Line Integrals

Line integrals are the computational backbone of classical mechanics and electromagnetism: the work done by a force field, the circulation of a fluid velocity field around a loop, and the electromotive force induced around a circuit (via $\oint \mathbf{E}\cdot d\mathbf{r}$ in Faraday's law) are all line integrals. In complex analysis, the contour integral $\oint_C f(z)\,dz$ is precisely a vector line integral in disguise (treating $\mathbb{C} \cong \mathbb{R}^2$), and Cauchy's integral theorem is the complex-analytic cousin of path-independence results that generalize Conservative Fields and Potentials. In computer graphics and geometric modeling, arc-length line integrals are used to reparametrize curves by arc length for smooth animation and to compute centers of mass of wireframe models. In statistics and probability, line integrals appear in the theory of stochastic integrals (Itô calculus) as limiting objects along random paths. In engineering, line integrals compute the work done moving a component along a specified track, and appear in the finite element method along boundary curves. The path toward Green's Theorem converts a line integral around a closed planar curve into a double integral over the enclosed region — the first deep bridge between line integrals and area integrals — and the question of when $\int_C \mathbf{F}\cdot d\mathbf{r}$ depends only on endpoints (not the path) is exactly the theory of Conservative Fields and Potentials, built directly on the machinery developed here.

:::quiz[For $\mathbf{r}(t) = (\cos t, \sin t)$, $t\in[0,2\pi]$, what is $\|\mathbf{r}'(t)\|$?]

• $\cos t$
• $1$ ✓
• $t$
• $\sin t + \cos t$

Ask the tutor: How does the line integral $\oint_C \mathbf{F}\cdot d\mathbf{r}$ around a closed curve relate to a double integral over the region it encloses?

Think of walking along a wire bent into the shape of $C$, carrying a scale that reads the value of $f$ at your current location. The scalar line integral $\int_C f\, ds$ is what you'd get by chopping the wire into tiny arcs of length $\Delta s$, multiplying each by the reading $f$ there, and summing — exactly the Riemann-sum philosophy of the ordinary integral, but with "$dx$" replaced by the arc-length element "$ds$." The factor $\|\mathbf{r}'(t)\|$ converts a small change $dt$ in the parameter into the actual length $ds$ traveled on the curve; this is the same "speed" idea used to compute arc length in single-variable calculus. The vector line integral tells a different story: it measures how much a field pushes along the direction of travel. If $\mathbf{F}$ is a force field and you move along $C$, only the component of $\mathbf{F}$ tangent to your motion does work on you; a force perpendicular to your path does nothing. This is exactly the content of the dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$: it isolates the tangential component of $\mathbf{F}$ and weights it by speed. Reverse the direction of travel, and the sign flips — walking against the wind takes different work than walking with it.

Example 1 (scalar line integral: mass of a wire on an asymmetric arc). A wire is bent into the curve

and has linear density $\delta(x,y) = 1 + y$ (mass per unit length). Find the total mass of the wire. Solution. Mass is $\displaystyle M = \int_C \delta\,ds = \int_0^2 \delta(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt$. Compute $\mathbf{r}'(t) = (2t,\, 2t^2)$, so

for $t \in [0,2]$ (where $t \geq 0$, so $|t| = t$). Next, $\delta(\mathbf{r}(t)) = 1 + \tfrac{2}{3}t^3$. So

Split into two integrals: $M = M_1 + M_2$ where

Computing $M_1$: substitute $w = 1+t^2$, $dw = 2t\,dt$; when $t=0$, $w=1$; when $t=2$, $w=5$:

Numerically $5\sqrt5 \approx 11.1803$, so $M_1 \approx \tfrac{2}{3}(10.1803) \approx 6.7869$.

Computing $M_2$: this integral does not reduce to an elementary closed form via elementary substitution alone in a short expression, so we evaluate it directly using the standard reduction for $\int t^4\sqrt{1+t^2}\,dt$ via hyperbolic substitution $t = \sinh\theta$, $dt = \cosh\theta\,d\theta$, $\sqrt{1+t^2} = \cosh\theta$:

Rather than carry the antiderivative symbolically, we evaluate $M_2$ numerically (Simpson's rule with $n=8$ subintervals on $[0,2]$, step $h = 0.25$), which is standard practice once a closed form is unwieldy. Let $\psi(t) = t^4\sqrt{1+t^2}$:

Simpson's rule:

Compute the odd-index sum: $\psi_1+\psi_3+\psi_5+\psi_7 = 0.00394 + 0.3536 + 2.6432 + 9.2049 = 12.2056$, times $4$: $48.8225$. Even-index (interior) sum: $\psi_2+\psi_4+\psi_6 = 0.0699 + 1.1180 + 5.2202 = 6.4081$, times $2$: $12.8162$. Endpoints: $\psi_0+\psi_8 = 0 + 14.7196 = 14.7196$. Total bracket: $14.7196 + 48.8225 + 12.8162 = 76.3583$. Multiply by $h/3 = 0.25/3 = 0.08333$:

So $M_2 \approx \tfrac{4}{3}(6.3632) \approx 8.4843$.

Total mass:

This example illustrates that setting up a line integral is always mechanical (parametrize, compute $\|\mathbf{r}'(t)\|$, substitute, integrate), but evaluation may require numerical methods once the algebra is not friendly — a completely typical and expected feature of real problems.

Example 2 (vector line integral: work along a helical-then-linear path). Let $\mathbf{F}(x,y,z) = (y, -x, z^2)$, a vector field on $\mathbb{R}^3$. Let $C$ be the piecewise-smooth curve consisting of:

• $C_1$: one quarter turn of a helix, $\mathbf{r}_1(t) = (\cos t, \sin t, t)$ for $t \in [0, \pi/2]$;
• $C_2$: the straight segment from $\mathbf{r}_1(\pi/2) = (0,1,\pi/2)$ to $(3, 1, \pi/2)$.

Compute $\displaystyle \int_C \mathbf{F}\cdot d\mathbf{r}$. Solution. By definition, $\int_C \mathbf{F}\cdot d\mathbf{r} = \int_{C_1}\mathbf{F}\cdot d\mathbf{r} + \int_{C_2}\mathbf{F}\cdot d\mathbf{r}$.

On $C_1$: $\mathbf{r}_1'(t) = (-\sin t, \cos t, 1)$. Evaluate $\mathbf{F}$ along the path:

Dot product:

So

Numerically, $(\pi/2)^3 = (1.5708)^3 \approx 3.8758$, so $\tfrac{1}{3}(3.8758) \approx 1.2919$, and $\pi/2 \approx 1.5708$. Thus

On $C_2$: parametrize the segment from $(0,1,\pi/2)$ to $(3,1,\pi/2)$ linearly:

Here $y$ and $z$ are constant along $C_2$. Evaluate $\mathbf{F}$:

Dot product with $\mathbf{r}_2'(s) = (3,0,0)$:

So

Total:

This example shows the two essential mechanics of vector line integrals in action: the trigonometric identity $\sin^2 t + \cos^2 t = 1$ collapsing the helix computation, and the fact that only the tangential direction of $\mathbf{r}_2'$ (purely in $x$) survives the dot product on the straight piece, silently discarding the $y$- and $z$-components of $\mathbf{F}$ entirely.