Line Integrals

A line integral computes the accumulation of a field along a curve.

Scalar line integral — Integrate a scalar field $f$ along curve $C$:

$$\int_C f\,ds = \int_a^b f(\vec{r}(t))\,\|\vec{r}'(t)\|\,dt$$

Vector line integral — Integrate a vector field $\vec{F}$ along $C$:

$$\int_C \vec{F}\cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt$$

This computes the work done by $\vec{F}$ along the path.

Conservative fields — A field $\vec{F}$ is conservative if $\vec{F} = \nabla\phi$ for some potential $\phi$. Then:

$$\int_C \vec{F}\cdot d\vec{r} = \phi(\vec{r}_B) - \phi(\vec{r}_A)$$

The integral depends only on the endpoints, not the path.

Test for conservativeness in $\mathbb{R}^{3}$:

$$\nabla\times\vec{F} = \vec{0}$$

Example — Work done by gravity moving from height $h_1$ to $h_2$:

$$W = \int_C \vec{F}_g\cdot d\vec{r} = -mg(h_2-h_1)$$

This is path-independent because gravity is conservative with $\phi = mgh$.

Key insight: If $\nabla\times\vec{F}=\vec{0}$ on a simply connected domain, then $\vec{F}$ is conservative and you can find a potential function instead of computing the integral directly.

Common pitfall: Path-independence is a privilege of conservative fields, not a general truth. Before replacing a line integral by endpoint values, verify $\vec{F} = \nabla f$ (e.g. check the curl vanishes on a simply connected region).