Evaluating line integral technique.

Asked Jul 23 '15 at 22:22

Active Jul 23 '15 at 22:22

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I'm trying to do a few questions set by a lecturer on line integrals.

I was struggling with a few of them and decided to look at the solutions:

Often when $$\int_C \vec{F}\cdot d\vec{r}$$ is tricky to evaluate he will say $$\int_C~ \vec{F}\cdot d\vec{r}=\int_C ~(\vec{F}-\nabla f)\cdot d\vec{r}$$ I have never seen this method before and don't understand what he is doing how the two are equal or even where the function $f$ comes from. Could anyone explain in the simplest way possible or give me something to research so I know what is going on.

Thanks.

asked Jul 23 '15 at 22:22

Lio Rampart

2

If $C$ is a closed contour, then since $df=\nabla \cdot d\vec r$,$$\oint_C \nabla f\cdot d\vec r=0$$ – Mark Viola Jul 23 '15 at 22:27
$C$ isn't closed when it is used. I know that you can just evaluate at the end points if $F$ is conservative but I don't think that's what's going on though. – Lio Rampart Jul 23 '15 at 22:40
Then the expression of interest is generally not true. – Mark Viola Jul 23 '15 at 22:44

Evaluating line integral technique.

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