Example 11. Calculate the integral $I=\oint_{L} \frac{y}{x^{2}+y^{2}} d x-\frac{x}{x^{2}+y^{2}} d y$, where $L$ is the circle:

a) $x^{2}+y^{2}=1$, b) $(x-1)^{2}+y^{2}=1$, c) $(x-1)^{2}+(y-1)^{2}=1$.
Solution. Let's check the conditions of Green's theorem.

$$
P=\frac{y}{x^{2}+y^{2}}, Q=-\frac{x}{x^{2}+y^{2}}, \frac{\partial P}{\partial y}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \frac{\partial Q}{\partial x}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}}
$$

If $(x, y) \neq(0,0)$, then the functions $P$ and $Q$ are continuous along with their partial derivatives and $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$.

a) In the circle $x^{2}+y^{2}=1$, the conditions of Green's theorem are not satisfied, since the functions $P$ and $Q$ are not defined at the point $O(0,0)$. Therefore, $I$ must be computed directly. We have: $x=\cos t, y=\sin t, 0 \leqslant t \leqslant 2 \pi ; d x=-\sin t d t$, $d y=\cos t d t, x^{2}+y^{2}=1$. Therefore,

$$
I=\int_{0}^{2 \pi}\left(-\sin ^{2} t-\cos ^{2} t\right) d t=-2 \pi
$$

b) In this case, Green's formula is not applicable: the circle $(x-1)^{2}+y^{2}=1$ passes through the origin, and the integrand is undefined at this point. Therefore, $I$ does not exist in this case.

c) In the circle $(x-1)^{2}+(y-1)^{2}=1$, the conditions of Theorem 5 are satisfied, and $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$. Therefore, $I=0$.

Answer. a) $-2 \pi ;$ b) does not exist; c) 0 .