Radial tensor moments in three dimensions ($\int F(r)x_i{x}_j{x}_k{x}_l{d}^{3}x=\frac{4\pi }{15}{\int }_0^{\infty }{r}^{6}F(r)dr({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})$)

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.