Integration of Real Vector Fields

Line Integrals

Definition: Vector Line Integral

Let $\mathbf{F}:{\mathcal{D}}_{\mathbf{F}}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field and let $\gamma :{\mathcal{D}}_{\gamma }\to {\mathbb{R}}^n$ be vector-valued with $\gamma ({\mathcal{D}}_{\gamma })\subseteq \mathcal{D}$ and such that $\gamma$ is differentiable on $(a;b)$.

The (vector) line integral of $\mathbf{F}$ over $\gamma$ is the integral

$${\int }_{{\mathcal{D}}_{\gamma }}\mathbf{F}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt},$$

where $\cdot$ denotes the dot product.

NOTATION

$${\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}$$

If $\gamma$ is a closed parametric curve, then we write

$${\oint }_{\gamma }\mathbf{F}\cdot \mathrm{ds}$$

Theorem: Vector Line Integral to Scalar Line Integral

Let $\mathbf{F}:\mathcal{D}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be vector-valued with $\gamma ([a;b])\subseteq \mathcal{D}$ and such that $\gamma$ is differentiable on $(a;b)$.

The vector line integral of $\mathbf{F}$ over $\gamma$ is equal to the scalar line integral of the dot product of $\mathbf{F}$ with the unit tangent vector of $\gamma$:

$${\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}={\int }_{\gamma }\mathbf{F}\cdot \mathbf{T}\mathrm{ds}$$

PROOF

$$\begin{array}{rl}{\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}& ={\int }_a^b\mathbf{F}(\gamma (t))\cdot {\gamma }^{\prime }(t)dt\\ {\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}& ={\int }_a^b\mathbf{F}(\gamma (t))\cdot \frac{{\gamma }^{\prime }(t)}{\Vert {\gamma }^{\prime }(t)\Vert }\Vert {\gamma }^{\prime }(t)\Vert dt\\ & ={\int }_a^b\left(\mathbf{F}(\gamma (t))\cdot \mathbf{T}(t)\right)\Vert {\gamma }^{\prime }(t)\Vert \mathrm{dt}\\ & ={\int }_{\gamma }(\mathbf{F}\cdot \mathbf{T})\mathrm{ds}\end{array}$$

Theorem: Linearity of the Vector Line Integral

Let $\mathbf{v}:{\mathcal{D}}_{\mathbf{v}}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ and $\mathbf{u}:{\mathcal{D}}_{\mathbf{u}}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector fields and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be vector-valued with $\gamma ([a;b])\subseteq {\mathcal{D}}_{\mathbf{u}}\cap {\mathcal{D}}_{\mathbf{v}}$ and such that $\gamma$ is differentiable on $(a;b)$.

For all $\lambda ,\mu \in \mathbb{R}$, the vector line integral is linear:

$${\int }_{\gamma }(\lambda \mathbf{u}+\mu \mathbf{v})\cdot \mathrm{ds}=\lambda {\int }_{\gamma }\mathbf{u}\cdot \mathrm{ds}+\mu {\int }_{\gamma }\mathbf{v}\cdot \mathrm{ds}$$

PROOF

TODO

Theorem: Vector Line Integrals and Equivalence

Let $\mathbf{F}:{\mathcal{D}}_{\mathbf{F}}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field and let $\gamma :{\mathcal{D}}_{\gamma }\subset \mathbb{R}\to {\mathbb{R}}^n$ and $\phi :{\mathcal{D}}_{\phi }\subset \mathbb{R}\to {\mathbb{R}}^n$ be vector-valued.

If $\gamma$ and $\phi$ are continuously differentiable and are equivalent up to a continuously differentiable reparametrization, then the line integrals of $\mathbf{F}$ along $\gamma$ and $\phi$

• are equal whenever $\gamma$ and $\phi$ have the same orientation

$${\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}={\int }_{\phi }\mathbf{F}\cdot \mathrm{ds}$$

• are opposite whenever $\gamma$ and $\phi$ have opposite orientations

$${\int }_{\gamma }\mathbf{F}\cdot \mathrm{ds}=-{\int }_{\phi }\mathbf{F}\cdot \mathrm{ds}$$

PROOF

TODO

Surface Integrals

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 such that $S({\mathcal{D}}_S)\subseteq {\mathcal{D}}_{\mathbf{F}}$.

The (vector) surface integral of $\mathbf{F}$ over $S$ is the double integral of the dot product $\mathbf{F}\circ S$ with the surface normal of $S$:

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

NOTATION

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

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 such that $S({\mathcal{D}}_S)\subseteq {\mathcal{D}}_{\mathbf{F}}$.

The surface integral of $\mathbf{F}$ over $S$ is equal to the 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 and Equivalence

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 $\psi :{\mathcal{D}}_{\psi }\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^n$ be parametric surfaces.

If $\varphi$ and $\psi$ are continuously differentiable and are equivalent up to a continuously differentiable reparametrization, then the surface integrals of $\mathbf{F}$ over $\varphi$ and $\psi$ are:

• equal when $\varphi$ and $\psi$ have the same orientation

$${\iint }_{\varphi }\mathbf{F}\cdot \mathrm{d\varphi }={\iint }_{\psi }\mathbf{F}\cdot \mathrm{d\psi }$$

• opposite when $\varphi$ and $\psi$ have opposite orientations

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

PROOF

TODO