 Solution

The inertia tensor about the centroid has entries $J_{ij}={\int }_B\rho (r^2{\delta }_{ij}-x_i{x}_j)dV$. Reflection symmetry makes every off-diagonal entry zero. Under $x=a\xi$, $y=a\eta$, $z=c\zeta$, the ellipsoid becomes the unit ball. Its three coordinate second moments are equal, andUsing the stated mass gives ${\int }_B\rho x^{2}dV=M{a}^2/5$, the same for $y$, and ${\int }_B\rho z^{2}dV=M{c}^2/5$. ThereforeFor the parallel coordinate axes based at $P=(0,0,c)$, the parallel axis theorem in tensor form is $J^{(P)}=J^{(G)}+M(|d{|}^{2}I-d{d}^T)$, where $d=P-G$. Thus