Line Integrals (Real Scalar Fields)#

Definition: Line Integral of a Real Scalar Field

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\gamma: I \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve which is differentiable on the interior of the interval \(I\) with \(\gamma(I) \subseteq \mathcal{D}_f\).

The line integral of \(f\) along \(\gamma\) is the (potentially improper) Riemann integral

\[\int_I f(\gamma(t)) ||\gamma'(t)||\, \mathrm{d}t,\]

provided that it exists.

Notation

We denote the line integral of \(f\) along \(\gamma\) as follows:

\[\int_{\gamma} f \,\mathrm{d}s\]

If \(\gamma\) is closed, we write:

\[\oint_{\gamma} f \,\mathrm{d}s\]

Example

Consider the real scalar field \(f: \mathbb{R}^2 \to \mathbb{R}\) defined as

\[f(x, y) = x\]

and the parametric curve \(\gamma: [0,1] \to \mathbb{R}^2\) defined as follows:

\[\gamma(t) = \begin{bmatrix}t \\ t^2\end{bmatrix}\]

It is continuous on \([0,1]\) and continuously differentiable on \((0,1)\):

\[\gamma'(t) = \begin{bmatrix} 1 \\ 2t\end{bmatrix}\]

Since \(f\) is also continuous, we know that \(f(\gamma(t))||\gamma'(t)||\) is continuous and thus Riemann-integrable. Therefore, we have

\[\begin{aligned}\int_{\gamma} f \,\mathrm{d}s & = \int_0^1 f(\gamma(t))||\gamma'(t)|| \,\mathrm{d}t \\ & = \int_0^1 t\sqrt{1+4t^2}\,\mathrm{d}t \\ & = \left.\frac{1}{12}(1+4t^2)^{\frac{3}{2}}\right\vert_0^1 \\ & = \frac{1}{12}(5^{\frac{3}{2}} - 1)\end{aligned}\]

for the line integral of \(f\) along \(\gamma\).

Theorem: Linearity of Line Integrals

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^n \to \mathbb{R}\) be real scalar fields and let \(\gamma: I \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve which is differentiable on the interior of the interval \(I\) with \(\gamma(I) \subseteq \mathcal{D}_f \cap \mathcal{D}_g\).

If the line integrals of \(f\) and \(g\) along \(\gamma\) exist, then so does the line Integral of \(\alpha f + \beta g\) for all \(\alpha, \beta \in \mathbb{R}\):

\[\int_{\gamma} \alpha f + \beta g\,\mathrm{d}s = \alpha \int_{\gamma} f \, \mathrm{d}s + \beta \int_{\gamma} g \, \mathrm{d}s\]

Proof

\[\begin{aligned}\int_{\gamma} \alpha f + \beta g\,\mathrm{d}s & = \int_I((\alpha f + \beta g) \circ \gamma)(t) ||\gamma'(t)||\,\mathrm{d}t \\ & = \int_I(\alpha f (\gamma(t)) + \beta g(\gamma(t)))||\gamma'(t)||\,\mathrm{d}t \\ & = \int_I\alpha f (\gamma(t))||\gamma'(t)|| \,\mathrm{d}t + \int_I\beta g(\gamma(t))||\gamma'(t)|| \,\mathrm{d}t \end{aligned}\]

Since the line integrals of \(f\) and \(g\) along \(\gamma\) exist, we know that

\[\int_If (\gamma(t))||\gamma'(t)|| \,\mathrm{d}t= \int_{\gamma} f \,\mathrm{d}s\]

and

\[\int_Ig (\gamma(t))||\gamma'(t)|| \,\mathrm{d}t = \int_{\gamma} g \,\mathrm{d}s\]

exist. Therefore:

\[\begin{aligned}\int_{\gamma} \alpha f + \beta g\,\mathrm{d}s & = \alpha \int_If (\gamma(t))||\gamma'(t)|| \,\mathrm{d}t + \beta \int_Ig (\gamma(t))||\gamma'(t)|| \,\mathrm{d}t \\ & = \alpha \int_{\gamma} f \, \mathrm{d}s + \beta \int_{\gamma} g \, \mathrm{d}s\end{aligned}\]

Theorem: Line Integrals under Reparametrization

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\gamma: I_{\gamma} \subseteq \mathbb{R} \to \mathbb{R}^n\) and \(\varphi: I_{\varphi} \subseteq \mathbb{R} \to \mathbb{R}^n\) be real parametric curve which are continuously differentiable on the interior of the intervals \(I_{\gamma}\) and \(I_{\varphi}\), respectively, with \(\gamma(I_{\gamma}) = \varphi(I_{\varphi}) \subseteq \mathcal{D}_f\).

If the line integrals of \(f\) along \(\gamma\) and \(\varphi\) exist and \(\gamma\) and \(\varphi\) are equivalent up to a continuously differentiable reparametrization, then those line integrals are equal:

\[\int_{\gamma} f \,\mathrm{d}s = \int_{\varphi} f \,\mathrm{d}s\]

Proof

TODO

Mean Value Theorem for Scalar Line Integrals

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\gamma: [a,b] \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve with \(\gamma([a,b]) \subseteq \mathcal{D}_f\).

If \(f\) is continuous on \(\gamma([a,b])\) and \(\gamma\) is continuous on \([a,b]\) and continuously differentiable on \((a,b)\), then there exists some \(t \in [a,b]\) such that the line integral of \(f\) along \(\gamma\) is equal to \(f(\gamma(t))\cdot L\), where \(L\) is the arclength traced by \(\gamma\):

\[\int_{\gamma} f \, \mathrm{d}s = f(\gamma(t))\cdot L\]

Proof

TODO