Why is this vector field non-conservative in the curve? The vector field F is as follows
$$
F(x,y) = (\frac{y}{(x-1)^2 + y^2 }   ,  \frac{1-x}{(x-1)^2 + y^2 }  )
$$
F(x,y) is clearly non-conservative, since it becomes undefined at P(1,0). Thing is, I'm also given the curve C1:
$$
C1 : (x,y)   \in R  |  (x-1)^2 + y^2 = 1 
$$
And I'm trying to judge whether this section is conservative or not. According to the solution, it isn't (and I don't understand why). 
C1 is the set of all points in R that satisfy (x-1)^2 + y^2 = 1. The point  (1,0) is not one of them. It goes even further: C1 puts a constant value in F's denominator = 1. Why is P(1,0) in C1's domain even though (1-1)^2 + 0^2 is not equal to one? Is there something I'm not considering?
 A: Conservative means that $ \int_a^b F \cdot dx $ is independent of path, or equivalently, that the integral around a closed curve always vanishes. Because this is quite difficult to check, Stokes's theorem allows you to reduce this condition to whether the curl is zero on the interior of any path.
If the curl is zero but the domain is not simply connected, as it is not here, there are closed paths for which the interior has holes. Hence Stokes's theorem does not apply, and you have to check manually what the integral along a curve surrounding the hole is. (You can apply Stokes's theorem to show that any curve that winds around the hole once will give the same answer.) If it gives zero, you can still say the field is conservative, but if it doesn't give zero, as here, the field is not conservative on any domain that contains a circle enclosing the singular point.
There are examples that have singular point, but nevertheless may be called conservative on domains that do contain curves enclosing a singular point: you can check that
$$ \int_{x^2+y^2=1} \left( \frac{-x^2+y^2}{(x^2+y^2)^2} , \frac{-2xy}{(x^2+y^2)^2} \right) \cdot (dx,dy) $$
is zero, for example. Indeed, it is the gradient of $x/(x^2+y^2)$, so this follows from the fundamental theorem of line integrals, and this vector field is conservative on $ \mathbb{R}^2 \setminus \{ 0\}$.
The moral of the story is that whether a field is conservative depends on the global condition of what the domain is as well as the local condition of having zero curl. Your $F$ is conservative on something like $\mathbb{R}^2 \setminus \{(x,y) : y=0 \wedge x<1 \}$, which avoids the possibility that a curve can enclose $(1,0)$.
