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.

We now examine the integral of a vector function. The first step is the line integral, which accumulates the contributions of the vector field at each point as we travel along a curve in the space \(\mathbb{R}^n\) where the field is defined. What is remarkable 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 Fundamental Theorem of Calculus.

Line Integral

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

\[\int_C f\dd{s} = \int_a^b f(\mathbf{r}(t))\lvert \mathbf{r}'(t)\rvert \dd{t}.\]

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

The above integral is independent of the parametrization of the curve, since its value is preserved under any \(C^1\) reparametrization by the change of variables formula. As a special case, if \(f \equiv 1\), then \(\int_C \dd{s}\) gives the length of the curve.

To extend 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 \(\mathbf{r}\colon [a, b] \rightarrow \mathbb{R}^n\) is

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

If \(\mathbf{F}\) represents a force field, the above integral gives the work done by that force on an object moving along the curve \(C\). This definition arises from the fact that the work done by a constant force \(\mathbf{F}\) through a displacement \(\mathbf{d}\) is \(\mathbf{F}\cdot \mathbf{d}\): we approximate the instantaneous displacement by \(\mathbf{r}'(t)\dd{t}\), take the dot product with the force at each instant, and sum these over the entire curve.

If \(\mathbf{r}\) is a regular curve, we may write it using the unit tangent vector \(\mathbf{T} = \mathbf{r}'/\lvert \mathbf{r}'\rvert\), and verify that

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (\mathbf{F}\cdot \mathbf{T})\dd{s}.\]

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

\[\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\dd{x} + Q\dd{y}\]

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

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 its two endpoints.

Proof

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

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

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

Theorem 4 Let \(\mathbf{F}\) be continuous on a connected open region \(D\). Then 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). Here, since \(\mathbf{F} = \nabla f\) is continuous, the potential \(f\) is automatically \(C^1\). \((3 \Leftrightarrow 2)\) follows from cutting a closed curve at a point into two paths, and noting that reversing one path yields a closed curve. The integral over this reversed path is the negative of the original integral by the change of variables \(t \mapsto a + b - t\), so the integral over a closed curve being zero is equivalent to the integrals over the two paths being equal.

Thus the key claim is \((3 \Rightarrow 1)\). For this, we must construct the 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 line integral of \(\mathbf{F}\) from \(\mathbf{x}_0\) to \(\mathbf{x}\). This would normally depend on the choice of curve \(\mathbf{r}\) joining \(\mathbf{x}_0\) and \(\mathbf{x}\), but since we are assuming the third condition, 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 \(h \rightarrow 0\) it converges to \(F_i(\mathbf{x})\), and therefore \(\partial f/\partial x_i = F_i\), 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 \((0,0)\) 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

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

Indeed, since \(\mathbf{F} = \nabla(xy)\), by Theorem 3 (Fundamental theorem for line integrals) we can recover the above computation by calculating 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 \(\mathbf{r}(t)=(t,t)\) (\(0 \leq t \leq 1\)), then

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

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

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

Meanwhile, in §Vector Fields, ⁋Proposition 6 we saw that a conservative field necessarily satisfies the irrotational condition. Theorem 4 reveals why this is not sufficient, in the language of path independence. Since being a conservative field is equivalent to the integral over every closed curve being zero, if there is even one closed curve whose integral is nonzero despite the field being irrotational, then the field is not conservative. Such examples actually arise when the domain has a hole, and the following example is precisely that.

Example 6 Consider the vector field 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

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

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

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.