We consider the space $D$ that we get if we remove from the square $[-2,7]\times [-3,6]$ the open discs with center the point $(0,0)$ and radius $1$ and with center $(3,3)$ and radius $2$.
Calculate $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )$$ where $\sigma_1, \sigma_2, \sigma_3$ are the boundary curves of $D$ with the positive orientation in relation to $D$.
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Can we calculate directly the curve integrals or do we have to apply Green's Theorem?
If we apply Green's Theorem we get the following: $$\sum_{j=1}^3\oint_{\sigma_j}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\iint_D\left (\frac{\partial{\left (\frac{x}{x^2+y^2}\right )}}{\partial{x}}-\frac{\partial{\left (-\frac{y}{x^2+y^2}\right )}}{\partial{y}}\right )dxdy=\iint_D\left (\frac{-x^2+y^2}{(x^2+y^2)^2}+\frac{x^2-y^2}{(x^2+y^2)^2}\right )dxdy=\iint_D\left (\frac{0}{(x^2+y^2)^2}\right )dxdy=0$$ right?
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EDIT:
$\sigma_1$ is the parametrization of the rectangle, $\sigma_2$ the parametrization of the small circle and $\sigma_3$ of the big circle.
We have that $\sigma_2=(\cos t,-\sin t), t\in [0, 2\pi]$ and $\sigma_3=(2\cos t+3,-2\sin t+3), t\in [0, 2\pi]$.
We consider for the reactangle the parametraization $\sigma_1$ as the union of the following: $$\rho_1=(t, -3), -2\leq t\leq 7 \\ \rho_2=(7, t), -3\leq t\leq 6 \\ \rho_3=(7-t, 6), 0\leq t\leq 9 \\ \rho_4=(-2, 6-t), 0\leq t\leq 9$$
For the curve integral over $\sigma_1$ I have found the following: $$\oint_{\sigma_1}=\oint_{\rho_1}+\oint_{\rho_2}+\oint_{\rho_3}+\oint_{\rho_4}=\ldots =\int_{-2}^7-\frac{3}{t^2+9}dt+\int_{-3}^6\frac{7}{49+t^2}dt+\int_0^9\frac{6}{(7-t)^2+36}dt+\int_0^9\frac{2}{4+(6-t)^2}dt$$
The curve integral over $\sigma_2$ is equal to $$\oint_{\sigma_2}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-(-\sin t)\cdot (-\sin t)+\cos t (-\cos t)\right )dt=\int_0^{2\pi}dt=2\pi$$
The curve integral over $\sigma_3$ is equal to $$\oint_{\sigma_3}\left (-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right )=\int_0^{2\pi}\left (-\frac{-2\sin t+3}{-12\sin t+12\cos t+22}\cdot (-2\sin t)+\frac{2\cos t+3}{-12\sin t+12\cos t+22}\cdot (-2\cos t)\right )dt=\int_0^{2\pi}\left (\frac{-4\sin^2 t+6\sin t-4\cos^2 t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=\int_0^{2\pi}\left (\frac{-4+6\sin t-6\cos t}{-12\sin t+12\cos t+22}\right )dt=0$$
Can that be correct?