Vector Line Integrals

Vector Line Integrals over Parametric Curves

Definition: Vector 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 a differentiable parametric curve whose image $\gamma ([a;b])$ is a subset of $\mathcal{D}$.

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

$${\int }_a^b\mathbf{F}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt},$$

where $\cdot$ denotes the dot product.

NOTATION

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

If the image of $\gamma$ is a closed curve, then we write

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

Theorem: Vector Line Integrals over Reparameterisations

Let $\mathbf{F}:D\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ and $\phi :[c;d]\to {\mathbb{R}}^n$ be equivalent, piecewise continuously differentiable parametric curves whose image is a subset of $\mathcal{D}$.

If $\gamma$ and $\phi$ have the same orientation, then the line integrals of $\mathbf{F}$ over $\gamma$ and $\phi$ are equal:

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

If $\gamma$ and $\phi$ have opposite orientations, then the line integrals of $\mathbf{F}$ over $\gamma$ and $\phi$ are equal in magnitude but have different algebraic signs:

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

PROOF

We will just show this for the case when $\gamma$ and $\phi$ are not piecewise, since the proof is easily generalisable - if $\gamma$ and $\phi$ are piecewise, then one can just apply the following proof to each of their partitions and obtain the same end result after summing the results from each partition.

Since $\gamma$ and $\phi$ parameterise the same curve $\mathcal{C}$, we can reparameterise one in the other. More specifically, there exists a bijective, continuously differentiable function $u:[a;b]\to [c;d]$ such that

$$\phi (u(t))=\gamma (t)$$

The chain rule gives us

$${\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_a^b\mathbf{v}(\phi (u(t)))\cdot {\phi }^{\prime }(u(t))u^{\prime }(t)\mathrm{dt}$$

If $\gamma$ and $\phi$ have the same orientation, then $u(a)=c$ and $u(b)=d$. Using the substitution $\mathrm{d}u=u^{\prime }(t)\mathrm{dt}$ we obtain

$${\int }_a^b\mathbf{v}(\phi (u(t)))\cdot {\phi }^{\prime }(u(t))u^{\prime }(t)\mathrm{dt}={\int }_c^d\mathbf{v}(\phi (u))\cdot {\phi }^{\prime }(u)\mathrm{du}$$ $${\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}={\int }_c^d\mathbf{v}(\phi (u))\cdot {\phi }^{\prime }(u)\mathrm{du}$$

If $\gamma$ and $\phi$ have the same orientation, then $u(a)=d$ and $u(b)=c$. Using the substitution $\mathrm{d}u=u^{\prime }(t)\mathrm{dt}$ we obtain

$${\int }_a^b\mathbf{v}(\phi (u(t)))\cdot {\phi }^{\prime }(u(t))u^{\prime }(t)\mathrm{dt}={\int }_d^c\mathbf{v}(\phi (u))\cdot {\phi }^{\prime }(u)\mathrm{du}$$ $${\int }_a^b\mathbf{v}(\gamma (t))\cdot {\gamma }^{\prime }(t)\mathrm{dt}=-{\int }_c^d\mathbf{v}(\phi (u))\cdot {\phi }^{\prime }(u)\mathrm{du}$$

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 a differentiable parametric curve whose image $\gamma ([a;b])$ is a subset of $\mathcal{D}$.

The line integral of $\mathbf{F}$ over $\gamma$ is equal to the line ntegral 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

The vector line integral is linear:

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

PROOF

TODO

Vector Line Integrals over Geometric Curves

Theorem: Line Integrals over Equivalent Parametrizations

Let $\mathcal{C}$ be a simple curve in ${\mathbb{R}}^n$, let $\gamma :[a;b]\to {\mathbb{R}}^n$ and $\phi :[c;d]\to {\mathbb{R}}^n$ be parametrizations of $\mathcal{C}$ and let $\mathbf{F}:\mathcal{D}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field.

If $\gamma$ and $\phi$ are continuously differentiable on $(a;b)$ and $(c;d)$, respectively, 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 equal in magnitude but opposite in sign whenever $\gamma$ and $\phi$ have opposite orientations

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

PROOF

TODO

The aforementioned theorem guarantees that the line integrals of a vector field $\mathbf{F}$ over continuously differentiable parametrizations of the same curve $\mathcal{C}$ which are equivalent up to a continuously differentiable reparametrization are either equal or equal in magnitude but opposite in sign. However, while some parametrizations may be equivalent amongst each other and some other parametrizations may also be equivalent amongst each other, this does not ensure that the two groups of parametrizations are all equivalent. If we want to define the notion of a line integral over a curve geometrically, we need to determine which equivalence class to consider. A very natural choice is based on the equivalence of regular injective parametrizations. Since these parametrizations are injective, the absolute value of the line integrals of $\mathbf{F}$ over them depend only on the length of $\mathcal{C}$ and are thus equal. However, each of these parametrizations still has one of two possible orientations. This leaves us with two options for defining the line integral of $\mathbf{F}$ over $\mathcal{C}$, none of which is objectively better.

Definition: Vector Line Integrals over Curves

Let $\mathbf{F}:\mathcal{D}\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^n$ be a real vector field and let $\mathcal{C}\subseteq \mathcal{D}$ be a simple curve in ${\mathbb{R}}^n$.

The line integral of $\mathbf{F}$ over $\mathcal{C}$ is defined as a line integral of $\mathbf{F}$ over any injective parametrization $\gamma :[a;b]\subset \mathbb{R}\to {\mathbb{R}}^n$ of $\mathcal{C}$ which is continuously differentiable on $(a;b)$ with a non-vanishing derivative:

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