Vector Surface Integral

Definition: Vector Surface Integral

Let $\mathbf{F}:{\mathcal{D}}_{\mathbf{F}}\subseteq {\mathbb{R}}^3\to {\mathbb{R}}^3$ be a real vector 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 ${\mathcal{D}}_{\mathbf{F}}$.

The (vector) surface integral of $f$ over $S$ is the double integral

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

where $\mathbf{N}$ is the normal vector of $S$ and $\cdot$ denotes the dot product.

NOTATION

$${\iint }_S\mathbf{F}{\iint }_S\mathbf{F}\cdot \mathrm{d}\mathbf{S}$$

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

Theorem: Vector Surface Integral to Scalar Surface Integral

Let $\mathbf{F}:{\mathcal{D}}_{\mathbf{F}}\subseteq {\mathbb{R}}^3\to {\mathbb{R}}^3$ be a real vector 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 ${\mathcal{D}}_{\mathbf{F}}$.

The vector surface integral of $\mathbf{F}$ over $S$ is equal to the scalar surface integral of the dot product between $\mathbf{F}$ and the unit surface normal of $S$.

$${\iint }_S\mathbf{F}\cdot \mathrm{dS}={\iint }_S\mathbf{F}\cdot \mathbf{n}\mathrm{dS}$$

PROOF

TODO

Theorem: Vector Surface Integrals of Reparameterisations

Let $\mathbf{F}:{\mathcal{D}}_{\mathbf{F}}\subseteq {\mathbb{R}}^3\to {\mathbb{R}}^3$ be a real vector 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 equivalent parametric surfaces whose image is a subset of ${\mathcal{D}}_{\mathbf{F}}$.

If $\varphi$ and $\psi$ have the same orientation, then the surface integrals of $\mathbf{F}$ over $\varphi$ and $\psi$ are equal:

$${\iint }_{\varphi }\mathbf{F}={\iint }_{\psi }\mathbf{F}$$

If $\varphi$ and $\psi$ have opposite orientations, then the surface integrals of $\mathbf{F}$ over $\varphi$ and $\psi$ are equal in magnitude but differ in algebraic sign:

$${\iint }_{\varphi }\mathbf{F}=-{\iint }_{\psi }\mathbf{F}$$

PROOF

TODO