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How do we find the function of radial distance $f(r)$ from an equation of the form $\int_a^b \nabla f(r)\cdot d\vec{r}=c$ for some constant $c$? $f(r)$ is radially symmetric.

Thank you.

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    @AlexBecker: Would the additional fact the $f(r)$ is radially symmetric help?2012-02-28

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Assuming that your function's sole independent variable is r we have:

$\int_a^b \nabla f(r)\cdot d\vec{r}=\int_a^b \frac{df(r)}{dr} \hat{r} \cdot d\vec{r}=\int_a^bdf(r)=f(b)- f(a)=c $.

So your equation becomes: $f(b)-f(a)=c \; \ (1)$.

So your equation has infinite solutions because if you find a function $f$ that satisfies the condition (1) every function of the form $h=f+ct$ will satisfy it too.

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    No problem, *ct=constant number*. I didn't use the *c* symbol to avoid any confusion with the *c* used in your equation. If the answer is suitable for you mark the answer as acceptable:)2012-02-28