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I'm trying to compute the moment of inertia of a 2D ellipse about the z axis, centered on the origin, with major/minor axes aligned to the x and y axes. My best guess was to try to compute it as:

$$4\rho \int_0^a \int_0^{\sqrt{b^2(1 - x^2/a^2)}}(x^2 +y^2)\,dydx$$

... I couldn't figure out how to integrate that. Is there a better way or a trick, or is the formula known? I'd also be happy with a good numerical approximation given a and b.

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Did you try polar coordinates? – leonbloy Jun 1 '12 at 0:27
Equivalently, if you work out the moment of inertia tensor of a circle, you should be able to get the one for an ellipse by a linear transformation of coordinates. – Alex R. Jun 1 '12 at 1:27
@Sam: It is not at all clear to me how you can use the formula for a circle to generate the formula for an ellipse. Can you elaborate? – copper.hat Jun 1 '12 at 5:48
up vote 5 down vote accepted

Use 'polar' coordinates, as in $\phi(\lambda, \theta) = (\lambda a \cos \theta, \lambda b \sin \theta)$, with $(\lambda, \theta) \in S = (0,1] \times [0,2 \pi]$. It is straightforward to compute the Jacobian determinant as

$$ J_{\phi}(\lambda, \theta) = |\det D\phi(\lambda, \theta)| = \lambda a b.$$

Let $E = \{ (x,y) \,|\, 0 <(\frac{x}{a})^2 + (\frac{y}{b})^2 \leq 1 \}$. (Eliminating $(0,0)$ makes no difference to the integral, and is a technicality for the change of variables below.) We have $E = \phi (S)$, and

$$\begin{align} I &= \rho \int_{\phi ( S)} (x^2+y^2) \, dx dy \\ &= \rho \int_{S} \lambda^2 (a^2 \cos^2 \theta+ b^2 \sin^2 \theta) \lambda a b \, d \lambda d \theta \\ &= \rho a b \int_{0}^1 \lambda^3 \, d \lambda \int_0^{2 \pi} a^2 \cos^2 \theta+ b^2 \sin^2 \theta \, d\theta \\ &= \rho \pi a b \frac{a^2+b^2}{4}. \end{align}$$

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I took the liberty of formatting your chain of equations for you. You can check the source to see how I did it. – Rahul Jun 1 '12 at 4:15
That is really clever! – MJD Jun 1 '12 at 4:18
Thanks @RahulNarain! – copper.hat Jun 1 '12 at 4:27
Awesome! Thank you. It's been a while since I've had to integrate anything nontrivial. – Ben Jones Jun 1 '12 at 15:26

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