= Solution
For positions measured from the chosen origin, the <inertia tensor> is
$$
\boxed{I_{ij}=\int\rho(\mathbf r)(r^2\delta_{ij}-x_ix_j)\,d^3r.}
$$
It relates <angular momentum> and <rotational kinetic energy> to <angular velocity>:
$$
\mathbf L=\mathbf I\boldsymbol\omega,\qquad T_{\rm rot}=\tfrac12\boldsymbol\omega\cdot\mathbf I\boldsymbol\omega.
$$
The real symmetric tensor has orthogonal principal axes, in which $\mathbf I=\operatorname{diag}(I_1,I_2,I_3)$. Its <eigenvalues> are nonnegative moments of inertia. Rotation about a <principal axis> has $\mathbf L\parallel\boldsymbol\omega$; a general rotation does not. The moment about a unit axis $\mathbf n$ is $I_{\mathbf n}=\mathbf n\cdot\mathbf I\mathbf n$.
Changing from the centre of mass to a parallel origin displaced by $\mathbf a$ gives the <tensor parallel-axis theorem>,
$$
\boxed{\mathbf I_O=\mathbf I_{\rm CM}+M(a^2\mathbf 1-\mathbf a\mathbf a^{\mathsf T}).}
$$
For example, a uniform solid sphere has $\mathbf I=(2MR^2/5)\mathbf1$, while a thin rod along $x$ has principal moments $0,Ml^2/12,Ml^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free <precession>; a single scalar inertia only suffices after an axis is specified.<Classical Dynamics by David Tong>[Sections 3.1–3.3: inertia tensor, principal axes and the parallel-axis theorem]{ref}
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