Line Integrals of Vector Fields
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.2: “Line Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-2-line-integrals |
| Textbook used in class | Stewart, Calculus, Section 16.2: “Line Integrals” (Examples 5, 6) |
Quick Reference
Line integral of $\mathbf{F}$ over $C$: $$\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.$$
Equivalent notation: $\int_C P\,dx + Q\,dy$ (in 2D) $= \int_a^b [P(x(t),y(t))x'(t) + Q(x(t),y(t))y'(t)]\,dt$.
Physical meaning: $\int_C \mathbf{F}\cdot d\mathbf{r}$ is the work done by the force field $\mathbf{F}$ in moving a particle along $C$.
Orientation: $\int_{-C}\mathbf{F}\cdot d\mathbf{r} = -\int_C\mathbf{F}\cdot d\mathbf{r}$ (reversing path negates).
Motivation
Work in physics is force times displacement when the force is constant and motion is linear. For a variable force field $\mathbf{F}$ and a curved path $C$, the work is the line integral $\int_C\mathbf{F}\cdot d\mathbf{r}$. The dot product $\mathbf{F}\cdot d\mathbf{r}$ picks out the component of force in the direction of motion, and the integral accumulates this over the entire path.
Key Concepts
1. The Formula
Substituting $\mathbf{r}(t) = \langle x(t), y(t), z(t)\rangle$: $$d\mathbf{r} = \mathbf{r}'(t)\,dt = \langle x'(t), y'(t), z'(t)\rangle\,dt.$$
$$\int_C\mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.$$
This is an ordinary single-variable integral in $t$.
2. Difference from $\int_C f\,ds$
The scalar line integral uses $|\mathbf{r}'(t)|$ (the speed) and is independent of orientation. The vector line integral uses $\mathbf{r}'(t)$ (the velocity, including direction) and depends on orientation. Reversing the direction of traversal flips the sign.
Worked Example
Evaluate $\int_C \mathbf{F}\cdot d\mathbf{r}$ where $\mathbf{F}(x,y,z) = \langle xy, y^2, z\rangle$ and $C$ is parametrized by $\mathbf{r}(t) = \langle t, t^2, t^3\rangle$ for $0 \leq t \leq 1$. (Stewart 16.2, Example 5.)
$\mathbf{r}'(t) = \langle 1, 2t, 3t^2\rangle$.
$\mathbf{F}(\mathbf{r}(t)) = \langle t\cdot t^2, t^4, t^3\rangle = \langle t^3, t^4, t^3\rangle$.
$\mathbf{F}\cdot\mathbf{r}' = t^3\cdot 1 + t^4\cdot 2t + t^3\cdot 3t^2 = t^3 + 2t^5 + 3t^5 = t^3 + 5t^5$.
$$\int_0^1(t^3+5t^5)\,dt = \left[\frac{t^4}{4}+\frac{5t^6}{6}\right]_0^1 = \frac{1}{4}+\frac{5}{6} = \frac{3}{12}+\frac{10}{12} = \frac{13}{12}.$$
$\int_C\mathbf{F}\cdot d\mathbf{r}$ and $\int_C f\,ds$ are computed the same way. The vector line integral uses $\mathbf{r}'(t)$ (not $|\mathbf{r}'(t)|$): you dot $\mathbf{F}$ with the velocity vector. The scalar line integral uses the speed $|\mathbf{r}'(t)|$: you multiply $f$ by the arc-length element. The dot product in the vector case introduces a sign that depends on whether $\mathbf{F}$ points with or against the direction of motion.
Common Misconceptions
$\int_C \mathbf{F}\cdot d\mathbf{r}$ measures the total magnitude of the force field along the curve.
This is the input-output-confusion error. The line integral $\int_C \mathbf{F}\cdot d\mathbf{r}$ measures work: it accumulates the component of $\mathbf{F}$ in the direction of motion, not the magnitude $|\mathbf{F}|$. A force perpendicular to the curve contributes zero to the integral even if $|\mathbf{F}|$ is large. For example, a constant field $\mathbf{F} = \langle 0, 1\rangle$ does zero work along the horizontal segment from $(0,0)$ to $(1,0)$, because the force is always perpendicular to the displacement.
Leveled Practice
Problem 1. Evaluate $\int_C y\,dx + x\,dy$ where $C$ is the line segment from $(0,0)$ to $(2,4)$.
Show answer
Parametrize: $x = 2t$, $y = 4t$, $t \in [0,1]$. $dx = 2\,dt$, $dy = 4\,dt$.
$\int_0^1(4t)(2) + (2t)(4)\,dt = \int_0^1(8t+8t)\,dt = \int_0^1 16t\,dt = 8$.