Line Integrals

An ordinary integral \int_a^b f\,dx sums a function along a straight stretch of the x-axis. A line integral does the same thing along a curve C winding through the plane. There are two flavours, and the difference is everything.

For a scalar field f(x, y) we sum its values against arc length ds — think of laying a wire along C whose density at each point is f, and weighing it:

\int_C f \, ds.

For a vector field field \mathbf{F} we sum the part of \mathbf{F} that points along the curve — its tangential component — against displacement d\mathbf{r}. This is the work the field does on a particle dragged along C:

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

Reducing a vector line integral to an ordinary integral

The trick that makes line integrals computable is parametrisation: describe the curve by a moving point \mathbf{r}(t) as t runs over an interval, and everything collapses to a single integral in t. Here is the reduction, line by line.

Step 1 — parametrise the curve. Write the curve as

\mathbf{r}(t) = \big(x(t),\; y(t)\big), \qquad a \le t \le b,

a single point sliding from the start \mathbf{r}(a) to the end \mathbf{r}(b).

Step 2 — express the displacement. A tiny step along the curve is the velocity times dt:

d\mathbf{r} = \mathbf{r}'(t)\, dt = \big(x'(t),\; y'(t)\big)\, dt.

Step 3 — substitute into the integral. Replace \mathbf{F} by its value on the curve and d\mathbf{r} by the expression above:

\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big) \cdot \mathbf{r}'(t)\, dt.

Step 4 — expand the dot product. With \mathbf{F} = (P, Q) the integrand is an ordinary scalar function of t:

= \int_a^b \Big[\, P\big(x(t), y(t)\big)\, x'(t) + Q\big(x(t), y(t)\big)\, y'(t) \,\Big]\, dt.

The vector line integral has become a plain single-variable integral — one you already know how to do.

Worked example: work along a path

Let the field be \mathbf{F}(x, y) = (y,\; x) and let C be the quarter-circle from (1, 0) to (0, 1). Find the work \int_C \mathbf{F}\cdot d\mathbf{r}.

Step 1 — parametrise. The unit circle is \mathbf{r}(t) = (\cos t,\; \sin t), and t runs from 0 (the point (1,0)) to \tfrac{\pi}{2} (the point (0,1)).

Step 2 — velocity.

\mathbf{r}'(t) = (-\sin t,\; \cos t).

Step 3 — field on the curve. Substitute x = \cos t, y = \sin t into \mathbf{F} = (y, x):

\mathbf{F}\big(\mathbf{r}(t)\big) = (\sin t,\; \cos t).

Step 4 — dot product.

\mathbf{F}\cdot\mathbf{r}' = (\sin t)(-\sin t) + (\cos t)(\cos t) = \cos^2 t - \sin^2 t = \cos 2t.

Step 5 — integrate.

\int_0^{\pi/2} \cos 2t \, dt = \left[ \tfrac12 \sin 2t \right]_0^{\pi/2} = \tfrac12(\sin\pi - \sin 0) = 0.

The total work is 0: along this path the field pushes forward as much as it pushes back. (As it happens \mathbf{F} = (y, x) = \nabla(xy) is a gradient field, and that is no accident — the next page explains why such fields care only about the endpoints.)

Scalar line integral. Along a curve \mathbf{r}(t), a \le t \le b, \int_C f \, ds = \int_a^b f\big(\mathbf{r}(t)\big)\,\big\|\mathbf{r}'(t)\big\|\, dt, with the arc-length element ds = \|\mathbf{r}'(t)\|\,dt. It is independent of the direction of travel.
Vector line integral (work). \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t)\, dt = \int_a^b \big(P\,x' + Q\,y'\big)\, dt. It reverses sign if you traverse C backwards.
In general the value depends on the path C, not just on its endpoints — two routes between the same two points can give different work.

The vector line integral is the mathematician's definition of work. A force \mathbf{F} dragging an object along C does work W = \int_C \mathbf{F}\cdot d\mathbf{r} — only the component of the force along the motion counts, which is exactly what the dot product \mathbf{F}\cdot\mathbf{r}' extracts. Push perpendicular to the path and you do no work; that is why carrying a suitcase horizontally costs (ideally) no work at all.

And the path genuinely matters. Take the rotation field \mathbf{F} = (-y, x) from (1,0) to (-1,0). Going over the top semicircle gives +\pi; going under gives -\pi; the straight chord through the origin gives 0. Three routes, three answers — because this field is not a gradient field. Friction is the everyday culprit: the longer the road, the more work lost, which is precisely path-dependence. When the field is a gradient field, this dependence miraculously vanishes — the cliff-hanger for the next page.

Watch the work accumulate

A particle (the dot) slides along the path C through the field's arrows as you sweep the parameter t. The little gauge shows the running total \int \mathbf{F}\cdot d\mathbf{r} so far: it climbs where the field points along the motion and falls where it opposes it.