Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 12C / c / i / Solution

Place the origin at the centre and the axes along the edges. Symmetry makes the off-diagonal entries of the inertia tensor zero. Uniform density gives $⟨x^2⟩=a^2/12$, $⟨y^2⟩=b^2/12$, $⟨z^2⟩=c^2/12$. Consequently the principal moments of inertia areFor example $I_1=\rho \int (y^2+z^2)dxdydz$ gives the first expression directly.