Can a vortex vector field be conservative?

Question

For the following vortex vector field $$F(x,y)=\left(\frac{2xy}{(x^2+y^2)^2},\frac{y^2-x^2}{(x^2+y^2)^2}\right)$$ If we apply the extended Green's Theorem for an arbitrary simple closed curve $C$ that doesn't pass through the origin and with a circular "hole" $C'$ with radius $a$ centered at the origin, we will get $$\iint_R\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)dA=\int_C\vec{F}\cdot d\vec{r}-\int_{C'}\vec{F}\cdot d\vec{r}$$ Since $$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=0, \int_{C'}\vec{F}\cdot d\vec{r}=0$$ We will get $$\int_C\vec{F}\cdot d\vec{r}=0$$ Does this means that this vortex vector field is conservative?

Answer 1

Notice that $(x^2+y^2)^2$ is in the denominator of both components, so by quotient rule a good guess for a potential function would be to assume that it takes the form of

$$\frac{f(x,y)}{x^2+y^2}$$

which sets up the following system of equations:

$$\begin{cases}(x^2+y^2)f_x - 2xf = 2xy \\ (x^2+y^2)f_y - 2yf = y^2-x^2 \end{cases}$$

And as a guess, $f(x,y) = -y$ satisfies both equations (though by no means is it the only $f$ to do so). Which means the potential function is given by

$$\frac{-y}{x^2+y^2}$$

which has an isolated singularity as opposed to a branch cut, so its gradient will be conservative everywhere on its domain.