6.2 Line Integrals

Big idea

A single integral adds up a function along an interval of the $x$-axis. A line integral does the same job along a curve that bends through the plane or through space. The domain is no longer a straight stretch of axis, so the little piece of domain is no longer $dx$ but $ds$, the element of arc length. Everything else about the definition is familiar: cut the curve into small pieces, sample the function on each piece, multiply by the length of the piece, add, and take a limit.

There are two different integrals under this one name, and keeping them apart is most of the work in this section. The first integrates a scalar function and asks how much of something is spread along the curve. Mass of a wire with variable density is the model case. The second integrates a vector field and asks how much the field pushes along the curve. Work done by a force is the model case, and it is the version the rest of the chapter is built on.

The mechanism in both cases is a parametrization. Choose $\mathbf{r}(t)$ tracing the curve for $a \le t \le b$, substitute it into the integrand, and convert the geometric piece into $t$. For arc length that conversion is $ds = \|\mathbf{r}'(t)\|\,dt$. For the vector version it is $d\mathbf{r} = \mathbf{r}'(t)\,dt$. Both turn a curve integral into an ordinary single integral in $t$, which you already know how to evaluate.

The two versions behave differently in one important respect. The scalar integral cannot tell which way you walked the curve, because arc length is positive no matter what. The vector integral can, because it measures a push in a direction, and walking backward turns every push into a resistance. That asymmetry is not a technicality; it is the reason orientation has to be specified in every statement in the rest of the chapter.

Decoder

Parametrize the curve, substitute the parametrization into the integrand, and replace the geometric element by its expression in the parameter: arc length becomes the speed times $dt$, and the displacement vector becomes the velocity times $dt$.

The word doing the work is “speed”. In $ds = \|\mathbf{r}'(t)\|\,dt$ the factor $\|\mathbf{r}'(t)\|$ is a length per unit parameter, so it corrects for the fact that the parameter may run through the curve quickly in some places and slowly in others. Forgetting that factor and integrating $f(\mathbf{r}(t))\,dt$ is the standard error; it computes an average weighted by parameter rather than by length.

For the vector version, $\mathbf{F} \cdot \mathbf{r}'(t)$ keeps only the component of the field that lies along the direction of travel. A field perpendicular to the curve contributes nothing at all, no matter how strong it is. Written with the unit tangent, the two integrals are visibly related: $\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (\mathbf{F}\cdot\mathbf{T})\,ds$, so the work integral is a scalar line integral of one particular scalar function, namely the tangential component of the field.

Neither integral depends on which parametrization you picked, as long as it traces the curve once. The scalar one does not even depend on direction. The vector one depends on direction only through a sign.

Definitions and results

Scalar line integral. For a scalar function $f$ and a smooth curve $C$ parametrized by $\mathbf{r}(t)$ on $[a,b]$, $$ \int_C f\,ds = \int_a^b f\big(\mathbf{r}(t)\big)\,\left\|\mathbf{r}'(t)\right\|\,dt $$ Taking $f = 1$ returns the arc length of $C$. Taking $f$ to be a linear density returns the mass of a wire shaped like $C$.

Line integral of a vector field. For a vector field $\mathbf{F}$ and the same curve, $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big)\cdot \mathbf{r}'(t)\,dt $$ If $\mathbf{F}$ is a force field and $C$ is the path of a particle, this number is the work done by the force on the particle.

Differential form. With $\mathbf{F} = \langle P, Q\rangle$ and $\mathbf{r}(t) = \langle x(t), y(t)\rangle$, expanding the dot product gives the notation used throughout the chapter: $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_C P\,dx + Q\,dy = \int_a^b \Big( P\,\frac{dx}{dt} + Q\,\frac{dy}{dt} \Big)\,dt $$ In space there is a third term $R\,dz$.

Orientation. An oriented curve carries a choice of direction of travel. Write $-C$ for the same curve traversed backward. Then $$ \int_{-C} \mathbf{F}\cdot d\mathbf{r} = -\int_C \mathbf{F}\cdot d\mathbf{r}, \qquad \int_{-C} f\,ds = \int_C f\,ds $$ The sign flips for the vector integral because the tangent direction reverses, and does not flip for the scalar one because $ds$ is a length.

Independence of parametrization. Any two parametrizations that trace $C$ once in the same direction give the same value for both integrals. The speed factor is exactly what cancels the difference between them.

