Find the moment of inertia of a homogeneous ellipse
Find the moment of inertia of a homogeneous ellipse with semi-axes $a,b$ ($\varrho=1$) with respect to its center.
Explanation of Method
The polar moment $I_0=\iint(x^2+y^2)\,dA$ splits into $I_x+I_y$, and the elliptic substitution $x=ar\cos\theta,\,y=br\sin\theta$ (Jacobian $abr$) reduces both pieces to elementary products of a radial and an angular integral.
- Set up $I_0=I_x+I_y$ and the elliptic substitution.
- Compute $I_y$ and $I_x$.
- Combine, and express in terms of the mass.
Step 1 · Setup
With $x=ar\cos\theta,\,y=br\sin\theta$ ($0\le r\le1$), Jacobian $abr$: $I_0=\iint_D(x^2+y^2)\,dA=I_x+I_y$.
Step 2 · Compute $I_y,I_x$
$$ I_y = \iint_Dx^2\,dA = a^3b\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^1r^3\,dr = a^3b\cdot\pi\cdot\frac14 = \frac{\pi a^3b}4, $$
and by the same computation with $\sin^2\theta$:
$$ I_x = \frac{\pi ab^3}4. $$
Step 3 · Combine
$$ I_0 = I_x+I_y = \frac{\pi ab(a^2+b^2)}4. $$
With total mass $m=\pi ab$: $I_0=\dfrac{m(a^2+b^2)}4$.
Additional Notes
- This derivation matches the source exactly. For a circle ($a=b=R$): $I_0=\dfrac{\pi R^4}2$, the standard polar moment of a disk.
- Verified numerically (e.g. $a=1.4,\,b=2.2$): $I_0\approx16.4494$, matching $\dfrac{\pi ab(a^2+b^2)}4$ exactly.
Result
The moment of inertia about the center is: