Line integral

A force field $F$ pushes on an object that moves along a path $C$ from A to B. Over each short step of length $ds$, only the part of the force along the direction of travel does work, $F\cdot T\,ds$, where $T$ is the unit tangent. Adding up these pieces gives the line integral of $F$ along $C$, the work $W$.

The yellow ticks along the path show $F\cdot T$, filled and pointing forward where the force helps, hollow and pointing back where it holds back. The graph under the view plots $F\cdot T$ against the distance $s$ walked from A. Counting the hatched area below the axis as negative, the area up to the red point S is the work done so far, and the whole area is $W$: the line integral has become an ordinary integral in $s$.

At S the force is split into a yellow part along the path and a faint part across it, which does no work. In the radial field $F = (x, y)$, around the circle of radius 2 about the origin, the force has length 2 but points straight out, so $F\cdot T = 0$ and $W = 0$. Work is not the size of the force times the length of the path. Reverse traces the same path backwards, turning $T$ into $-T$ and $W$ into $-W$.

In a gradient field $F = \nabla\varphi$, the chain rule makes $F\cdot T\,ds$ the change in $\varphi$ over the step. The work up to S is $\varphi(S) - \varphi(A)$, and when $\nabla\varphi$ is continuous, the work along any piecewise smooth path depends only on its ends:

Contour lines shows why without a formula. It draws the lines where $\varphi$ is an integer. From one line up to the next, $\varphi$ rises by 1, so a path that climbs between them does one unit of work, gathered smoothly on the way. However a path winds, its extra crossings come in pairs, one up and one down, and cancel. Every path from A to B therefore crosses the same net number of lines, and that number, with the parts of a unit at the two ends, adds up to $\varphi(B) - \varphi(A)$.

A field whose work depends only on the ends of the path is called conservative. In such a field every closed path does zero work, because it ends where it starts.

The rotation field $F = (-y, x)$ has no potential. From $(2, 0)$ to $(-2, 0)$, the top half circle has $F\cdot T = 2$ along its length of $2\pi$, so $W = 4\pi \approx 12.57$. Along the x-axis the force is across the path and $W = 0$. Over the top and back along the axis is a closed path that does $4\pi$ of work.

The spiral staircase $F = (-y, x)/(x^2 + y^2)$ lies in between. Its curl is zero wherever it is defined, and near any point other than the origin it is the gradient of the polar angle $\theta$. But no single angle function works all the way round, because one counterclockwise turn about the origin adds $2\pi$. Its contour lines are rays, and the work is the change in $\theta$ followed along the path: $2\pi \approx 6.28$ once counterclockwise around the origin, and 0 around a circle that does not enclose it.