Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 6 / d

For $i\ne j$, the inertia tensor component is $I_{ij}=-\rho {\int }_V{x}_i{x}_{j}dV$. Reflect either coordinate appearing once: the cylindrical domain and density are unchanged, while the integrand changes sign. Thus every off-diagonal component vanishes. Rotational symmetry also implies $I_{11}=I_{22}$.

Use cylindrical coordinates $x_1=r\cos \theta$, $x_2=r\sin \theta$, $x_3=z$, with volume element $rdrd\theta dz$. The axial moment of inertia isThe two contributions to a transverse component areThereforeThe axial value $M{R}^2/2$ and transverse value $M(3R^2+h^2)/12$ for $R=1$, $h=4$ give an independent check on both diagonal entries.