= Solution
Take the centre of the ball as the origin. An orthogonal change of variables leaves both this domain and its volume measure invariant. Consequently its Cartesian moment integrals are <isotropic tensors>. Equivalently, inversion $x\mapsto-x$ immediately makes every odd moment zero.
There is no nonzero invariant rank-one <vector>, so
$$
\boxed{\int_Vx_i\,dV=0.}
$$
An <isotropic tensor> of rank two is a scalar multiple of the <Kronecker delta>, giving $\int_Vx_ix_j\,dV=\alpha\delta_{ij}$. Contract the indices and use <spherical coordinates>:
$$
3\alpha=\int_Vr^2\,dV=4\pi\int_0^a r^4\,dr=\frac{4\pi a^5}{5}.
$$
Thus the <Cartesian moments of a three-dimensional ball> give
$$
\boxed{\alpha=\frac{4\pi a^5}{15},\qquad\int_Vx_ix_j\,dV=\frac{4\pi a^5}{15}\delta_{ij},\qquad\int_Vx_ix_jx_k\,dV=0.}
$$
For the last identity, one may also use the general rotation-invariant rank-three tensor $c\epsilon_{ijk}$: the moment is fully symmetric in its indices, whereas the <Levi-Civita symbol> is antisymmetric, forcing $c=0$.
The origin convention is necessary. For a ball centred at a nonzero vector $c$, its first moment would instead be $(4\pi a^3/3)c_i$, so the displayed claims implicitly refer to coordinates measured from its centre.
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