Axisymmetry of the inertia tensor ($\mathbf{I}=\mathrm{diag}(I_{\perp },I_{\perp },I_{\parallel })$)

If a rigid body's mass density is invariant under every rotation about an axis through its centre of mass, its inertia tensor in any orthonormal basis with third vector on that axis has the displayed form. In particular all transverse directions are principal axes with equal principal moments of inertia.

For a proof, let that axis be $z$. In the defining integrals for the inertia tensor, a half-turn $(x,y,z)\mapsto (-x,-y,z)$ makes the $xz$ and $yz$ products change sign, so their integrals vanish. A quarter-turn $(x,y,z)\mapsto (-y,x,z)$ makes the $xy$ integral vanish and equates the integrals of $x^2$ and $y^2$. Therefore all off-diagonal components vanish and $I_{11}=I_{22}$. The values are $I_{\perp }=\int \rho [(x^2+y^2)/2+z^2]d^{3}r$ and $I_{\parallel }=\int \rho (x^2+y^2)d^{3}r$. No reflection symmetry in $z$ is needed.