A force pushes an object along a curved path. Only the part of the force along the direction of motion does work, and the force may change from point to point. The line integral of a vector field adds up that work. It builds on line integrals of functions and on the dot product.
Idea clave
For a continuous field $\mathbf{F}$ and a smooth oriented curve $C$ given by $\mathbf{r}(t)$, $a \le t \le b$:
$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_C \mathbf{F} \cdot \mathbf{T}\,ds = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt,$$
where $\mathbf{T}$ is the unit tangent in the direction of travel and $d\mathbf{r} = \mathbf{r}'(t)\,dt$. Reversing the direction changes the sign: $\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}$.
Por qué funciona
Where the formula comes from: over a short piece of path the force is nearly constant and the displacement is $\Delta\mathbf{r} \approx \mathbf{r}'(t)\,\Delta t$, so the work is about $\mathbf{F} \cdot \Delta\mathbf{r}$ (force times displacement, counting only the component along the motion). Add and take the limit. In terms of arc length, $\mathbf{r}'(t)\,dt = \mathbf{T}\,|\mathbf{r}'(t)|\,dt = \mathbf{T}\,ds$: the work integral is the scalar line integral of the tangential component $\mathbf{F} \cdot \mathbf{T}$.
Why the sign flips: walking the other way reverses $\mathbf{T}$ at every point, so $\mathbf{F} \cdot \mathbf{T}$ changes sign while $ds$ does not.
Why the sign flips: walking the other way reverses $\mathbf{T}$ at every point, so $\mathbf{F} \cdot \mathbf{T}$ changes sign while $ds$ does not.
Worked examples
Ejemplo resuelto: Along a segment
Find $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle x^2, xy \rangle$ along the segment from $(0, 0)$ to $(1, 2)$.
- $\mathbf{r}(t) = \langle t, 2t \rangle$ and $\mathbf{r}'(t) = \langle 1, 2 \rangle$.
- $\mathbf{F}(\mathbf{r}(t)) = \langle t^2, 2t^2 \rangle$, so $\mathbf{F} \cdot \mathbf{r}' = t^2 + 4t^2 = 5t^2$.
- $\int_0^1 5t^2\,dt = \frac{5}{3}$.
Respuesta: $\frac{5}{3}$
Ejemplo resuelto: A half circle, both ways
Find the work done by $\mathbf{F} = \langle -y, x \rangle$ along the upper half of $x^2 + y^2 = 4$ from $(2, 0)$ to $(-2, 0)$, and then from $(-2, 0)$ to $(2, 0)$.
- $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$, ${}0 \le t \le \pi$; $\mathbf{r}'(t) = \langle -2\sin t, 2\cos t \rangle$.
- $\mathbf{F} \cdot \mathbf{r}' = 4\sin^2 t + 4\cos^2 t = 4$, so the work is $\int_0^\pi 4\,dt = 4\pi$.
- The field pushes along the motion all the way. In the other direction it pushes against it: $-4\pi$.
Respuesta: ${}4\pi$ and $-4\pi$
Ejemplo resuelto: In space
Find $\int_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle z, x, y \rangle$ along $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$, ${}0 \le t \le 1$.
- $\mathbf{F}(\mathbf{r}(t)) = \langle t^3, t, t^2 \rangle$ and $\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle$.
- $\mathbf{F} \cdot \mathbf{r}' = t^3 + 2t^2 + 3t^4$.
- $$\int_0^1 (t^3 + 2t^2 + 3t^4)\,dt = \frac{1}{4} + \frac{2}{3} + \frac{3}{5} = \frac{91}{60}.$$
Respuesta: $\frac{91}{60}$
Error común
Ignoring orientation. Unlike $\int_C f\,ds$, the integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ depends on the direction of travel. A problem that says "from $A$ to $B$" fixes it. If you parametrize from $B$ to $A$, either redo the parametrization or flip the sign of your answer. In the second example the two directions give ${}4\pi$ and $-4\pi$.