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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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Not enough information is given to uniquely specify the function. Also, $d\vec{r}$ indicates we are ranging over a several dimensions yet we integrate over just one. – Alex Becker Feb 28 '12 at 19:12
@AlexBecker: Would the additional fact the $f(r)$ is radially symmetric help? – Nicola Feb 28 '12 at 19:19
up vote 1 down vote accepted

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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THanks, chemeng, what is $t$ in $h=f+ct$? – Nicola Feb 28 '12 at 19:51
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:) – chemeng Feb 28 '12 at 19:54

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