## Showing that f is the potential function of F

Hey guys.

I'm trying to solve an exercise, and the solution is proberbly simpler than what I'm trying to do.

We are given

$\displaystyle F = (x^2 + y^2)^{-1} \binom{-y}{x}$

The curve $\displaystyle C$ is given by the parametrization

$\displaystyle \mathbf{r}(t) = (r(t)\cos(\theta (t)) , r(t) \sin (\theta (t)))$

where $\displaystyle r$ and $\displaystyle \theta$ both have continous differentials and $\displaystyle r(t) > 0$

Then I have shown that
$\displaystyle F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \theta'(t)$ and $\displaystyle \int _C F \cdot d\mathbf{r} = \theta(b)-\theta(a)$.

For the part of the excercise I'm having issues with, we are given

$\displaystyle f(x,y) = 2 \mbox{Arctan} \frac{y}{x + \sqrt{x^2 + y^2}}$.

Now I need to find the domain of $\displaystyle f$, and show that $\displaystyle f$ is the potential function of $\displaystyle F$.

I understand that the domain is $\displaystyle \{(x,y) | x + \sqrt{x^2+y^2} > 0\}$, but can I write it in a prettier way?

I'm getting a hint, that I should convert to polar coordinates as soon as possible, because the calculations will be easier. But I cant really figure out how to solve the excercise. Anyone have some hints? Especially on how, when and why I'm supposed to convert to polar coordinates.

Thanks a million.