Area element of a surface ($dA$)

In a positively oriented surface chart,The coordinate-change Jacobian and the metric determinant transform inversely, so this defines a global area form.

The surface area of a measurable region $U\subseteq S$ is ${\int }_{U}dA$. For a parametrization $X(u,v)$ it is $\int |X_u\times X_v|dudv$.

For projection of a planar region with unit normal $\mathbf{n}$ onto a plane with unit normal $\mathbf{k}$, the projected area is $|\mathbf{n}\cdot \mathbf{k}|$ times the original area. Indeed, the signed projected area of the parallelogram spanned by tangent vectors $\mathbf{u},\mathbf{v}$ is $(\mathbf{u}\times \mathbf{v})\cdot \mathbf{k}$, while $\mathbf{u}\times \mathbf{v}$ is parallel to $\mathbf{n}$. Taking magnitudes proves the factor. The same pointwise formula applies to a parametrized surface; adding projected patches counts multiplicity unless their images overlap only in sets of zero area.

The area of a planar oriented region with a sign determined by orientation. For a triangle with cyclically ordered points $A,B,C$, it is $\frac{1}{2}((B-A)\times (C-A))$, using the two-dimensional cross product; reversing the order reverses the sign.

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.