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I'm not sure if this is the right name for it but with the theorem:

Let $f:\sigma \rightarrow \mathbb{R} $ be a smooth scalar field and assume $r: [a,b] \rightarrow \mathbb {R}^n$ is a piecewise parametrisation of a path $C$ whose image is included in $\sigma $. Then $$\int \limits_C (\nabla f) \cdot dr =f(r(b))-f(r(a))$$

Can someone give some examples of when this theorem can and cannot be used please.

Or is it the case of when you can or cannot easily find a scalar field s.t. the gradient of the scalar field is the vector field.

Ivo Terek
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snowman
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Remember that if $\bf F$ is a vector field which domain is simply connected, then a potential will exist if and only if $\nabla \times {\bf F} = {\bf 0}$. So if you have a integral $$\int_C {\bf F} \cdot {\rm d}{\bf r}$$ and $\bf F$ admits a potential $f$, then you can find $f$ and use the FTC. You would have to analyze what is easier: finding the potential to use the FTC, or computing the integral directly?

For specific examples I think you will profit more looking at some book (e.g., Leithold's Calculus with Analytic Geometry, vol. $2$).

Ivo Terek
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  • So would you advise me to do the following: Check if curl is $0$. If yes then see if it is easy to make up a scalar field such that when you find the gradient of it, it would give vector field $F$. If it is hard to find this then do it directly? – snowman Apr 26 '15 at 21:42
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    Pretty much it. – Ivo Terek Apr 26 '15 at 21:42
  • Just curious, is there an explicit method to finding the scalar field of any vector field? I don't really mind if there isn't lol. I can always do it directly. – snowman Apr 26 '15 at 21:45
  • Actually I just remembered that there is. Just integrate each component and then match them right? – snowman Apr 26 '15 at 21:49
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    Yes, integrate and compare. Keep in mind that when you integrate a partial derivative wrt to a variable, the "constant" of integration is actually a function of the other variables. – Ivo Terek Apr 26 '15 at 21:50
  • Yes that nice then. Just forgot this since I haven't done it in a long time. – snowman Apr 26 '15 at 21:51