Green's Theorem and Conservative Fields I came to a question on Green's Theorem involving the Vector Field $$F(x,y)=\left(-\frac{y}{x^2 + y^2} , \frac{x}{x^2+y^2}\right) ,$$
which equated the work done around the ellipse $3x^2 + y^2 = 1$ equal to $2\pi$ by "removing" the origin (not continuous or differentiable here) and evaluating Green's theorem using a new domain, which lead me to a conceptual problem.
If i evaluate the work integral around any circular path, I get work done equal to $2\pi$ using parametrisation. However, the curl of this vector field is the 0 vector $(0,0,0)$. Doesn't this mean that the work done should be zero because the work done only depends on the endpoint which are two of the same? 
Thanks for any solutions or ideas. 
 A: You are using the statement:

If $curl F = 0$ then the work done by $F$ around any closed curve is zero.

But this statement is not true in general.  The following two statements are true:


*

*If $F$ is conservative, meaning $F = \nabla f$ for some smooth function $f(x,y)$, then the work done by $F$ around any closed curve is zero.

*Let $D$ be a simply connected open set with oriented boundary $C$.  If $F$ is smooth on a neighborhood of $D$ and the curl of $F$ is zero then $F$ is conservative.
Now look carefully at the setup in your problem.  The ellipse $3x^2 + y^2 = 1$ is the boundary of a simply connected open set, namely $3x^2 + y^2 < 1$, but $F$ is not smooth on this open set: it is discontinuous at $(0,0)$.  It is smooth on the open set obtained by removing $(0,0)$ from the interior of the ellipse, but the interior of the ellipse with a point removed is not simply connected.
So statement 2 above does not apply, and we can't conclude that $F$ is conservative even though its curl is zero.  Indeed, your computation shows that it can't be conservative: if it were then the path integral would be zero by statement 1.
