Vector area element ($d\mathbf{S}$)

For an oriented parametrized surface $\mathbf{r}(u,v)$,

For a regular implicit surface $f(x,y,z)=0$ with $f_z\ne 0$, the implicit function theorem gives a local graph $z=h(x,y)$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Differentiating $f(x,y,h(x,y))=0$ yields $h_x=-f_x/f_z$, $h_y=-f_y/f_z$, proving the formula. Division by $f_z$ makes the final component positive independently of the sign of the original gradient. At a point where $f_z=0$, another projection or parametrization is needed.

The vector area of an oriented surface is the integral of its unit normal vector times its area element. If its oriented boundary is a closed curve $C$, Stokes theorem givesConsequently it depends only on the oriented boundary, not on the spanning surface. For a planar surface it is signed area times a unit normal vector. A current loop has magnetic dipole moment $\mathbf{m}=I\mathbf{S}$; the scalar area of a nonplanar spanning surface is not interchangeable with this vector.

For the faces of a closed polyhedron, take outward unit normal vectors. Applying the divergence theorem to an arbitrary constant vector field $c$ givesSince $c$ 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.

For an oriented closed polygon, triangulate with a common vertex $P$. Its total vector area is $\frac{1}{2}{\sum }_i(B_i-P)\times (B_{i+1}-P)$; the terms involving $P$ telescope, giving the displayed boundary-only formula. Its signed projected area along a unit vector $V$ is $V\cdot \mathbf{A}$. The maximum is $|\mathbf{A}|$, in direction $\mathbf{A}/|\mathbf{A}|$ if nonzero, and zero projections have directions perpendicular to $\mathbf{A}$. If $\mathbf{A}=0$, every direction has zero signed projection despite possible nonzero unsigned surface area.