Scalar surface integral

For a smooth parameterized surface $S$ with parameterization $\mathbf{r}(u,v)$ over a parameter domain $D$, and a continuous scalar function $f$ on $S$, the scalar surface integral is

${\iint }_{S}fdS={\iint }_{D}f(\mathbf{r}(u,v))|{\mathbf{r}}_u\times {\mathbf{r}}_v|dudv.$

The factor $|{\mathbf{r}}_u\times {\mathbf{r}}_v|$ converts the parameter-area $dA=dudv$ to surface-area $dS$. It is the surface analog of the speed $|{\mathbf{r}}^{\prime }(t)|$ converting $dt$ to $ds$ in a scalar line integral.

Computing surface area

Set $f\equiv 1$:

$\mathrm{Area}(S)={\iint }_{S}1dS={\iint }_D|{\mathbf{r}}_u\times {\mathbf{r}}_v|dudv.$

Orientation-independence

The scalar surface integral is independent of parameterization and independent of orientation — same number whichever way you parameterize, whichever side you call “outward.” This matches the corresponding property of scalar line integrals. The integrand $f$ and the differential $dS$ are both scalars; no signed information.

Worked example: surface area of a sphere

Sphere of radius $a$, parameterized as

$\mathbf{r}(\theta ,\varphi )=⟨a\sin \theta \cos \varphi ,a\sin \theta \sin \varphi ,a\cos \theta ⟩.$

Partials: ${\mathbf{r}}_{\theta }=⟨a\cos \theta \cos \varphi ,a\cos \theta \sin \varphi ,-a\sin \theta ⟩$, ${\mathbf{r}}_{\varphi }=⟨-a\sin \theta \sin \varphi ,a\sin \theta \cos \varphi ,0⟩$.

Cross product (after computation): ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_{\varphi }=⟨a^2{\sin }^2\theta \cos \varphi ,a^2{\sin }^2\theta \sin \varphi ,a^2\sin \theta \cos \theta ⟩$.

Magnitude: $a^2\sin \theta$.

$\mathrm{Area}={\int }_0^{2\pi }{\int }_0^{\pi }{a}^2\sin \theta d\theta d\varphi =a^2\cdot 2\pi \cdot 2=4\pi a^2.✓$

Graph shortcut

For a surface that’s the graph $z=f(x,y)$ over a region $D$ in the $xy$-plane, $|{\mathbf{r}}_x\times {\mathbf{r}}_y|=\sqrt{f_x^2+f_y^2+1}$, so

${\iint }_{S}gdS={\iint }_{D}g(x,y,f(x,y))\sqrt{f_x^2+f_y^2+1}dxdy.$

Physical meanings

If $f$ is a surface density (mass per unit area), ${\iint }_{S}fdS$ is total mass. If $f\equiv 1$, total area.

In context

Scalar surface integrals are less central than flux integrals (which use a vector field and an orientation) in Vector Calculus and Complex Analysis. The major theorems — Stokes’ theorem, Divergence theorem — both use flux integrals, not scalar ones. But the scalar version is needed for computing surface areas and is the simplest example of integration over a parameterized surface.