Prove that the sum of the moments of inertia of a plane
Prove that the sum of the moments of inertia of a plane figure with respect to any pair of mutually perpendicular axes lying in the plane of the figure and passing through a fixed point $O$ is a constant (independent of the orientation of the pair).
Explanation of Method
Writing the distances from a point $(x,y)$ to two perpendicular axes through $O$ at angle $\varphi$ as linear combinations of $x,y$, and summing the two squared distances, the cross terms cancel identically, leaving $x^2+y^2$ — independent of $\varphi$.
- Set up the two perpendicular axes at angle $\varphi$.
- Write the distance from a point to each axis.
- Expand and sum the two moments of inertia.
- Conclude that the sum equals the polar moment $I_O$, independent of $\varphi$.
Step 1 · Setup
Let the figure occupy region $D$ with density $\varrho(x,y)$, and let $\ell_1,\ell_2$ be perpendicular axes through the origin $O$, with $\ell_1$ at angle $\varphi$ to the $x$-axis and $\ell_2$ at angle $\varphi+\pi/2$.
Step 2 · Distances to the axes
$$ d_1 = |{-x\sin\varphi+y\cos\varphi}|, \qquad d_2 = |x\cos\varphi+y\sin\varphi|. $$
Step 3 · Sum of moments
Expanding and adding the two squares, the cross terms $-2xy\sin\varphi\cos\varphi$ and $+2xy\cos\varphi\sin\varphi$ cancel:
$$ (-x\sin\varphi+y\cos\varphi)^2+(x\cos\varphi+y\sin\varphi)^2 = x^2(\sin^2\varphi+\cos^2\varphi)+y^2(\cos^2\varphi+\sin^2\varphi) = x^2+y^2. $$
Step 4 · Conclusion
Therefore:
$$ I_{\ell_1}+I_{\ell_2} = \iint_D(x^2+y^2)\varrho\,dA = I_O, $$
which depends only on the figure, its density, and the point $O$ — not on $\varphi$. In the standard orientation ($\varphi=0$) this reads $I_x+I_y=I_O$, the familiar form of the perpendicular axis theorem.
Additional Notes
- This derivation matches the source exactly.
Conclusion
For any pair of mutually perpendicular axes through a fixed point $O$ in the plane of a figure: