I am trying to calculate the integral of the vector field $F = \begin{bmatrix}-y & x\end{bmatrix}^T$ in the point set $S = \{(x, y) \in \mathbb{R}^2\mid 1 \leq x^2 + y^2 \leq 4\}$ in two ways with Green's formula. Currently I am having both computational and conceptual issues.
The computational issue to which I ran into is that if I am not mistaken, $\int\int_S(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})dA = \int_1^2\int_1^{\sqrt{4 - x^2}}2dydx = \int_1^2\sqrt{4 - x^2} - 1dx$. From this I don't really see any other way forward than to substitute $x = 2\sin(u)$ and recalculate the limits. Is this really the only way forward, or is there a better trick to this?
The conceptual issue I am having is that how do even start to compute the boundary integral? Sure, $\partial S = C_1 \cup C_4$, where $C_1 = \{(x, y) \in \mathbb{R}^2\mid x^2 + y^2 = 1\}$ and $C_4 = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 4\}$. How can I parametrize $\partial S$? I am bit rusty with path integrals, so does it hold that $\int_{\partial S}F_1 dx + F_2 dy = (\int_{C_1}F_1 dx + F_2 dy) + (\int_{C_4}F_1 dx + F_2 dy)$?
Edit: The question I am solving is that Calculate the integrals of Green's theorem w.r.t vectorfield $F$ and pointset $S$.
Your bounds for the region is not correct. The region is the area between two circles $x^2+y2 = 1$ and $x^2+y^2 = 4$. If you do it in cartesian coordinates, you will have to split your integral. It is easier to do it in polar coordinates, the limits should be $1 \leq r \leq 2, 0 \leq \theta \leq 2 \pi$ and $dA = r \ dr \ d\theta$.
As far as the question, I think it could be written more clearly.
To apply Green's Theorem, the curve has to be positively oriented, piecewise smooth, simple and closed.
So if we define curve $C$ as going from point $A$ along $C_1$ in anti-clockwise direction to point $B$ and then to point $C$, along $C_2$ in clockwise direction (region continues to be on your left) back to point $C$, then to point $B$ and then along $C_1$ in anticlockwise direction back to point $A$, we have a region that meets all the conditions we spoke about earlier to apply Green's theorem.
When we are doing line integral, note that the line integral along $BC$ and $CB$ will cancel out.
So the line integral along $C$ is equivalent to line integral along $C_1 \cup C_2$ where $C_1$ is anti-clockwise along $r = 2$ and $C_2$ is clockwise along $r = 1$.
You can parametrize $C_1$ as $(2 \cos t, 2 \sin t), 0 \leq t \leq 2\pi$ and $C_2$ as $(\cos t, \sin t), 2\pi \leq t \leq 0$.