Calculate the line integral $ \oint_c \vec F \: d \vec s $ , where $$ \vec F(x,y) = \Biggl( \frac{\partial}{\partial x} \biggl( \frac{x}{x^2+y^2} \biggr)+1 \:, \; \frac{\partial}{\partial y} \biggl( \frac{x}{x^2+y^2} \biggr) \Biggr) \; , \; (x,y) \neq \vec 0 $$ and $ \pmb c $ is a random simple closed curve in $ \mathbb R^2 $, which doesn't pass through $ (0,0) $.
This question was solved in a lesson in two steps, considering whether $ \pmb c $ encloses $ (0,0) $ or not.
a)
If $ \pmb c $ does not enclose $ (0,0) $, then applying Green's theorem we have $$ \oint_c \vec F \: d \vec s = \iint_D (Q_x-P_y) \: dxdy = \; ... \;= 0 $$ , where $ \vec F(x,y) = \Bigl(P(x,y) \;, \; Q(x,y) \Bigr) $ and $ \pmb c = \partial \pmb D $.
b)
If $ \pmb c $ encloses $ (0,0) $ (Green's theorem cannot be applied), then $$ \oint_c \vec F \: d \vec s = \oint_{γ} \vec F \: d \vec s= \int_0^{2π} \vec F \bigl(\vec r(t) \bigr) \vec r'(t) \: dt =\; ... \;=0 $$ , where $ \pmb {γ} $ is a path given by $ \vec r(t) = (\cos t , \sin t) \; , \; t \in [0,2π] $.
So, in any case $ \oint_c \vec F \: d \vec s = 0 $.
However, we can notice that $\vec F = \vec \nabla f $ , where $ f(x,y) = \cfrac{x}{x^2+y^2}+x \; , \; (x,y) \neq \vec 0 $.
Could we bypass steps a, b and say that: $$ \oint_c \vec F \: d \vec s = \oint_c \vec \nabla f \: d \vec s = f \bigl (\vec c(t_0) \bigr)-f \bigl (\vec c(t_0) \bigr) = 0 \quad \forall \; t_0 : \vec c(t_0) \neq \vec 0 $$
From the hypothesis, $ \pmb c $ doesn't pass through $ (0,0) $, so $ f \bigl (\vec c(t) \bigr) $ can be defined for every $t$, such that $ \vec c(t)$ is defined. So, $ \oint_c \vec F \: d \vec s = 0 $.
Is the second method correct ?
If there were more points, except for $ (0,0) $, that did not belong to the domain of the potential function $f$, would the second method be wrong ?
Thanks in advance
After some research on related questions,
For the first question:
Yes, the gradient theorem can be applied, as long as $ \vec F = \vec \nabla f $ is true $ \; \forall \; (x,y) \in D_{\vec F} $ ( field's domain). So, we get that the line integral equals zero, for any simple closed curve which doesn't pass through $ (0,0) $. So, (assuming that the definition of a conservative field requires a simply connected domain) we can see that the gradient theorem can sometimes be applied on non-conservative fields, as $ \mathbb{R^2} \setminus\{ (0,0)\} $ in my example is not simply connected.
For details, check the answer of the following question:
Can I apply the gradient theorem for a field with not simply connected domain?
For the second question:
If there was no function $f$, such that $ \vec F = \vec \nabla f \; \; \forall \; (x,y) \in D_{\vec F} $, then the second method would be wrong, because for some paths it could lead us to wrong results. A common example for this, is the vector field $ \vec G(x,y)=\cfrac{(-y,x)}{x^2+y^2} $.
For details, check the answer of the following question:
Why is this vector field not conservative, even though it has a potential? (what is the actual definition of a conservative vector field?)