Vector surface area element ($\mathbf{n}dS=\pm ({\mathbf{X}}_u\times {\mathbf{X}}_v)dudv$)

A vector area element combines the scalar surface area element with a chosen unit normal vector. Reversing the order of the two parameter directions reverses its sign. Flux through an oriented surface is the integral of the dot product of a vector field with this vector element. For a graph $z=f(x,y)$, the upward choice is $(-f_x,-f_y,1)dxdy$.

For a closed polyhedron, let $A_F$ be a face area and ${\mathbf{n}}_F$ its outward unit normal vector. The sum of the face area vectors is zero. To prove it, fix any constant vector $\mathbf{q}$ and apply the divergence theorem to the constant vector field $\mathbf{q}$: its divergence vanishes, while its outward flux is $\mathbf{q}\cdot {\sum }_F{A}_F{\mathbf{n}}_F$. This is zero for every $\mathbf{q}$, so the sum vanishes. For a tetrahedron, the same identity follows directly by expanding the four oriented face cross products; multiplying every face vector by two leaves the identity intact.