Line Integrals

A path through a current

Each gray arrow is the current at that point. Here it is the same everywhere: 1 m/s to the right, F = ⟨1, 0⟩.

The dots A and B are the ends of a straight path C, and a canoe sits partway along it. The dark arrow at the canoe is F. The thick arrow is the part of F that points along the path, F · T. It is green when the current runs with you and orange when it runs against you.

The line integral ∫_C F · dr adds up that along-the-path part over the whole trip.

Rests on A1 A2 A3 D1

Try itDrag A and B around the grid, or grab the path itself and slide it. When does the path turn orange? Can you make the answer exactly 0?

How the number is computed

The computation panel works it out the way you would by hand:

• write the path as r(t) = A + t(B − A) for 0 ≤ t ≤ 1,
• differentiate to get r′(t),
• plug the path into F, dot with r′(t), and integrate from 0 to 1.

The graph at the bottom of the panel is the integrand F(r(t)) · r′(t). Its signed area is the answer. The t slider moves the canoe, and the darker shading is the running total so far.

Try itSlide t from 0 to 1 and watch the running total under the graph build up to the final answer.

Fields that change from place to place

A constant field is the easy case. The field menu now has three more:

• Radial ⟨x, y⟩ points away from the origin and grows as you go out.
• Rotational ⟨−y, x⟩ swirls counterclockwise.
• Shear ⟨y, 0⟩ runs right above the x-axis and left below it.

Now the field changes along the path, so the integrand is no longer a constant. Part of a path can be green and part orange, and the pieces can cancel.

Try itTry each field in the menu. In Rotational, find a straight path with zero flow. Hint: aim it through the origin.

Bend the path

The path is now curved, and a square handle M has appeared. The path is the parabola

r(t) = (1 − t)²A + 2t(1 − t)M + t²B

which starts at A, ends at B, and is pulled toward M. With M at the midpoint of AB it is the straight line again. The dashed line is the straight path from A to B, kept for comparison.

Same endpoints, different route. Does the answer change? The blue box under the result compares the two.

Try itDrag M to bend the path in the rotational field. Then switch to Radial and bend it again. Which one cares about the route?

Along the path, or across it?

So far every number has been a flow: the part of F that runs along the path, ∫_C F · T ds, which is ∫_C F · dr written another way.

Switch Measure to Flux across and the tool adds up the part that crosses the path instead: ∫_C F · n ds, where n points to your right as you travel from A to B.

With F = ⟨P, Q⟩, the two integrands use the same pieces:

flow: P x′ + Q y′     flux: P y′ − Q x′

In flux mode, green means the current crosses from your left to your right.

Rests on D2

Try itSwitch between Flow along and Flux across. In the constant field, find a path with all flow and no flux, then one with all flux and no flow.

Which one ignores the route?

Back at stop 4, bending the path changed the flow in the rotational field but not in the radial one. Flux is the mirror image:

• Flow ignores the route when the field has no curl (no spin). Then F = ∇f and the flow is f(B) − f(A).
• Flux ignores the route when the field has no divergence (nothing spreads out or piles up). Whatever crosses one route between A and B must also cross the other.

The rotational field spins but does not spread, and the radial field spreads but does not spin.

Try itWith Flux across on, bend the path in Rotational: the flux holds still. Now try Radial.

Force or velocity?

Measure now has a third choice, Work. The computation is identical to flow. What the number means is not.

A river current is a velocity field, so its line integral is a flow, not work.

Rests on D1 D3

Try itSwitch between Flow along and Work. The number stays put; the wording and units change.

A whirlpool

This current swirls counterclockwise around the origin. It is still at the very center, fastest about 0.7 m out, and nearly gone a couple of meters away:

F = 3e^{−(x² + y²)}⟨−y, x⟩

After you plug in a path there is no tidy antiderivative, so step 5 switches to Simpson's rule. It runs twice, at two resolutions, and reports whether the answers agree.

A path that arcs over the top runs against the spin, and one that dips underneath runs with it. A path that stays far away barely notices.

Rests on A2 A5

Try itDrag M to bend the path over the whirlpool, then under it. Then slide the whole path far away.

A river channel

Now a river between two banks. The current is fastest down the middle and stops at the banks:

F = ⟨1 − y²/9, 0⟩

Paddle straight across and the current is perpendicular to you the whole way. The flow along your path is zero, but the flux across it is not: the whole river passes through your line.

The menu also has River with a whirlpool: this channel with an eddy turning near (0.5, −1).

Rests on A2 A4 A6 D2

Try itThe path starts straight across: zero flow. Drag it to run downstream through the middle, then along a bank. Put it back across the river and switch to Flux across: 4 m²/s, however you bend it between the banks. Then pick River with a whirlpool from the menu.

What the line integral leaves out

When F is a force, this tool computes the work that one force does along C. It adds up only the part of the force that points along your path. It is not a model of everything happening to the canoe.

Two other things are going on in a real crossing, and they are separate ideas:

• Drag from the water. Whenever the canoe moves relative to the water, the water pushes back on the hull. That push depends on how fast and which way the boat is moving, not only on where it is, so it is not a vector field of position like the ones here. It is what carries you downstream if you stop correcting.
• Your paddling. To hold a straight line across, you angle the bow upstream and keep pushing against the drift. That effort comes from you, not from the field.

So a straight crossing can have zero flow along the path and still wear you out. The current points across your path, not along it, so it adds nothing to the integral. You still burn energy fighting the drift, the same way holding a heavy box still is tiring even though you do no work on the box.

Work also depends on who measures it. To someone standing on the bank, you paddle straight across. To someone drifting on a raft with the current, you are paddling upstream at an angle, and they would measure a different amount of work.

Rests on A3 A7 D3

Try itThe path runs straight across in Work mode. Read F as a force that pushes downstream: it does zero work on this crossing. Your arms would disagree.

Every route, compared

Keep A and B fixed and bend the path. For the constant and radial fields, the flow never changes: they are gradients, so only the endpoints matter. For the rotational, shear, whirlpool and river fields, it does.

On the river, a route through the fast middle and a route along the slow bank get different answers.

Everything is unlocked now: every field, both path shapes, and flow, flux and work. Play as long as you like, then scroll on to finish.

Try itKeep A and B near the slow bank and pull M through the fast middle, then back toward the bank.