Work, Circulation, and Vector Line Integrals, multivariable and vector calculus lesson

Push a shopping cart across a parking lot while the wind blows. Where the wind pushes along your direction of travel it helps you, where it pushes against you it costs you effort, and where it blows straight across your path it does nothing for or against you. Adding up that helpful or harmful part of the wind along the whole route is exactly a vector line integral.

Key terms

Oriented curves and their parametrizations

Definition An oriented curve is a piecewise regular curve $C$ together with a chosen direction of travel, from an initial point to a terminal point, or a chosen sense of travel if $C$ is closed. A regular parametrization $\mathbf{r}(t)$ , $a \le t \le b$ , of a regular piece is compatible with the orientation when $\mathbf{r}(t)$ moves in the chosen direction as $t$ increases. On such a piece the unit tangent vector in the chosen direction is

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lvert \mathbf{r}'(t) \rvert}.

In plain words The same set of points can be walked two ways. An orientation picks one, and a compatible parametrization is a way of walking it in that direction.

Example The segment from $(1, 0)$ to $(0, 1)$ has the compatible parametrization $\mathbf{r}(t) = \langle 1 - t, t \rangle$ , $0 \le t \le 1$ , with unit tangent

\mathbf{T} = \frac{1}{\sqrt{2}} \langle -1, 1 \rangle.

The parametrization $\langle t, 1 - t \rangle$ , $0 \le t \le 1$ , traces the same points from $(0, 1)$ to $(1, 0)$ , so it belongs to the opposite orientation.

Vector line integral

Definition Let $C$ be an oriented regular curve with compatible parametrization $\mathbf{r}(t)$ , $a \le t \le b$ , and let $\mathbf{F}$ be a vector field that is continuous on $C$ . The line integral of $\mathbf{F}$ along $C$ is the scalar line integral of the tangential component $\mathbf{F} \cdot \mathbf{T}$ ,

\begin{aligned} \int_C \mathbf{F} \cdot d\mathbf{r} &= \int_C \mathbf{F} \cdot \mathbf{T} \, ds \\[0.4em] &= \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt. \end{aligned}

The second form holds because the speeds cancel in $\mathbf{T} \, ds = \mathbf{r}'(t) \, dt$ . For an oriented piecewise regular curve $C = C_1 \cup \cdots \cup C_m$ , each piece carrying the orientation of $C$ , the integral is the sum of the integrals over the pieces. Any two compatible regular parametrizations that trace each piece exactly once give the same value.

In plain words At each point keep only the part of the field that points along the curve, with a sign, and add it up along the length. A field perpendicular to the path contributes nothing.

Example Let $\mathbf{F}(x, y) = \langle y, x^2 \rangle$ and let $C$ be the parabola $y = x^2$ from $(0, 0)$ to $(1, 1)$ , with $\mathbf{r}(t) = \langle t, t^2 \rangle$ , $0 \le t \le 1$ . Substitute to get $\mathbf{F}(\mathbf{r}(t)) = \langle t^2, t^2 \rangle$ and $\mathbf{r}'(t) = \langle 1, 2t \rangle$ . Then

\begin{aligned} \int_C \mathbf{F} \cdot d\mathbf{r} &= \int_0^1 (t^2 + 2t^3) \, dt \\[0.4em] &= \frac{1}{3} + \frac{1}{2} = \frac{5}{6}. \end{aligned}

The differential dr

Definition For a parametrization $\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle$ the vector differential is

d\mathbf{r} = \mathbf{r}'(t) \, dt = \langle dx, dy, dz \rangle,

where $dx = x'(t) \, dt$ , $dy = y'(t) \, dt$ and $dz = z'(t) \, dt$ . Its length is $\lvert d\mathbf{r} \rvert = \lvert \mathbf{r}'(t) \rvert \, dt = ds$ when $dt > 0$ , and its direction is the direction of travel, so $d\mathbf{r} = \mathbf{T} \, ds$ .

In plain words Think of $d\mathbf{r}$ as a tiny arrow along the curve, pointing the way you walk, whose length is the tiny distance $ds$ .

Example On the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ ,

d\mathbf{r} = \langle -\sin t, \cos t, 1 \rangle \, dt.

At $t = 0$ this arrow points along $\langle 0, 1, 1 \rangle$ , and its length is $\sqrt{2} \, dt = ds$ .

Coordinate forms

Definition Writing $\mathbf{F} = \langle P, Q, R \rangle$ with $P$ , $Q$ , $R$ continuous on $C$ , the line integral in differential form is

\begin{aligned} &\int_C P \, dx + Q \, dy + R \, dz \\[0.4em] &\quad = \int_a^b \big( P x'(t) + Q y'(t) \\[0.4em] &\qquad\quad + R z'(t) \big) \, dt, \end{aligned}

with $P$ , $Q$ , $R$ evaluated at $\mathbf{r}(t)$ . In the plane drop the $R$ term. A single piece such as $\int_C P \, dx$ is the case $\mathbf{F} = \langle P, 0, 0 \rangle$ .

In plain words Replace $x$ , $y$ , $z$ by their formulas in $t$ everywhere, and replace $dx$ , $dy$ , $dz$ by the derivatives times $dt$ . Then it is a one-variable integral.

Example Compute $\int_C x^2 \, dx + xy \, dy$ along the segment from $(0, 0)$ to $(1, 2)$ . Take $x = t$ , $y = 2t$ , $0 \le t \le 1$ , so $dx = dt$ and $dy = 2 \, dt$ . The integrand becomes $t^2 + 2t^2 \cdot 2 = 5t^2$ , and

\int_0^1 5t^2 \, dt = \frac{5}{3}.

