Line Integrals of Scalar Functions
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 1, 2, 3) |
Quick Reference
Line integral of $f$ over curve $C$ parametrized by $\mathbf{r}(t)$, $a \leq t \leq b$: $$\int_C f\,ds = \int_a^b f(\mathbf{r}(t))|\mathbf{r}'(t)|\,dt.$$
Special case $f = 1$: $\int_C 1\,ds = \int_a^b|\mathbf{r}'(t)|\,dt = $ arc length of $C$.
In 2D with $ds$: $ds = \sqrt{(x')^2+(y')^2}\,dt$.
Motivation
A line integral generalizes ordinary integration to integration along a curve. If $f(x,y)$ is the linear density of a wire and $C$ is its shape, then $\int_C f\,ds$ gives the total mass of the wire. If $f = 1$, the integral returns the length of the wire. The “integral along the curve” is computed by parametrizing $C$ and using $ds = |\mathbf{r}'(t)|\,dt$ to convert the arc-length element.
Key Concept
The parametrization does not affect the value. Two different parametrizations of the same curve give the same value for $\int_C f\,ds$ (as long as they traverse the curve the same direction, for appropriate curves). This is because $|\mathbf{r}'(t)|\,dt = ds$ always measures actual arc-length, regardless of the speed of parametrization.
Worked Example
Evaluate $\int_C x y^4\,ds$ where $C$ is the right half of the circle $x^2+y^2 = 1$ from $(0,-1)$ to $(0,1)$. (Stewart 16.2, Example 2.)
Parametrize: $\mathbf{r}(t) = \langle \cos t, \sin t\rangle$, $t \in [-\pi/2, \pi/2]$ (right half of the unit circle).
$\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$, $|\mathbf{r}'(t)| = 1$.
$f(\mathbf{r}(t)) = \cos t\cdot\sin^4 t$.
$$\int_C xy^4\,ds = \int_{-\pi/2}^{\pi/2}\cos t\sin^4 t\,dt = \left[\frac{\sin^5 t}{5}\right]_{-\pi/2}^{\pi/2} = \frac{1}{5} - \left(-\frac{1}{5}\right) = \frac{2}{5}.$$
$\int_C f\,ds$ depends on the speed of the parametrization. The line integral $\int_C f\,ds$ is independent of the parametrization (for the same curve and direction), because $ds = |\mathbf{r}'|\,dt$ automatically accounts for speed. If you traverse the curve faster (larger $|\mathbf{r}'|$) but over a shorter $t$-interval, the product $|\mathbf{r}'|\,dt$ remains the same. Only the line integral $\int_C \mathbf{F}\cdot d\mathbf{r}$ (next skill) is sensitive to orientation.
Common Misconceptions
the line integral $\int_C f\,ds$ measures the area under the curve $f$ along the $x$-axis, just like a regular definite integral.
This is the input-output-confusion error. The ordinary integral $\int_a^b f(x)\,dx$ accumulates values of $f$ along the $x$-axis. The line integral $\int_C f\,ds$ accumulates values of $f$ along an arbitrary curve $C$ in space, using the arc-length element $ds = |\mathbf{r}'(t)|\,dt$. For a vertical curve or a curve in the $xy$-plane, the two objects differ in domain, element of integration, and geometric meaning: $\int_C f\,ds$ gives the mass of a wire shaped like $C$ with linear density $f$, not an area under a graph.
Leveled Practice
Problem 1. Find the mass of a wire in the shape of $y = x^2$ from $(0,0)$ to $(1,1)$ with linear density $\rho(x,y) = y$.
Show answer
Parametrize: $x = t$, $y = t^2$, $t \in [0,1]$. $\mathbf{r}'(t) = \langle 1, 2t\rangle$, $|\mathbf{r}'| = \sqrt{1+4t^2}$.
$\rho(\mathbf{r}(t)) = t^2$.
$m = \int_0^1 t^2\sqrt{1+4t^2}\,dt$.
(This integral requires substitution $u = 1+4t^2$ and is best left in setup form for this exercise.)