Line Integrals

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Now we examine the integral of vector functions. The first step toward this is the line integral, which accumulates the contributions of the vectors at each point along a curve defined in the space \(\mathbb{R}^n\) where the vector field is defined. What is interesting is that if the vector field is conservative, this integral becomes independent of the path and depends only on the endpoints; this can be regarded as a higher-dimensional version of the §The Fundamental Theorem of Calculus.

Line Integral

Definition 1 The line integral of a continuous scalar field \(f\) over a \(C^1\) curve \(\mathbf{r}\colon [a, b] \to \mathbb{R}^n\) is

\[\int_C f\mathop{ds} = \int_a^b f(\mathbf{r}(t))\lvert \mathbf{r}'(t)\rvert \mathop{dt}\]

Here, \(ds = \lvert \mathbf{r}'(t)\rvert \mathop{dt}\) is the arc length element.

By definition, the above integral is the value using an arc-length parametrization, so it does not depend on the parametrization of the curve. In the special case where \(f \equiv 1\), the integral \(\int_C \mathop{ds}\) gives the length of the curve.

To lift this to the integral of a vector function, we must take the direction of the curve into account and define it as follows.

Definition 2 The line integral of a continuous vector field \(\mathbf{F}\) over a \(C^1\) curve \(\mathbf{r}\colon [a, b] \to \mathbb{R}^n\) is

\[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\mathop{dt}\]

Writing this using the unit tangent vector \(\mathbf{T} = \mathbf{r}'/\lvert \mathbf{r}'\rvert\), we can verify that

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (\mathbf{F}\cdot \mathbf{T})\mathop{ds}\]

In particular, in the plane, if \(\mathbf{F} = (P, Q)\) and \(\mathbf{r}(t) = (x(t), y(t))\), the notation

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\mathop{dx} + Q\mathop{dy}\]

is also commonly used. Also, the symbol \(\oint\) is sometimes used to denote the integral along a closed curve, but this is merely a matter of notation and no essentially new content is added.

Fundamental Theorem for Line Integrals

Our main theorem is, as foreshadowed above, that the line integral of a conservative field reduces to the difference of the function values at the endpoints.

Theorem 3 (Fundamental theorem for line integrals) If \(f\) is \(C^1\) and \(C\) is a \(C^1\) curve from \(\mathbf{r}(a) = \mathbf{A}\) to \(\mathbf{r}(b) = \mathbf{B}\), then

\[\int_C \nabla f \cdot d\mathbf{r} = f(\mathbf{B}) - f(\mathbf{A})\]

In particular, the line integral of a conservative field depends only on the two endpoints.

Proof

By the multivariable chain rule (§Multivariable Functions and Partial Derivatives, ⁋Theorem 6 (Multivariable Chain Rule)), we have \(\frac{d}{dt} f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\). Therefore, applying §The Fundamental Theorem of Calculus, ⁋Theorem 4,

\[\int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\mathop{dt} = \int_a^b \frac{d}{dt} f(\mathbf{r}(t))\mathop{dt} = f(\mathbf{r}(b)) - f(\mathbf{r}(a))\]

Theorem 3 (Fundamental theorem for line integrals) says that the line integral of a conservative field is independent of the path. Surprisingly, the converse also holds.

Theorem 4 When \(\mathbf{F}\) is continuous on a connected open region \(D\), the following are equivalent.

1. \(\mathbf{F}\) is a conservative field on \(D\).
2. For every closed curve \(C\) in \(D\), \(\oint_C \mathbf{F} \cdot d\mathbf{r} = 0\).
3. \(\int_C \mathbf{F}\cdot d\mathbf{r}\) depends only on the two endpoints of \(C\) and is independent of the path.

Proof

\((1 \Rightarrow 3)\) is Theorem 3 (Fundamental theorem for line integrals). \((3 \Leftrightarrow 2)\) is obvious: cutting a closed curve at one point gives two paths, and joining one path in reverse gives a closed curve.

Thus the core claim is \((3 \Rightarrow 1)\). For this we must construct a potential directly. Fix a base point \(\mathbf{x}_0 \in D\), and for any \(\mathbf{x}\in D\) define \(f(\mathbf{x})\) to be the value of the line integral of \(\mathbf{F}\) from \(\mathbf{x}_0\) to \(\mathbf{x}\). Originally this would depend on the choice of curve \(\mathbf{r}\) joining \(\mathbf{x}_0\) and \(\mathbf{x}\), but we are assuming the third condition, so this definition is well-defined. Now the average rate of change in the coordinate direction \(\mathbf{e}_i\),

\[\frac{f(\mathbf{x} + h \mathbf{e}_i) - f(\mathbf{x})}{h}\]

is the integral over the straight line segment from \(\mathbf{x}\) to \(\mathbf{x} + h \mathbf{e}_i\) divided by \(h\), so as \(h \to 0\) it converges to \(F_i(\mathbf{x})\); hence \(\partial f/\partial x_i = F_i\), that is, \(\nabla f = \mathbf{F}\).

Let us verify this in the following example.

Example 5 (Example of a conservative field) Let us integrate \(\mathbf{F} = (y, x)\) along the parabola \(\mathbf{r}(t) = (t, t^2)\) (\(0 \leq t \leq 1\)) from the point \((0,0)\) to \((1,1)\).

\[\mathbf{F}(\mathbf{r}(t)) = (t^2, t),\qquad \mathbf{r}'(t) = (1, 2t)\]

so

\[\mathbf{F}\cdot \mathbf{r}' = t^2 + 2t^2 = 3t^2\]

and therefore integrating this gives

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 3t^2\mathop{dt} = 1\]

Indeed, since \(\mathbf{F} = \nabla(xy)\), we can recover the above calculation by Theorem 3 (Fundamental theorem for line integrals) as the difference of the endpoint values of \(xy\), namely \(1\cdot 1 - 0\cdot 0 = 1\). This depends only on the endpoints; for instance, if we take \(\mathbf{r}(t)=(t,t)\) (\(0 \leq t \leq 1\)),

\[\mathbf{F}(\mathbf{r}(t))=(t,t),\qquad \mathbf{r}'(t)=(1,1)\]

so \(\mathbf{F}\cdot \mathbf{r}'=2t\), and we can verify that

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 2t\mathop{dt} = 1\]

Meanwhile, in §Vector Fields, ⁋Proposition 6 we saw that a conservative field has the necessary condition of being irrotational. Theorem 4 reveals, in the language of path independence, why this is not sufficient. Since being conservative is equivalent to the integral over every closed curve being \(0\), if there is even one example where the closed curve integral is not \(0\) despite being irrotational, then it is not conservative. Such an example actually arises when the domain has a hole; the following example is exactly that.

::: Example 6 Consider the vector field defined on the plane with the origin removed, \(\mathbb{R}^2 \setminus \{0\}\),

\[\mathbf{F} = \left(\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}\right)\]

Differentiating this directly,

\[\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} = \frac{y^2 - x^2}{(x^2+y^2)^2}\]

so this vector field is irrotational. However, traversing the unit circle \(\mathbf{r}(t) = (\cos t, \sin t)\) once, we have \(\mathbf{F}(\mathbf{r}(t)) = (-\sin t, \cos t) = \mathbf{r}'(t)\), so

\[\oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} (\sin^2 t + \cos^2 t)\mathop{dt} = 2\pi \neq 0\]

By Theorem 4, \(\mathbf{F}\) is not conservative on this region. The reason is that locally this vector field can be expressed as the gradient of the polar angle \(\theta = \arctan(y/x)\), but the polar angle increases by \(2\pi\) when going around the origin and thus cannot be defined as a single-valued function.