Hi all,

Can I use the Gauss-Green Theorem to calculate the moment of inertia of
a 2D closed bezier curve ? Let me explain myself:

Let g(t) be a piecewise differentiable curve, with 0 <= t <= 1. g(t) is
oriented clockwise and g(1) = g(0).

Let
F(x,y) = [-x²y , xy²]

Then:
div(F(x, y)) = x²+y²

Now the divergence theorem gives you the area inside the closed curve
g(t) as a line integral along the curve:

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

perp(x, y) = (-y, x)

where int is for integration, d/dt for differentiation and dot for dot
product. The integration has to be pieced to the parts corresponding to
the smooth curve segments.

Did I do any mistake ( I calculated the integral expression with Maple )
? The numerical results are wrong..

I used the same approach to evaluate the area of the curve ( F(x,y) =
[x,-y]/2 and calculated with Maple too )and works great..

=?ISO-8859-1?Q?Fernando_Nadal_Mart=EDnez?= <fnm@uma.es> writes:

Hi all,

Can I use the Gauss-Green Theorem to calculate the moment of inertia of
a 2D closed bezier curve ? Let me explain myself:

Let g(t) be a piecewise differentiable curve, with 0 <= t <= 1. g(t) is
oriented clockwise and g(1) = g(0).

Let
F(x,y) = [-x²y , xy²]

Then:
div(F(x, y)) = x²+y²

No, div F = 0. Perhaps you want [x y^2, x^2 y].

Now the divergence theorem gives you the area inside the closed curve
g(t) as a line integral along the curve:

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

perp(x, y) = (-y, x)

where int is for integration, d/dt for differentiation and dot for dot
product. The integration has to be pieced to the parts corresponding to
the smooth curve segments.

Did I do any mistake ( I calculated the integral expression with Maple )
? The numerical results are wrong..

I used the same approach to evaluate the area of the curve ( F(x,y) =
[x,-y]/2 and calculated with Maple too )and works great..

Divergence of that one is 0.