Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 9E / a / Solution

Put the cylinder's centre at the origin, with $-h/2\le z\le h/2$, $0\le r\le a$, and uniform density $\rho =M/(\pi a^{2}h)$. Symmetry makes the inertia tensor diagonal and gives $I_1=I_2$. Direct integration yieldsFor a transverse principal moment of inertia, the two contributions areTherefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),I_3=M{a}^2/2}$ are the three principal moments of inertia.