It can be shown that if F has continuous 2nd partials, then div curl F = 0. According to Stoke's Theorem, the work done around a closed path C is equal to the flux integral of curl F on a surface sigma that has C as its boundary in positive orientation. However, this integral is equal to the volume triple integral of div curl F. But if F has continuous 2nd partials, then div curl F = 0, and hence the work around C must be 0. Doesn't this show that F is conservative if F has continuous 2nd partials?(adsbygoogle = window.adsbygoogle || []).push({});

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# Continuous 2nd Partials a Substantial Requirement for Conservative Field?

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