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Assume you have a triangle in the plane defined by its three vertices at $(x_0,y_0)$, $(x_1,y_1)$, and $(x_2,y_2)$. Is there a general expression for the moments of the triangle, where the moments are defined as averages by the following expression:

$$\langle x^n y^m \rangle = \frac{1}{A}\int_\triangle x^n y^m \, dA $$

Integrating manually to compute the first few moments, one can obtain:

$$\langle x \rangle = \frac{1}{3} \left(x_0+x_1+x_2\right) $$

$$\langle x^2 \rangle = \frac{1}{6} \left(x_0^2+x_1^2+x_2^2+x_0 x_1+x_0 x_2+x_1 x_2\right) $$

$$\langle x y \rangle = \frac{1}{12}\left(2 x_0 y_0+x_1 y_0+x_2 y_0+x_0 y_1+2 x_1 y_1+x_2 y_1+x_0 y_2+x_1 y_2+2 x_2 y_2 \right)$$

Is it possible to obtain a closed form expression for general $n$ and $m$ using combinations of the coordinates?

EDIT:

Can the same be done for line segments defined by their endpoints $(x_0,y_0)$ and $(x_1,y_1)$? Again the first few moments in this case are:

$$\langle x \rangle = \frac{1}{2} \left(x_0+x_1\right) $$ $$ \langle x^2 \rangle = \frac{1}{3} \left(x_0^2+x_0 x_1+x_1^2\right) $$ $$ \langle x^3 \rangle = \frac{1}{4} \left(x_0^3+x_0^2 x_1+x_0 x_1^2+x_1^3\right) $$ $$ \langle x y \rangle = \frac{1}{6} \left(2 x_0 y_0+x_1 y_0+x_0 y_1+2 x_1 y_1\right) $$ $$ \langle x^2 y \rangle = \frac{1}{12} \left( 3 x_0^2 y_0+2 x_0 x_1 y_0+x_1^2 y_0+x_0^2 y_1+2 x_0 x_1 y_1+3 x_1^2 y_1\right)$$

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    Note that the formula for the area of a triangle, given the three coordinates, is $\frac12\begin{vmatrix}x_0&y_0&1\\\\x_1&y_1&1\\\\x_2&y_2&1\end{vmatrix}$2010-09-04
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    http://dx.doi.org/10.1016/0031-3203(95)00147-6 and http://dx.doi.org/10.1007/BF02020802 might also be useful.2010-09-04
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    The final integral I settled with is $\int_0^1\int_0^1 2u\left((1-u)x_0+u\left((1-v)x_1+vx_2\right)\right)^p\left((1-u)y_0+u\left((1-v)y_1+vy_2\right)\right)^q\mathrm{d}v\mathrm{d}u$ which evaluates to a hypergeometric mess in *Mathematica*; I doubt you'd want that.2010-09-04

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