Q.State and prove the perpendicular axis theorem.

Step 1. Statement of the theorem.

The moment of inertia of a plane laminar body (a flat body of negligible thickness) about an axis perpendicular to its own plane equals the sum of its moments of inertia about any two mutually perpendicular axes lying IN the plane of the body, provided all three axes pass through one common point.

Step 2. Setting up the proof.

Let the $X$ and $Y$ axes lie in the plane of the lamina, and the $Z$ axis be perpendicular to that plane, with all three axes intersecting at a common origin $O$. Consider a representative particle of the lamina, of mass $m$, located at coordinates $(x,y)$ within the plane.

Step 3. Distance of the particle from each axis.

Since the lamina lies entirely in the $XY$-plane, every particle has $z=0$. The particle's distance from the $Z$-axis (which passes through $O$ perpendicular to the plane) is $r=\sqrt{x^2+y^2}$. Its perpendicular distance from the $X$-axis would, for a general 3-dimensional body, be $\sqrt{y^2+z^2}$ — but because $z=0$ here, this reduces to simply $y$. Likewise, its distance from the $Y$-axis reduces to simply $x$. This reduction is exactly why the theorem is restricted to flat, plane laminar bodies: it depends on the body having no extent perpendicular to its own plane.

Step 4. Contribution of the particle to $I_Z$.

Step 5. Summing over the whole lamina. …

Direct summation using $r^2=x^2+y^2$ for a particle confined to the plane ($z=0$), valid only …

• Applying the theorem to a genuinely 3-dimensional body (a solid sphere, cylinder, or cube) — it holds ONLY for a flat lamina of negligible thickness; for a real 3D body the distances from the $X$ and $Y$ axes involve the extra $z$-coordinate ($\sqrt{y^2+z^2}$ and $\sqrt{x^2+z^2}$) and do not reduce to plain $y$ and $x$. …