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Currently going through some line integral problems and in the worked example provided, the line integral evaluated to become zero.

The curve C that they have provided was a simple closed curve (it was a half-circle).

Does the answer being zero have anything at all to do with the fact that C is closed and simple? Or was it merely a coincidence that it turned out to be zero?

Trogdor
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  • If the field is conservative, then the closed loop integral is $0$. This is a known fact. If not, then it probably is a coincidence. – IAmNoOne Apr 10 '14 at 08:00
  • I jumped a few pages to the notes on conservative vector fields, and one of the propositions stated was that a vector field is conservative if and only if $\oint f.dx = 0$, so I guess that it is impossible to have a coincidence. – Trogdor Apr 10 '14 at 08:10
  • You can have an integral around a closed loop being 0 without the field being conservative. You just can't have it being 0 for all closed loops (so it can be a coincidence for one loop). – Paul Apr 10 '14 at 09:03
  • @Paul But why you can't have it for all loops? Maybe the closed integrals along any path are zero and only some integral along a curve are non-zero. How can we deduce from testing only the closed integrals that the field is conservative? Thanks in advance. – user599310 Mar 22 '20 at 20:50

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Yes if $C$ is a simple closed curve, $\textbf{v}:\mathbb{R}^m\rightarrow \mathbb{R}^m$ and$$\oint_{C} \textbf{v}\;d\textbf{r}=0,$$ then it follows that $\textbf{v}=\nabla f$ for some scalar function $f:\mathbb{R}^m\rightarrow \mathbb{R}$.

George1811
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