Piecewise smooth curves. If $C$ is built from finitely many smooth pieces joined end to end, integrate over each piece and add. Corners cause no trouble as long as each piece is smooth.

Closed curves. A curve whose endpoints coincide is closed, and the integral over it is often written $\oint_C$. A curve that does not cross itself is simple. The circulation of $\mathbf{F}$ around a closed curve is $\oint_C \mathbf{F}\cdot d\mathbf{r}$.

Tangential form. Since $\mathbf{r}'(t) = \|\mathbf{r}'(t)\|\,\mathbf{T}(t)$, the two definitions are linked by $\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (\mathbf{F}\cdot\mathbf{T})\,ds$.

Worked examples

Mass of a straight wire

Compute $\int_C (x+y)\,ds$ where $C$ is the segment from $(0,0)$ to $(1,2)$.

Parametrize with $\mathbf{r}(t) = \langle t, 2t\rangle$ for $0 \le t \le 1$. Then $\mathbf{r}'(t) = \langle 1, 2\rangle$ and $\|\mathbf{r}'(t)\| = \sqrt{5}$, a constant, since the segment is traced at constant speed. The integrand becomes $t + 2t = 3t$, so $$ \int_C (x+y)\,ds = \int_0^1 3t\,\sqrt{5}\,dt = \sqrt{5}\left[\frac{3t^2}{2}\right]_0^1 = \frac{3\sqrt{5}}{2} $$ Check the size: the segment has length $\sqrt{5}$ and $x+y$ runs from $0$ to $3$ linearly, so its average along the wire is $3/2$. Length times average is $\frac{3\sqrt5}{2}$, which agrees.

Work along a parabola

Compute $\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F}(x,y) = \langle xy, \; y^2\rangle$ along the arc of $y = x^2$ from $(0,0)$ to $(1,1)$.

Use $\mathbf{r}(t) = \langle t, t^2\rangle$ on $[0,1]$, so $\mathbf{r}'(t) = \langle 1, 2t\rangle$. On the curve, $\mathbf{F}(\mathbf{r}(t)) = \langle t\cdot t^2, \; (t^2)^2\rangle = \langle t^3, t^4\rangle$. The dot product is $t^3 + 2t^5$, so $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 \left(t^3 + 2t^5\right)dt = \frac{1}{4} + \frac{2}{6} = \frac{7}{12} $$

What reversing the curve does

Take the same field and the same arc, now traversed from $(1,1)$ back to $(0,0)$. Parametrize by $\mathbf{s}(u) = \langle 1-u, (1-u)^2\rangle$ for $0 \le u \le 1$, so $\mathbf{s}'(u) = \langle -1, \; -2(1-u)\rangle$. Writing $w = 1-u$, the field on the curve is $\langle w^3, w^4\rangle$ and the dot product is $-w^3 - 2w^5$. Substituting $w = 1-u$ and changing variables, $$ \int_{-C} \mathbf{F}\cdot d\mathbf{r} = \int_0^1 \left(-w^3 - 2w^5\right)dw = -\frac{7}{12} $$ The sign flipped and nothing else changed. Now run the scalar integral both ways: $\int_C 1\,ds$ and $\int_{-C} 1\,ds$ both equal the length of the arc, a positive number. The speed factor $\|\mathbf{r}'\|$ is a magnitude, so it cannot detect direction.

Circulation around a circle

Compute $\oint_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle -y, x\rangle$ around the unit circle traversed counterclockwise.

Use $\mathbf{r}(t) = \langle \cos t, \sin t\rangle$ on $[0, 2\pi]$, so $\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$ and $\mathbf{F}(\mathbf{r}(t)) = \langle -\sin t, \cos t\rangle$. The two are identical, so their dot product is $\sin^2 t + \cos^2 t = 1$ and $$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} 1\,dt = 2\pi $$ That is not an accident: this field is tangent to the circle everywhere, with magnitude $1$ on the unit circle, so the integrand had to be $1$ and the answer had to be the circumference. A nonzero value around a closed loop means the field genuinely drives circulation, and the next two sections are about measuring exactly that.

Practice

Begin with the scalar version. These ask for integrals with respect to arc length along segments, circles and simple space curves, including arc length and wire mass.

Then the vector version. Here you compute work and circulation, handle the $P\,dx + Q\,dy$ notation, and track the sign that orientation controls.

Quiz

Five items on arc length integrals, work integrals, parametrizing a curve, and the effect of reversing orientation.