Line Integrals of Scalar Fields

Scalar Line Integrals over Parametric Curves

Definition: Line Integral of a Scalar Field

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be a parametric curve which is continuously differentiable on $(a;b)$ and whose image $\gamma ([a;b])$ is a subset of $\mathcal{D}$.

The line integral of $f$ along $\gamma$ is the integral

$${\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}$$

NOTATION

$${\int }_{\gamma }f{\int }_{\gamma }f\mathrm{ds}$$

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

$${\oint }_{\gamma }f{\oint }_{\gamma }f\mathrm{ds}$$

Properties

Theorem: Linearity of the Scalar Line Integral

The scalar line integral is linear:

$${\int }_{\gamma }(\lambda f+\mu g)\mathrm{ds}=\lambda {\int }_{\gamma }f\mathrm{ds}+\mu {\int }_{\gamma }g\mathrm{ds}$$

PROOF

$$\begin{array}{rl}{\int }_{\gamma }(\lambda f+\mu g)\mathrm{ds}& ={\int }_a^b[\lambda f(\gamma (t))+\mu g(\gamma (t))]||\dot{\gamma }(t)||\mathrm{dt}\\ & ={\int }_a^b\lambda f(\gamma (t))||\dot{\gamma }(t)||+\mu g(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}\\ & =\lambda {\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}+\mu {\int }_a^{b}g(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}\\ & =\lambda {\int }_{\gamma }f\mathrm{ds}+\mu {\int }_{\gamma }g\mathrm{ds}\end{array}$$

Mean value theorem for Scalar Line Integrals

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be a continuously differentiable parametric curve whose image $\gamma ([a;b])$ is a subset of $\mathcal{D}$.

If $f$ is continuous, then there exists some $\mathbf{p}\in \gamma ([a;b])$ such that the line integral of $f$ along $\gamma$ is

$${\int }_{\gamma }f=f(\mathbf{p})\cdot L,$$

where $L$ is the length of $\gamma$.

PROOF

TODO

Scalar Line Integrals over Geometric Curves

The following theorem makes it possible to provide an unambiguous definition of the line integral of a scalar field over a curve in ${\mathbb{R}}^n$.

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 $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar 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 $f$ along $\gamma$ and $\phi$ are equal.

$${\int }_{\gamma }f={\int }_{\phi }f$$

PROOF

We will just show this for the case when $\gamma$ and $\phi$ are not piecewise, since the proof is easily generalizable - 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 bijection, continuously differentiable function $u:[a;b]\to [c;d]$ such that

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

The then gives us

$${\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}={\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))u^{\prime }(t)||\mathrm{dt}={\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))|||u^{\prime }(t)|\mathrm{dt}$$

If $\gamma$ and $\phi$ have the , then $u(a)=c$ and $u(b)=d$. Since $a<b,c<d$ and $u$ is bijective, $u$ must be strictly increasing, i.e. $|u^{\prime }(t)|=u^{\prime }(t)$. We thus have

$${\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))|||u^{\prime }(t)|\mathrm{dt}={\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))||u^{\prime }(t)\mathrm{dt}$$

By applying the substitution $\mathrm{d}u=u^{\prime }\mathrm{dt}$, we obtain

$${\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))||u^{\prime }(t)\mathrm{dt}={\int }_c^{d}f(\phi (u))||\dot{\phi }(u)||\mathrm{du}$$ $${\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}={\int }_c^{d}f(\phi (u))||\dot{\phi }(u)||\mathrm{du}$$

We are done with the case where $\gamma$ and $\phi$ have the same orientation. Now, if $\gamma$ and $\phi$ have , then $u(a)=d$ and $u(b)=c$. Since $a<b,c<d$ and $u$ is bijective, $u$ must be strictly decreasing, i.e. $|u^{\prime }(t)|=-u^{\prime }(t)$. We thus have

$${\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))|||u^{\prime }(t)|\mathrm{dt}=-{\int }_a^{b}f(\phi (u(t)))||\dot{\phi }(u(t))||u^{\prime }(t)\mathrm{dt}$$

By applying the substitution $\mathrm{d}u=u^{\prime }\mathrm{dt}$, we obtain

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

And so the proof is complete.

The aforementioned theorem guarantees that line integrals of a real scalar field over continuously differentiable, equivalent parametrizations of the same simple curve $\mathcal{C}$ are equal. 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. But then how do we know which group to choose? Well, a very natural choice is based on the equivalence of regular injective parametrizations. Since these parametrizations are injective, scalar line integrals over them depend only on the intrinsic length of $\mathcal{C}$ and not on the way these parametrizations trace out $\mathcal{C}$.

Definition: Scalar Line Integrals over Curves

The line integral of a real scalar field $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ over a simple curve $\mathcal{C}\subseteq \mathcal{D}$ is defined as the line integral of $f$ 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:

$${\int }_{\gamma }f$$

NOTATION

$${\int }_{\mathcal{C}}f{\int }_{\mathcal{C}}f\mathrm{ds}$$

Note: Generalization to Non-Simple Curves

If $\mathcal{C}$ is not simple but can be represented as the union $\mathcal{C}={\mathcal{C}}_1\cup {\mathcal{C}}_2\cup \cdots \cup {\mathcal{C}}_{p-1}\cup {\mathcal{C}}_p$ of finitely many simple curves ${\mathcal{C}}_1,\dots ,{\mathcal{C}}_p$ such that ${\mathcal{C}}_k$ and ${\mathcal{C}}_{k+1}$ share exactly one point and this point is an endpoint for both ${\mathcal{C}}_k$ and ${\mathcal{C}}_{k+1}$, then we define the line integral of $f$ over $\mathcal{C}$ as the sum of the line integrals of $f$ over ${\mathcal{C}}_1,\dots ,{\mathcal{C}}_p$:

$${\int }_{\mathcal{C}}f=\sum _{i=1}^p{\int }_{{\mathcal{C}}_i}f$$

Notation: Line Integrals over Closed Curves

If $\mathcal{C}$ is closed, we denote the line integral of $f$ over $\mathcal{C}$ as one of the following:

$${\oint }_{\mathcal{C}}f{\oint }_{\mathcal{C}}f\mathrm{ds}$$