Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.