Vector function on non-simply-connected set. Is it conservative? Probably not.

Question

I was given the following question in my Calculus class and I'm having a little trouble.

Let $\Omega=\mathbb{R^2} \setminus\{(x,0)\in\mathbb{R^2} : |x|\le1\}$ and $F:\Omega \to \mathbb{R^2}$ be given by $$ F(x,y) = \left(\frac{-y}{(x-1)^2+y^2}+\dfrac{y}{(x+1)^2+y²},\frac{x-1}{(x-1)^2+y^2}-\dfrac{x+1}{(x+1)^2+y²} \right)$$ Determine if F is a conservative field on $\Omega$

My guess is that it is not, here is what I've tried:

1. I showed that the curl of $F$ is zero. But that's actually useless because $\Omega$ is not simply connected. I thought that maybe I could "extend" $F$ to all of $\mathbb{R^2}$ and show that it's conservative there (and then conclude that it must be conservative in any subset) but it's undefined at the points $(1,0)$ and $(-1,0)$ so that wouldn't work.

2. Then I tried finding a potential. I found that $$ P(x,y) = -arctan\left(\frac{x-1}{y} \right)+arctan\left(\frac{x+1}{y} \right)$$ works in the sense that $\nabla P=F$ but $P$ isn't defined for $y=0$ so it's not a proper potential function for $F$.

So now I reckon that $F$ isn't a conservative field but I've tried giving a closed path that isn't zero but I can't come up with one. I'm pretty sure that if such a path exists it must cross (or at least touch) the $x$-axis because that's where $P$ has trouble.

I would really appreciate it if anyone could point me at a hint or mistake in my reasoning. Thanks a bunch.

Answer 1

You can "cross" the line $y=0$ if you use the function $\text{atan2}(y,x)$ or simply $\tan(x,y)$. $\arctan(y/x)$ is clearly insuficient in cases like $x=0,y=1$ where the angle is obviously $\pi/2$. So, the potential would be:

$$P(x,y) = -\text{atan2}\left(x-1,y \right)+\text{atan2}\left(x+1,y \right)$$

I think the field is conservative (but obviously, a path cannot cross the point where the field is not defined)

(Later I'd try to follow a path surrounding the point $(1,0)$ and passing by the origin. Maybe this path is special)

Added I calculated the integral along the square of vertex $A(2,-1),B(2,1),C(0,1),D(0,-1)$ anticlockwise and the circulation is not zero. The field is not conservative.