Scalar Surface Integral

Definition: Scalar Surface Integral

Let $f:{\mathcal{D}}_f\subseteq {\mathbb{R}}^3\to \mathbb{R}$ be a real scalar field and let $S:{\mathcal{D}}_S\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^3$ be a differentiable parametric surface whose image $S({\mathcal{D}}_S)$ is a subset of ${\mathcal{D}}_f$.

The (scalar) line integral of $f$ over $S$ is the double integral

$${\iint }_{{\mathcal{D}}_S}f(S(x,y))||\mathbf{N}(x,y)||\mathrm{d}{\mathcal{D}}_S,$$

where $\mathbf{N}$ is the normal vector of $S$.

NOTATION

$${\iint }_{S}f{\iint }_{S}f\mathrm{dS}$$

If $S({\mathcal{D}}_S)$ is a closed surface, then a circle can be put through the two integral signs.

Theorem: Surface Integrals of Scalar Fields

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^3\to \mathbb{R}$ be a real scalar field and let $\varphi :{\mathcal{D}}_{\varphi }\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^3$ and $\psi :{\mathcal{D}}_{\psi }\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^3$ be differentiable parametric surfaces whose images are subsets of $\mathcal{D}$.

If $\varphi$ and $\psi$ are smooth reparameterisations and $f$ is continuous, then the surface integrals of $f$ over $\varphi$ and $\psi$ are equal.

$${\iint }_{\varphi }f={\iint }_{\psi }f$$

PROOF

TODO