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.” Same as scalar line integrals. The integrand $f$ and the differential $dS$ are both scalars, so there’s 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 show up less than flux integrals (which use a vector field and an orientation). Stokes’ theorem and the Divergence theorem both use flux integrals, not scalar ones. The scalar version is still what you need for surface areas, and it’s the simplest case of integrating over a parameterized surface.