# Volume of a figure

For 2D figures in coordinate system there exists a simple formula for the area (it is shown here: http://mathworld.wolfram.com/PolygonArea.html ). Is there exist a similar formula for volume of a figure in 3D?

• How would you order your vertices in 3D space? – Matthias Mar 24 '16 at 20:22

Yes, there is an almost direct generalization by dissecting your volume (even non convex) into tetrahedra $A_0A_lA_mA_n$, with a common fixed origin $A_0$. The oriented volume of each tetrahedron is

$$(1/3!) det(\overrightarrow{A_0A_l},\overrightarrow{A_0A_m},\overrightarrow{A_0A_n})$$

"Oriented" means that one has to give the vertices $A_lA_mA_n$ in such a way that the triangular faces are oriented in such a way that vectors $\overrightarrow{A_0A_l},\overrightarrow{A_0A_m},\overrightarrow{A_0A_n}$ form a direct basis. The sum of all these volumes is the volume of the polyhedron.

It is not necessary that $A_0$ is a vertex of the polyhedron.

If $A_0$ is exterior to the polyhedron, one might think that the computed volume is larger than its true volume. This is not the case because the volumes of the tetrahedra that are "directly visible" from the origin come in substraction, not in addition.

The way you proceed the faces cannot be guided by a "canonical" ordering of the faces... (see the remark of @Matthias).

Remark: we have been speaking about triangulations: all polyedra can be triangulated, even if some of their faces have more than 3 vertices; in such cases, the face is divided into triangles.

• By direct basis you mean a reference with a given orientation? Is this the usual convention? Thanks. – Edu Dec 6 '16 at 18:23