Simply connected Let $F$ be a vector field of class $C^1$ defined in a simply connected domain $U \subset \mathbb{R}^2$ the following conditions are equivalent:
(1) $\oint_cF.dr=0$ for every closed loop $C \subset U$.
(2)...
What is $U$? Is the domain of $F$? So in case $F=(\frac{1}{x+y},\frac{1}{x+y})$, $U=\mathbb R^2-{(0,0)}$?
The reason I ask this is because the following problem asks for all possible values of the integral, where $C$ is a closed loop that doesn't cross the origin. $$\oint_C\frac{(x+y)\,dx+(y-x)\,dy}{x^2+y^2}$$
The solution is $0$ and $-2\pi$ I understand the process of how to get $-2\pi$ but I can't understand this part in the resolution:
"If the simply connected domain that contains $C$ doesn't contain the origin, then: $\oint_CF.dr=0$"
But if $U$ doesn't contain the origin, how can it be simply connected? It has a hole in it...
 A: May not answer your question, but too long for a comment:


*

*If $F$ is a vector field with natural domain $X$ (that is, if the component functions of $F$ are both defined throughout $X$ and in no larger subset), then $U$ is some open subset of $X$, which must be specified explicitly if we're to agree that "$F$ is defined". In the same manner, the domain of a mapping is an essential datum, in addition to the formula(s) defining the mapping.

*The natural domain of the vector field $F(x, y) = (\frac{1}{x + y}, \frac{1}{x + y})$ is the complement of the line $y = -x$, not the complement of the origin.

*The natural domain of $\frac{(x + y, y - x)}{x^{2} + y^{2}}$ is the complement of the origin, but if $F(x, y) = \frac{(x + y, y - x)}{x^{2} + y^{2}}$ in an open set $U$, you are not automatically guaranteed that $U$ is equal to the complement of the origin, only that $U$ is some open subset.

*I don't understand the claim that the possible values of the integral are $0$ and $-2\pi$: If $F$ integrates to $-2\pi$ over some closed path (which is correct in this example), then for every integer $n$, there exists a path over which $F$ integrates to $2\pi n$. Even if "closed loop" refers to a simple closed loop, there exists a loop over which the integral of $F$ is $2\pi \neq -2\pi$.

*A "simply-connected subset of the punctured plane" could be, as Daniel Littlewood says, an open disk of radius $r > 0$ centered at a point at distance at least $r$ from the origin. The set $U$ could also be a slit plane, the complement of a ray starting at the origin, or any of a vast number of other examples.
