Inertia tensor of a displaced uniform ball ($I(P)=M[(2R^2/5+|X{|}^2)I-X{X}^T],X=P-G$)

For a uniform ball of radius $R>0$ and positive mass $M$, let $G$ be its centre of mass. Spherical symmetry gives the second-moment matrix $\int u{u}^{T}dm=(M{R}^2/5)I$, so the central inertia tensor is $(2M{R}^2/5)I$. The tensor parallel-axis theorem gives the displayed shifted tensor. For $X\ne 0$ its eigenvector along $X$ has eigenvalue $2M{R}^2/5$, while the two-dimensional orthogonal complement has repeated eigenvalue $M(2R^2/5+|X{|}^2)$. Thus it is an isotropic tensor exactly at $P=G$. Displacement along a chosen axis does not change that axis's moment; the transverse moments increase by $M|X{|}^2$.