[Math] Calculating moment of inertia in 2d planar polygon

calculusvector analysis

I've derived equations for a 2D polygon's moment of inertia using Green's Theorem (constant density $\rho$)

$$I_y = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 + x_i x_{i+1} + x_{i+1}^2 ) ( x_i y_{i+1} – x_{i+1} y_i )$$

$$I_x = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( y_i^2 + y_i y_{i+1} + y_{i+1}^2 ) ( x_{i+1} y_i – x_i y_{i+1} )$$

And I'm trying to add them up for calculating $I_0 = I_x + I_y$.

$$I_0 = \frac{\rho}{12}\sum_{i=0}^{i=N-1} ( x_i^2 – y_i^2 + x_i x_{i+1} – y_i y_{i+1} + x_{i+1}^2 – y_{i+1}^2 ) ( x_i y_{i+1} – x_{i+1} y_i )$$

But I found a different(?) equation for $I_0$ on the internet. and many people say the equation given below is correct.

$$I_0 = \frac{\rho}{6} \frac{ \sum_{i=0}^{i=N-1} ( x_i^2 + y_i^2 + x_i x_{i+1} + y_i y_{i+1} + x_{i+1}^2 + y_{i+1}^2 ) ( x_i y_{i+1} – x_{i+1} y_i ) }{ \sum_{i=0}^{i=N-1} ( x_i y_{i+1} – x_{i+1} y_i ) }$$

So I'm confused now. I think my equations for $I_x$ and $I_y$ are correct.
But how am I gonna calculate $I_0$ (moment of inertia with respect to origin axis)?
I couldn't prove both equations are equal.

Could you help me out please ?

(This post has been cross-posted at MathOverflow)

Best Answer

Your moments don't pass two straightforward tests: They should be invariant under reversal of the vertex order (instead they change sign); and they should be quadratic under scaling (instead they scale with the fourth power). The expression you quote from the net passes both tests, so there's a good chance it's correct.

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