Prove $\oint_\Gamma\vec\nabla f\cdot d\vec{r}=0$ when $\Gamma$ is the unit circle I have encountered a statement in my book which didn't seem quite right to me. It was written exactly like this:

Let $f(x,y):\mathbb{R}^2\to\mathbb{R}$ be a differentiable function such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous functions for all $(x,y)\neq(0,0)$, and let $\Gamma$ be the unit circle, centered at $(0,0)$. Then:
$$\oint_\Gamma\vec\nabla f\cdot d\vec{r}=0$$

I tried to sit down and think why is this statement true. You all must have encountered before the infamous vector field:
$$\vec{F}=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$$
This vector field is the gradient of the function $f(x,y)=\arctan(\frac yx)$ if I remember correctly. This function is of course problematic on the $y$ axis. However, I assume you can define $f$ on the $y$ axis, apart from the point $(0,0)$, such that the partial derivatives of $f$ would exist and also be continuous there (I'm not sure about that, but the fact that the parital derivatives are clearly continuous for every point but $(0,0)$ has led me to that assumption).
So $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are indeed continuous for all $(x,y)\neq(0,0)$, but $\displaystyle \oint_\Gamma\vec\nabla f\cdot d\vec{r} $ would be equal to $\pm2 \pi$ (depends on the orientation of $\Gamma$).
To conclude, I would like to know whether my example disproves the above statement, or maybe it is wrong. If it's not correct, I would be glad to hear why, and in addition, have a proof of the statement. I thought of many ways to prove it - using Green's Theorem and the fact that $\vec\nabla\times\vec\nabla f\equiv0$, stating $\vec\nabla f$ is a conservative vector field (since it is derived from a gradient of a scalar potential function) and more; But of course I wouldn't go there before I know why my example is incorrect.
Thanks!
 A: Lucky you, the appearance of a totally wrong answer persuaded me to post a  complete solution.


Fundamental Theorem of Calculus for line integrals: Suppose $\Omega\subset\Bbb R^2$ is open and $f\in C^1(\Omega)$. If $\gamma:[0,1]\to\Omega$ is a $C^1$ curve  then $\int_\gamma\nabla f\cdot dr=f(\gamma(1))-f(\gamma(0))$.


(Hence if $\gamma$ is a closed curve (meaning $\gamma(1)=\gamma(0)$) the integral is $0$. With $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $\gamma(t)=(\cos(2\pi t),\sin(2\pi t))$ this proves the result you ask about; in particular what happens at the origin doesn't  matter.)
Proof, even though it must be in the book: By definition, if $F$ is a vector field then $$\int_\gamma F\cdot dr=\int_0^1F(\gamma(t))\cdot\gamma'(t)\,dt.$$Now suppose that $F=\nabla f$, and set $$g(t)=f(\gamma(t)).$$The chain rule shows that $$F(\gamma(t))\cdot \gamma'(t)=g'(t),$$so $$\int_\gamma F\cdot dr=\int_0^1g'(t)\,dt=g(1)-g(0).$$


Corollary: If $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $F$ is the infamous vector field mentioned in the question there does not exist $f\in C^1(\Omega)$ with $F=\nabla f$. 


(Indeed, in the language used in calculus books, $F$ is the standard example of a vector field which is closed but not exact.)
A: As $f(x,y) = \arctan \frac{y}{x}$ is not defined along the y-axis, the function is not continuous and the partial derivatives do not exist when $x = 0.$ 
Even if the equations $\nabla f = (\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2})$ don't indicate that there is a problem, the partial derivatives do not exist when $x = 0$, and they are not, in fact, continuous.
