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### #Actualmax343

Posted 29 November 2012 - 03:59 AM

Alright.

Take the vector field: F=(x-x0)/3, x and x0 are in R^3.
Its divergence with respect to x is 1.

So: Volume = Int[Div(F)*dV] = Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3
Where: A_i = Area of triangle. With its plane equation <n_i,x>=d_i, with unit normal.
If you choose x0=0, then you get Sum[A_i*d_i/3]
For unit n_i, d_i is exactly the distance of the plane from the origin. So what is written in the sum is exactly the volume of the tetrahedron.

### #1max343

Posted 29 November 2012 - 03:43 AM

Alright.

Take the vector field: F=(x-x0)/3, x and x0 are in R^3.
Its divergence is 1.

So: Volume = Int[Div(F)*dV] = Int[<F,dS>] = Sum[A_i * (d_i - <n_i,x_0>)]/3
Where: A_i = Area of triangle. With its plane equation <n_i,x>=d_i, with unit normal.
If you choose x0=0, then you get Sum[A_i*d_i/3]
For unit n_i, d_i is exactly the distance of the plane from the origin. So what is written in the sum is exactly the volume of the tetrahedron.

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