= Vector areas of a closed polyhedron sum to zero
{title2=$\sum_f A_f\hat{\mathbf n}_f=0$}
For the faces of a closed <polyhedron>, take outward <unit normal vectors>. Applying the <divergence theorem> to an arbitrary constant <vector field> $c$ gives
$$
c\cdot\sum_f A_f\hat{\mathbf n}_f
=\int_{\partial\Omega}c\cdot\hat{\mathbf n}\,dA
=\int_\Omega\nabla\cdot c\,dV=0.
$$
Since $c$ is arbitrary, the <vector areas> sum to zero. This is an oriented statement; the positive scalar <surface areas> do not cancel. The same proof works for every piecewise smooth closed boundary to which the <divergence theorem> applies.
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