= Axisymmetry of the inertia tensor
{title2=$\mathbf I=\operatorname{diag}(I_\perp,I_\perp,I_\parallel)$}
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 $I_\perp=\int\rho[(x^2+y^2)/2+z^2]\,d^3r$ and $I_\parallel=\int\rho(x^2+y^2)\,d^3r$. No reflection symmetry in $z$ is needed.
Back to article page
