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Moment of Inertia of a Closed Bezier Curve

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Fernando Nadal Martínez

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May 17, 2007, 4:42:56 AM5/17/07
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I have another question: can I use the same approach explained in my
recent post "Area of a Closed Bezier Point" to calculate the moment of
inertia of the closed bezier curve ?

In the same way, let:

F(x,y) = [-x²y,xy²]

Then:

div(F(x,y)) = x²+y²

The integral (moment of inertia) results:

I = int( dot( F(g(t)) , perp(g'(t) ) ) dt =
= int( -gx(t)²gy(t)d(gy(t))/dt + gx(t)gy(t)²d(gx(t))dt ) dt

Where:
g(t) = [gx(t),gy(t)]

Did I do any mistake ?

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