= 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 $\rho=M/(\pi a^2h)$. Symmetry makes the <inertia tensor> diagonal and gives $I_1=I_2$. Direct integration yields
$$
I_3=\rho\int_{-h/2}^{h/2}\!dz\int_0^a\!r^3dr\int_0^{2\pi}\!d\theta=\frac{Ma^2}{2}.
$$
For a transverse <principal moment of inertia>, the two contributions are
$$
\int y^2\,dm=\rho h\frac{a^4}{4}\int_0^{2\pi}\sin^2\theta\,d\theta=\frac{Ma^2}{4},\qquad
\int z^2\,dm=\rho\pi a^2\frac{h^3}{12}=\frac{Mh^2}{12}.
$$
Therefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),\quad I_3=Ma^2/2}$ are the three <principal moments of inertia>.
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