Total Vector Area of Polyhedron is Zero

Theorem

Let $P$ be a polyhedron.

Let the positive direction be defined as outward.

Let $\mathbf T$ be the total vector area of all the faces of $P$.

Then:

$\mathbf T = \mathbf 0$

Proof

$P$ can be geometrically divided into a finite number of tetrahedra.

Every face of these tetrahedra which are internal to $P$ appears twice: once with a positive vector area, and once with a negative normal.

Hence for any polyhedron, the total vector area is the sum of the vector areas of all the tetrahedra.

The result follows from Total Vector Area of Tetrahedron is Zero.

$\blacksquare$

Sources

• 1951: B. Hague: An Introduction to Vector Analysis (5th ed.) ... (previous) ... (next): Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area