Curl-free vector field on a convex set is conservative

Statement

Let $U \subseteq \mathbb{R}^3$ be open and convex, and let $F \colon U \to \mathbb{R}^3$ be a vector field of class $C^1$ with $\nabla \times F = 0$ on $U$. Then $F$ is conservative on $U$. Explicitly, for a fixed $p \in U$, the function $$\phi(x) = \int_0^1 F\big(p + t (x - p)\big) \cdot (x - p) \, dt$$ satisfies $\nabla \phi = F$ on $U$.

Remarks

The integral $\phi(x)$ is the line integral of $F$ along the segment from $p$ to $x$. Convexity makes sure that this segment lies in $U$. To prove $\nabla \phi = F$, differentiate under the integral sign and use $\partial_j F_i = \partial_i F_j$, which is the condition $\nabla \times F = 0$. The same proof works in $\mathbb{R}^n$ under the condition $\partial_j F_i = \partial_i F_j$ for all $i, j$, and on star-shaped open sets. The result holds more generally on simply connected open sets. It fails on some other open sets. With the converse, it gives: on a convex open set, a $C^1$ field is conservative if and only if its curl is zero.

Explanations

L1 human

On a convex set, a field with no swirl is the gradient of a potential. You build the potential by computing the work of the field along straight segments from one base point.