= Solution
The <parallel axis theorem> states $I=I_G+Md^2$ for parallel axes separated by $d$, with the reference axis through the <centre of mass>. The <perpendicular axis theorem> states $I_z=I_x+I_y$ for a planar lamina in the $xy$ plane, about mutually perpendicular axes through one point. It does not apply directly to a three-dimensional solid.
For the <transverse moment of inertia of a uniform solid cylinder>, divide the cylinder into thin disks at axial positions $z\in[-h/2,h/2]$. Symmetry and the <perpendicular axis theorem> give the diametral <moment of inertia> $r^2\,dM/4$ of each disk; the <parallel axis theorem> adds $z^2\,dM$. Hence
$$
I_\perp=\frac Mh\int_{-h/2}^{h/2}\left(\frac{r^2}{4}+z^2\right)dz,\qquad
\boxed{I_\perp=\frac M{12}(3r^2+h^2).}
$$
The thin-disk limit is $Mr^2/4$, while the slender-rod limit is $Mh^2/12$.
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