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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^3$ 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 $U$.\nThen $F$ is [[conservative|05R9NF2J49]] on $U$.\nExplicitly, for a fixed $p \\in U$, the function\n$$\\phi(x) = \\int_0^1 F\\big(p + t (x - p)\\big) \\cdot (x - p) \\, dt$$\nsatisfies $\\nabla \\phi = F$ on $U$.\n\n## Remarks\n\nThe integral $\\phi(x)$ is the [[line integral|0YS8RX87F4]] of $F$ along the segment from $p$ to $x$. Convexity makes sure that this segment lies in $U$.\nTo prove $\\nabla \\phi = F$, [[differentiate under the integral sign|0AR06W1E1B]] and use $\\partial_j F_i = \\partial_i F_j$, which is the condition $\\nabla \\times F = 0$.\nThe 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.\nThe result holds more generally on simply connected open sets. It [[fails on some other open sets|0GRM3BS0DD]]. With the [[converse|0M2JRW1J0N]], it gives: on a convex open set, a $C^1$ field is conservative if and only if its curl is zero.\n\n## Explanations\n\n### L1 human\n\nOn a convex set, a field with no swirl is the gradient of a potential.\nYou build the potential by computing the work of the field along straight segments from one base point.\n",
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        "url": "https://en.wikipedia.org/wiki/Conservative_vector_field"
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  "name": "Curl-free vector field on a convex set is conservative",
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