The paper Volumes From Overlaying 3-D Triangulations
in Parallel by Franklin and Kankanhalli presents a mysterious formula that I have corroborated numerically in several cases.
It can be used to reduce an integral over the interior of an n-dimensional polyhedron, to a sum over its vertices.
The Volume a Polyhedron in n-space Bounded by (n−1)-Simplexes
Choose a point P in space (call in an origin if you wish).
For each oriented bounding simplex number the vertices (independently) from 0 to n−1.
Let vi be the vector from P to vertex i.
For 0<i<n let xi = vi − v0
and x0 = v0 − P.
For the bounding simplex let V = det(x)/n!.
The volume of the polyhedron is ΣV summed over the bounding simplexes.
Insight: V for a bounding (n−1)-simplex, is the signed volume of the convex hull of the simplex and P (which is an n-simplex).
det(v) = det(x) but det(v) usually requires less precision.
For a performance gain one can choose a nearby plane PL and compute the content of the projection of a bounding simplex onto PL and multiply by the signed distance of the simplex centroid to that plane.
The sum of these products is also the volume of the polyhedron.
This is a computational gain if PL is parallel to one of the coordinate axis planes because a row of zeros is introduced into the determinant.
The same plane must be used for each bounding simplex.