Why isn't this closed loop curve integral $0$? I had to solve a problem where I had to calculate the work done by the force field given by:
$$ \vec{F} = \frac{(-y,x)}{x^2+4y^2}, (x,y) \neq (0,0)$$
where we travel along the whole unit circle in a positive orientation.
I managed to find the potential function given by: $\phi = \frac{-1}{2}\arctan(\frac{x}{2y})$. Since there exists a potential function for which $\vec{F} = \nabla(\phi)$, we have that the force field is conservative. We also note that the potential function isn't defined at $y = 0$.
This means I can't use the fundamental theorem of line integrals directly. Maybe I should extend my $\phi$ in so that I get a continous function by computing the limit as we tend to $0$ from the left and the right.
Then I'm also thinking that our curve we are walking along is a closed loop, and then indeed, for a conservative force field, this has to become $0$. Instead, my answer sheet tells me it's $\pi$.
I'd be thankful if you could describe to me what went wrong in my solution, and why that's the case.
 A: Note that your potential is not defined for $y=0$ and you cannot apply it to the line integral along the unit circle.
The force field $\vec{F}(x,y)=\frac{(-y,x)}{x^2+4y^2}$ is not conservative in $D=\mathbb{R}^2\setminus\{(0,0)\}$: by evaluating the line integral of $\vec{F}$ along the ellipse $\gamma$, $x^2+4y^2=4$ counterclockwise, we find a non-zero result:
$$\int_{\gamma}\vec{F}\cdot d\vec{s}=\int_0^{2\pi}\frac{(-\sin(t),2\cos(t))}{4}\cdot (-2\sin(t),\cos(t))\,dt=\pi$$
where $\gamma(t)=(2\cos(t),\sin(t))$ with $t\in [0,2\pi]$.
In order to find the line integral along the unit circle $C_1$ counterclockwise, we may apply the Green theorem to the domain $E=\{(x,y): x^2+y^2\geq 1, x^2+4y^2\leq 4\}$:
$$\iint_E \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)\,dxdy=\int_{\partial E}\vec{F}\cdot d\vec{s}
$$
Since $\partial E=\gamma \cup C_1^-$ and $\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}=0$, i.e. $\vec{F}$ is irrotational in $E$, it follows
$$0=\iint_E 0\,dxdy=\int_{\gamma}\vec{F}\cdot d\vec{s}+\int_{C_1^-}\vec{F}\cdot d\vec{s}=\int_{\gamma}\vec{F}\cdot d\vec{s}-\int_{C_1}\vec{F}\cdot d\vec{s}$$
and therefore
$$\int_{C_1}\vec{F}\cdot d\vec{s}=\int_{\gamma}\vec{F}\cdot d\vec{s}=\pi.$$
