Let $F$ be the vector field defined as $$ F = (x^2 - y, xy) $$ and $D$ the domain $$ D = \left\{(x,y)\in\mathbb{R}^2: x\leq x^2 + y^2 \leq 2x, y \geq 0\right\} $$ calculate the outward flux from the edge $\partial_+ D$ of $D$. I don't have the solution of this problem so I don't know if the result is the right one, but I'd like to know whether the method is right. So, I've split the edge of the domain in three curves $\gamma_1, \gamma_2, \gamma_3$: $$ \gamma_1 = \begin{cases}x(t) = \cos{t} + 1\\ y(t) = \sin{t}\end{cases}\qquad {\rm~for~~~} t\in[0,\pi] $$ $$ \gamma_2 = \begin{cases}x(t) = \frac{1}{2}\cos{t} + \frac{1}{2}\\ y(t) = \frac{1}{2}\sin{t}\end{cases}\qquad {\rm~for~~~} t\in[\pi,0] $$ $$ \gamma_3 = \begin{cases}x(t) = t\\ y(t) = 0\end{cases}\qquad {\rm~for~~~} t\in[1,2] $$ Now, for the divergence theorem $$ \iint_D (\nabla\cdot F) dxdy = \int_{\partial_+ D}F\cdot Nds $$ where $\partial_+ D = \gamma_1 + \gamma_2 + \gamma_3$ and $N$ is the vector normal to $\gamma_i$. Thus the flux becomes $$ \left[\int_0^{\pi}\cos{t}((\cos{t} + 1)^2-\sin{t})+(\cos{t} + 1)(\sin^2{t})dt + \int_{\pi}^0\frac{1}{2}\cos{t}\left(\left(\frac{1}{2}\cos{t} + \frac{1}{2}\right)^2-\frac{1}{2}\sin{t}\right) + \left(\frac{1}{4}\sin^2{t}\right)\left(\frac{1}{2}\cos{t} + \frac{1}{2}\right)dt + \int_1^2 (t^2) \cdot 0 - 0 dt \right]= 2\pi $$
Is this a licit method? I'm a little bit confused because most of the books I have report the divergence theorem only for vector fields$:\mathbb{R}^3\rightarrow\mathbb{R}^3$ and here I have a$:\mathbb{R}^2\rightarrow\mathbb{R}^2$ one.