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Find the potential function of a conservative vector field $\mathbf{F}(r,\phi)=r\sin(2\phi)\mathbf{\hat{r}}+r\cos(2\phi)\mathbf{\hat{\phi}}$

Does my solution seem right:

So if $\mathbf{F}(r,\phi)=r\sin(2\phi)\mathbf{\hat{r}}+r\cos(2\phi)\mathbf{\hat{\phi}}=\nabla\phi(r,\phi)$ then,

$$\frac{\partial{g}}{\partial{r}}=r\sin(2\phi)$$ and $$\frac 1r\frac{\partial{g}}{\partial{\phi}}=r\cos(2\phi)$$

When we integrate the first equation with respect to $r$ we get $$g=\frac{r^2}{2}\sin{2\phi}+ C(\phi)$$

Now when derivate this with respect to $\phi$ we get $$r^2\cos(2\phi)+C'(\phi)$$ Now when we insert this to $\dfrac 1r\dfrac{\partial{g}}{\partial{\phi}}=r\cos(2\phi)$ we get $$\begin{align}r\cos(2\phi)+C'(\phi)&=r\cos(2\phi)\\C'(\phi)&=0\\C(\phi)&=C \end{align}$$

So the potential function $g$ is $$g=\frac{r^2}{2}\sin{2\phi}+C$$

Did I get it right?

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  • $\begingroup$ Looks good to me. $\endgroup$ – David H Mar 2 '14 at 19:52
  • $\begingroup$ damn, but did I make a mistake where I was suppose to multiply $\frac 1r$ with the $\partial{g}/\partial{\phi}$. The integration constant $C'(\phi)$ should also be multiplied by $\frac 1r$. That doesn't change the result tho.. $\endgroup$ – ELEC Mar 2 '14 at 20:23

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