Homotopy invariance of line integrals of closed forms

Statement

Let $U \subseteq \mathbb{R}^3$ be open, and let $F \colon U \to \mathbb{R}^3$ be a vector field of class $C^1$ with $\nabla \times F = 0$ on $U$. Let $c_0, c_1 \colon [0, 1] \to U$ be piecewise smooth closed curves. Suppose that there is a piecewise smooth map $H \colon [0, 1] \times [0, 1] \to U$ with $H(t, 0) = c_0(t)$, $H(t, 1) = c_1(t)$ and $H(0, s) = H(1, s)$ for all $s, t \in [0, 1]$. Then $$\oint_{c_0} F \cdot d\mathbf{r} = \oint_{c_1} F \cdot d\mathbf{r}.$$

Remarks

Proof: apply Stokes' theorem to the surface $H([0, 1]^2)$. The curl term is zero, and the integrals along the curves $s \mapsto H(0, s)$ and $s \mapsto H(1, s)$ cancel. If $c_1$ is a constant loop, then $\oint_{c_0} F \cdot d\mathbf{r} = 0$. On a simply connected domain, each loop deforms to a point, and a smoothing argument gives a piecewise smooth $H$. So curl-free $C^1$ fields on such domains are conservative. The condition $\nabla \times F = 0$ means that the 1-form $F_1 \, dx + F_2 \, dy + F_3 \, dz$ is closed. This explains the name of the theorem.

Explanations

L1 human

For a field with no swirl, the work around a loop does not change when you slide and bend the loop continuously inside the domain. A loop that can shrink to a point therefore has zero work.