Line Integrals of Vector Fields

Main idea

The work a force field does on you along a path is $\int_C\vec F\cdot d\vec r$ . Only the part of the force pointing along your motion counts: a tailwind helps, a headwind hurts and a crosswind does nothing.

\int_C\vec F\cdot d\vec r = \int_a^b\vec F\big(\vec r(t)\big)\cdot{\htmlData{sym=0}{\vec r\,'(t)}}\,dt = \int_C {\htmlData{sym=1}{P\,dx + Q\,dy}}

What the symbols mean

$\vec r\,'(t)$
Velocity along the path
Dotting the force with it keeps only the part of the force pointing along the motion: the part that does work.
$P\,dx + Q\,dy$
Component form
The same integral written with the field's components.

Explanation

A line integral of a vector field is simply the total push a field gives you as you travel a path. Picture biking a winding route in the wind. At each moment, the wind either helps (blowing the way you're going), hurts (blowing against you) or does nothing (blowing straight across). The integral $\int_C\vec F\cdot d\vec r$ adds up all that help and hurt along the route. It's made of familiar pieces: $\vec F$ (the force where you are), $d\vec r$ (a tiny step along the path) and the dot product between them (how much of the force lines up with the step).

Where it comes from. Think back to work in physics: a constant force $\vec F$ over a straight displacement $\Delta\vec r$ does work $\vec F\cdot\Delta\vec r$ . The dot product is what keeps only the part of the force along the motion. On a curved path in a changing field, chop the path into tiny nearly straight steps $d\vec r = \vec r\,'(t)\,dt$ , use the force at each step, and add. That's the integral $\int_a^b\vec F(\vec r(t))\cdot\vec r\,'(t)\,dt$ .

Recipe: parametrize $C$ , substitute into $\vec F$ , dot with $\vec r\,'(t)$ , integrate.

Direction matters. This is the big difference from the last lesson. Reversing the path flips the sign: $\int_{-C}\vec F\cdot d\vec r = -\int_C\vec F\cdot d\vec r$ . It makes sense: a tailwind on the way out is a headwind on the way back. A wire's mass doesn't care which way you go; work does.

Interpretation. For a force field it's work. For a velocity field around a closed loop, $\oint_C\vec F\cdot d\vec r$ is the circulation: how much the flow pushes you around the loop. The little circle on $\oint$ just means the path ends where it started.

A common mistake. It's tempting to use $\|\vec r\,'(t)\|$ here, as in the last lesson. Don't: the dot product with $\vec r\,'(t)$ already accounts for both direction and length. Throwing in the speed as well counts the length twice.

Notice that a closed loop can give nonzero work. Whether it does says something deep about the field, and the next few lessons are about exactly that.

See it

A sailboat runs a counterclockwise loop in a wind that blows east and gets stronger to the north, F = ⟨1 + 0.3y, 0⟩. As you can see, the wind helps along the bottom, where the boat heads east, and fights it along the top, where it heads west. The top has the stronger wind, so the loop nets about −2.41: back at the start, the wind has done negative work.

Worked examples

Example 1

Evaluate $\int_C\vec F\cdot d\vec r$ for $\vec F = \langle 8x^2yz, 5z, -4xy\rangle$ along $\vec r(t) = \langle t, t^2, t^3\rangle$ , $0 \le t \le 1$ .

1
Substitute $x = t$ , $y = t^2$ , $z = t^3$ into the field: $\vec F(\vec r(t)) = \langle 8t^7, 5t^3, -4t^3\rangle$ . The velocity is $\vec r\,' = \langle 1, 2t, 3t^2\rangle$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Evaluate $\int_C\vec F\cdot d\vec r$ for $\vec F = \langle y, x\rangle$ along $\vec r = \langle t, t\rangle$ , $0 \le t \le 1$ .

Problem 2+5 XP

Find the circulation of $\vec F = \langle -y, x\rangle$ counterclockwise around the unit circle.

Problem 3+5 XP

Evaluate the example: $\vec F = \langle 8x^2yz, 5z, -4xy\rangle$ along $\langle t, t^2, t^3\rangle$ .

Problem 4+5 XP

If $\int_C\vec F\cdot d\vec r = 5$ , what is the integral along $C$ in the reverse direction?

Problem 5+5 XP

Find the work done by $\vec F = \langle x^2, 0\rangle$ moving along the $x$ -axis from $0$ to $3$ .

Problem 6+5 XP

A wind pushes on a cart with a constant force $\vec F = \langle -2, -1\rangle$ N while the cart rolls straight from $(0, 0)$ to $(4, 3)$ (meters). How much work does the wind do, in joules?

Problem 7+5 XP

Evaluate $\int_C x\,dy - y\,dx$ along the upper half of the circle of radius 2, from $(2, 0)$ to $(-2, 0)$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the work done by $\vec F = \langle y^2, x\rangle$ from $(0, 0)$ to $(1, 1)$ along two paths: the line $y = x$ , and the parabola $y = x^2$ . Enter both, line first.

Where it's used

The energy a motor spends moving a robot arm through a magnetic field, the work done by wind on a sailboat along its course, and the voltage around a circuit loop are line integrals of vector fields.

Flying a Loop Through the WindCalc III · Robotics

Line Integrals of Functions The Fundamental Theorem for Line Integrals