Prove that the moment of inertia of a plane figure

Prove that the moment of inertia of a plane figure with respect to any axis equals $Md^2+I_c$, where $M$ is the total mass, $d$ is the distance from the axis to the centroid, and $I_c$ is the moment of inertia about the parallel axis through the centroid.

Step 1 · Setup

Place the centroid $C$ of the figure at the origin, with the central axis $\ell_c$ along the $y$-axis, and the given axis $\ell$ the line $x=d$.

Step 2 · Expand $I_\ell$

The distance from $(x,y)$ to $\ell$ is $|x-d|$, so:

$$ I_\ell = \iint_D(x-d)^2\varrho\,dA = \iint_Dx^2\varrho\,dA - 2d\iint_Dx\varrho\,dA + d^2\iint_D\varrho\,dA. $$

Step 3 · Identify each term

The first term is $I_c=\iint_Dx^2\varrho\,dA$, the moment about the central axis. The second term is $2dM\bar x$; since the centroid was placed at the origin, $\bar x=0$, so this term vanishes. The third term is $Md^2$. Hence:

$$ I_\ell = I_c + Md^2. $$

Step 4 · Consequence

Since $Md^2\ge0$ with equality only when $d=0$, the central axis (through the centroid) gives the smallest moment of inertia among all axes parallel to a given direction; any other parallel axis exceeds it by $Md^2$.

Additional Notes

• This derivation matches the source exactly. This theorem (Steiner's theorem) has already been used repeatedly earlier in this collection as a cross-check (e.g. for the moment of inertia of a disk about a tangent line or about a boundary point).