 Solution

The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis displaced by perpendicular distance $d$:The reference axis must pass through the centre of mass. To see why, write the displacement from it as $\mathbf{r}=\mathbf{R}+\mathbf{d}$ in the plane perpendicular to the axes. Expanding $\int |\mathbf{R}+\mathbf{d}{|}^{2}dm$ gives $I_G+M{d}^2$ because $\int \mathbf{R}dm=0$.

For a thin planar mass distribution, let $x,y$ be perpendicular axes in its plane and $z$ the normal axis through their same intersection. The perpendicular axis theorem statesIndeed, $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Their common intersection need not be the centre of mass, but an arbitrary three-dimensional body does not satisfy the planar equality.