= Transverse moment of inertia of a uniform solid cylinder
{title2=$I_{\perp,G}=m(l^2+3r^2)/12$}
For a uniform solid cylinder of length $l$, radius $r$ and <mass> $m$, a thin circular slice at axial coordinate $z$ has transverse <moment of inertia> $(z^2+r^2/4)dm$ about a diametral axis through the cylinder's midpoint. Integrating from $-l/2$ to $l/2$ gives $I_{\perp,G}=ml^2/12+mr^2/4$. The <parallel axis theorem> gives $I_{\perp,\rm end}=ml^2/3+mr^2/4$ about a parallel axis through an end-face centre. The transverse and symmetry-axis <moments of inertia> coincide exactly when $l=\sqrt3r$.
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