Does Greens Theorem apply to the annulus? I realize that it does, but I can't prove it. If the region is of the form: $$ D = \{ (x,y) ) \ | \ x \in [a,b] , \ \mu(x) \le y \le v(x) \}$$ 
or, with $y$ and $x$ changing place (and especially if it holds both ways), then I can do it on my own..... but what about when we've got a shape like $$E = \{ (x,y) \in \mathbb{R}^2 \ | \ 1 \le x^2 + y^2 \le 2 \}$$ 
How would we 'cut' it, and why would that 'cut' work? Generally, I'm quite interested in how the proof where $E$ is a 'simple region/domain' can then be applied to more complicated regions. 
 A: Consider $E_+=\{ (x,y)\in E : x\geq 0\}$ and $E_-=\{ (x,y)\in E | x\leq 0\}$. You can now use Green's theorem on each of these domains and then add up the result. 
$$\int _{\partial E_+} F\cdot dl + \int _{\partial E_-} F\cdot dl=\int_{E_+}(Q'_x-P'_y) \cdot dx \cdot dy + \int_{E_-}(Q'_x-P'_y) \cdot dx \cdot dy = \int_{E}(Q'_x-P'_y) \cdot dx \cdot dy$$
The union of the boundaries of $E_+$ and $E_-$ gives you the boundary of $E$ plus the two lines where we cut the annulus, namely $l=\{(x,0) | 1\leq x^2 \leq 2\}$. Since we use the anticlockwise direction for both boundaries, we get that for $E_+$ the direction on $l$ is from left to right, and on $E_-$ is from right to left. It follows that the two integrals on $l$ cancel each other and you are left only with the integral on the boundary of $E$ (this is easily seen if you draw this). Note also that if you set $\Gamma_r$ to be the curve of the circle of radius $r$ around the origin rotating anticlockwise, then you get that
$$\int_{E}(Q'_x-P'_y) \cdot dx \cdot dy = \int_{\Gamma_\sqrt{2}} F\cdot dl - \int_{\Gamma_1} F\cdot dl$$
or in other words on the boundary on the "outside" you go anticlockwise and on the "inside" you go in the clockwise direction.
Basically, this can be extended to all the "nice" regions you usually encounter. Just cut this regions to simple regions, use the theorem for each of these simple regions and then add up. The integrals on the curves where you cut will always cancel each other.
