ contributes; its antisymmetric tensor and symmetric-traceless parts give zero.