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.
You are using the statement:
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.