Proving that a vector field is conservative using only Green Theorem Consider a vector field $F: \mathbb{R}^2/\{\vec{0}\} \to \mathbb{R}^2$
$$F=(F_1(x,y),F_2(x,y))$$
$\mathbb{R}^2/\{\vec{0}\}$  is not a simply connected domain . Suppose also that $F$ is irrotational.
Can the following procedure be valid to prove that $F$ is conservative?


*

*I find one closed curve $\gamma_1$ that goes around the origin and such that $$\oint_{\gamma_1} F \cdot ds=0$$
(assuming that such curve exists).

*If I consider any other closed curve $\gamma_2$ that goes around the origin, then I can use Green Theorem and see the union of $\gamma_2 \cup \gamma_1$ as the border of a regular domain in $\mathbb{R}^2$. 
Let's say this domain is $D$: then its border is $\partial D=\gamma_1 \cup \gamma_2$. I have to choose the positive orientation for $\partial D$ but suppose that $\gamma_1$ and $\gamma_2$ already have the right orientations in order to have $+\partial D$ (I think that this is not restrictive). Therefore I can finally say that:
$$\oint_{+\partial D} F \cdot ds=\oint_{\gamma_1 \cup \gamma_2 } F \cdot ds=\oint_{\gamma_1  } F \cdot ds+\oint_{\gamma_2  } F \cdot ds= \int \int_{D} \partial_{x} F_2- \partial_{y} F_1=0$$ $$\implies \oint_{\gamma_1  } F \cdot ds=-\oint_{\gamma_2  } F \cdot ds$$
Since $F$ is irrotational. But $\oint_{\gamma_1} F \cdot ds=0$, therefore
$$\oint_{\gamma_2  } F \cdot ds=0$$
This is valid for any curve $\gamma_2$ that goes around the origin. (indipendently from the fact that $\gamma_1$ and $\gamma_2$ are indeed oriented in the proper way to have $+\partial D$).


*

*For any curve $\gamma_3$ that does not go around the origin it is possible to find a subset $A \subset \mathbb{R}^2/\{\vec{0}\}$ simply connected such that $\gamma_3 \subset A$, and $F$ is irrotational, $F$ is conservative in $A$ and $$\oint_{\gamma_3} F \cdot ds=0$$

*So in conclusion we have 
$$\oint_{\gamma} F \cdot ds=0 \,\,\, \forall \gamma \subset \mathbb{R}^2$$ and $F$ is conservative.



Is the previous proof valid or are there any mistakes?
I think it should be valid, but it looks strange because if it is, I would conclude that $F$ is conservative looking at one only curve $\gamma_1$, which seems a bit reductive I think?
 A: This is a very good question. Your conjecture is correct: 
Theorem.  If the field $F=(P,Q)$ defined in $\Omega:={\mathbb R}^2\setminus\{0\}$ has vanishing curl: $Q_x-P_y\equiv0$, and if $\int_{\gamma_*}F\cdot dz=0$ for a single generating cycle $\gamma_*$, then $F$ is conservative. 
In order to prove this theorem you have to prove that $\int_\gamma F\cdot dz=0$ for all closed curves $\gamma\subset\Omega$. For this you cannot directly appeal to Green's theorem, because such a curve $\gamma$ may have selfintersections, or go around the origin several times, etc. In any case, you cannot assume that $\gamma$ and $\gamma_*$ together form the boundary of a nice domain $D$.
Instead we proceed by constructing first a function $f:\>\Omega\to{\mathbb R}$ with $\nabla f=F$ on $\Omega$, i.e., a potential of $F$. For this construction we shall only need Green's theorem for rectangles. For simplicity assume that you can choose four points $z_i=(x_i,y_i)$ $(i\ {\rm mod}\ 4)$ on your "witness curve" $\gamma_*$, such that $z_i$ is in the open quadrant $Q_i$, and such that the part  $\gamma_i$ of $\gamma$ connecting $z_{i-1}$ with $z_i$ completely lies in the  halfplane containing $Q_{i-1}\cup Q_i$.
Choose an arbitrary $c_0\in{\mathbb R}$ and define a function $f_R$ in the right halfplane $H_R$ by
$$f_R(x,y):=c_0+\int_{x_0}^x P(x',y_0)\>dx'+\int_{y_0}^y Q(x,y')\>dy'\ .$$
It follows that ${\partial f_R\over \partial y}=Q$ on $H_R$. Applying Green's theorem to the obvious rectangle we see that we also have
$$f_R(x,y)=c_0+\int_{y_0}^y Q(x_0,y')\>dy'+\int_{x_0}^x P(x',y)\>dx'\ ,$$
and this shows that ${\partial f_R\over \partial x}=P$ on $H_R$  as well. It follows that
$$\nabla f_R(x,y)=F(x,y)\qquad\bigl((x,y)\in H_R\bigr)\ .$$
We have $f(z_0)=c_0$ and, using a standard fact about potentials, 
$$f(z_1)=c_0+\int_{\gamma_1} F\cdot dz=:c_1\ .$$
In the same way we construct a potential $f_U$ in the upper halfplane requiring $f_U(z_1)=c_1$. Since $f_U(z_1)=f_R(z_1)$ we can conclude that $f_U$ and $f_R$ coincide in the intersection $Q_1$ of their domains; furthermore $$f_U(z_2)=c_1+\int_{\gamma_2} F\cdot dz=:c_2\ .$$
In the same way we construct a potential $f_L$ in the left halfplane requiring 
$f_L(z_2)=c_2$, and then a potential $f_B$ in the bottom halfplane requiring
$$f_B(z_3)=f_L(z_3)=c_2+\int_{\gamma_3} F\cdot dz=:c_3\ .$$
The function $f_B$ is also defined in the fourth quadrant $Q_4$,where we already have $f_R$. Since $\nabla f_B-\nabla f_R=F-F=0$ in $Q_4$ we know that $f_B$ and $f_R$ differ by a constant there. We therefore compute 
$$f_B(z_0)=c_3+\int_{\gamma_4}F\cdot dz=\ldots= c_0+\int_{\gamma_*}F\cdot dz=c_0\ .$$
This shows that in fact $f_B=f_R$ in $Q_4$, so that $f_R$, $F_U$, $f_L$, $f_B$ together constitute a globally defined potential function $f:\>\Omega\to{\mathbb R}$ of $F$. Therewith it is proven that $F$ is conservative on $\Omega$.