Work

Definition If $\mathbf{F}$ is a continuous force field and an object moves along an oriented piecewise regular curve $C$ , the work done by $\mathbf{F}$ on the object is

W = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C \mathbf{F} \cdot \mathbf{T} \, ds.

Each small step contributes $\mathbf{F} \cdot \mathbf{T} \, \Delta s$ , which is positive where the force has a component along the motion, negative where it opposes the motion and zero where it is perpendicular. In general the value depends on the path, not only on its endpoints.

In plain words Work is force along the path times distance, added up with signs. Two routes between the same points can cost different amounts.

Example Let $\mathbf{F}(x, y) = \langle -y, x \rangle$ , a field that circles the origin counterclockwise. Along the segment from $(1, 0)$ to $(0, 1)$ with $\mathbf{r}(t) = \langle 1 - t, t \rangle$ ,

\begin{aligned} &\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \\[0.4em] &\quad = \langle -t, 1 - t \rangle \cdot \langle -1, 1 \rangle \\[0.4em] &\quad = t + 1 - t = 1, \end{aligned}

so $W = \int_0^1 1 \, dt = 1$ . Along the quarter circle $\langle \cos t, \sin t \rangle$ , $0 \le t \le \pi/2$ , between the same points, the integrand is $\sin^2 t + \cos^2 t = 1$ and $W = \pi/2$ . Both are positive since the field pushes along both paths, but the values differ, so the work depends on the route.

Circulation

Definition If $C$ is an oriented closed piecewise regular curve, the line integral of $\mathbf{F}$ around $C$ is called the circulation of $\mathbf{F}$ around $C$ and is written

\oint_C \mathbf{F} \cdot d\mathbf{r}.

For a simple closed curve in the plane, positive orientation means counterclockwise, with the enclosed region on the left.

In plain words Circulation measures how much the field pushes you around the loop overall. A positive value means the field tends to carry you in the chosen direction.

Example For $\mathbf{F} = \langle -y, x \rangle$ around the unit circle counterclockwise, $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ , $0 \le t \le 2\pi$ , the integrand is $\sin^2 t + \cos^2 t = 1$ , so

\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 1 \, dt = 2\pi.

The field points along the circle at every point, so every bit contributes positively. Clockwise the circulation is $-2\pi$ .

Reversing orientation

Definition Let $-C$ denote $C$ with the opposite orientation. For $\mathbf{F}$ continuous on the oriented piecewise regular curve $C$ ,

\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}.

On a regular piece, the reversed parametrization $\mathbf{r}^-(t) = \mathbf{r}(a + b - t)$ has velocity $-\mathbf{r}'(a + b - t)$ , and the substitution $u = a + b - t$ turns the integral into the negative of the original. Equivalently, $\mathbf{T}$ changes sign while $ds$ does not. The same holds for each coordinate form, for example $\int_{-C} P \, dx = -\int_C P \, dx$ .

In plain words Walk the path backwards and every tangent arrow flips, so help becomes hindrance and the total changes sign. Scalar integrals with $ds$ do not do this.

Example Reverse the parabola from the vector line integral example. With $\mathbf{r}^-(t) = \langle 1 - t, (1 - t)^2 \rangle$ , $0 \le t \le 1$ , going from $(1, 1)$ to $(0, 0)$ , the substitution gives

\begin{aligned} &\int_0^1 \big( -(1 - t)^2 - 2(1 - t)^3 \big) \, dt \\[0.4em] &\quad = -\frac{1}{3} - \frac{1}{2} = -\frac{5}{6}, \end{aligned}

the negative of $5/6$ , as predicted.

Check yourself

Find the work done by $\mathbf{F}(x, y) = \langle x + y, y \rangle$ on an object that moves from $(0, 0)$ to $(1, 0)$ and then straight up to $(1, 1)$ .

Show the answer

On the first piece, $\mathbf{r}(t) = \langle t, 0 \rangle$ , the integrand is $\langle t, 0 \rangle \cdot \langle 1, 0 \rangle = t$ , giving $1/2$ . On the second, $\mathbf{r}(t) = \langle 1, t \rangle$ , the integrand is $\langle 1 + t, t \rangle \cdot \langle 0, 1 \rangle = t$ , giving $1/2$ . The total work is $1$ .

Find the circulation of $\mathbf{F}(x, y) = \langle y, 0 \rangle$ around the unit circle traversed counterclockwise. Explain the sign.

Show the answer

With $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ the integrand is $\sin t \cdot (-\sin t) = -\sin^2 t$ , so

\oint_C \mathbf{F} \cdot d\mathbf{r} = -\int_0^{2\pi} \sin^2 t \, dt = -\pi.

The field points right above the $x$ -axis and left below it, which is clockwise, against the direction of travel.

The work done by a field $\mathbf{F}$ along an oriented curve $C$ is $7$ . What is the work along $-C$ , and what happens to $\int_C \lvert \mathbf{F} \rvert \, ds$ when the orientation is reversed?

Show the answer

The work along $-C$ is $-7$ , because vector line integrals change sign with the orientation. The integral of $\lvert \mathbf{F} \rvert$ with respect to arc length is a scalar line integral, so it does not change.

Work, Circulation, and Vector Line Integrals

Key terms

Oriented curves and their parametrizations

Vector line integral

The differential dr

Coordinate forms

Work

Circulation

Reversing orientation

Check yourself

How this lesson connects

Resources for this lesson

Vector field explorer