How can one calculate the hypervolume of a convex n-dimensional polyhedron given the coordinates of its...

I can give a full and complete answer to this question.

            n
Working in R  , let  v , i=1,...N be points in this vector space. N >= n
                      i

v = {x  , x   ,x  , ... x  }
 i    0i   1i   2i       ni

Assume that this set of points forms a convex hull.

Then the volume of the hull is:

                |v |
                | 1|
                |  |
                |v |
                | 2|
                |  |
  1/N!  dyad    |v |
                | 3|
                |  |
                |. |
                |. |
                |v |
                | N|

This is a dyadic tensor, not a determinant since the matrix above may
not be square. As an example let n=2, N=3 (a triangle) with the following
points:
(1,0), (3,0), (2,4). The area is then:

1/2! dyad(  (1,0), (3,0), (2,4)).

To find the dyadic in two space write down:

1 0
3 0
2 4
1 0

The fourth row is formed by adding the top one at the bottom. Add up the
diagonal products (1*0 + 3*4 + 2*0) subtract the other diagonals and
divide by 2!.

It works the same in n dimensions except one needs generalized dyadics,
not 2D.

-- Bob Silverman, in reply to Renwill Yang <ryang@watdcsu.waterloo.edu>

[ Editor: the example divides by 2! = n!, not by N! = 3! as in the
general formula; with 2! it gives the correct area,
((0 + 12 + 0) - (0 + 0 + 4))/2 = 4. ]