Line Integral $\int_{C} \frac{x dy - y dx}{x^{2}+y^{2}}$ Find $$\int_{C} \frac{x dy - y dx}{x^{2}+y^{2}}$$ along the oriented broken line $C$ with vertices $(2,-2)$, $(4,4)$, $(-5,5)$ oriented counterclockwise. 
I noted that $C$ is a closed curve which passes through the origin, so Green's theorem cannot be applied here. Also, the vector field is not conservative, so the integral is nonzero. 
 A: Hint: If you take $C$ as a circle centered at origin with radius $r$, the result of integral does not change. (Why ?)
A: Set $F(x,y)=\left(\frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right)$, therefore
$$...=\int_C F\cdot d\ell\underset{green}{=}\iint_{\text{int}(C)}\text{rot}(F)dxdy$$
which it work because $F$ is $\mathcal C^1(\text{int}(C))$
A: Let $C_1$ be the segment from $(2,-2)$ to $(4,4)$. Let $C_2$ be the segment from $(4,4)$ to $(-5,5)$. Let $C_3$ be the segment from $(-5,5)$ to $(2,-2)$. 
Over the simply connected region $\{(x,y) \in \mathbb{R}^2 \ | \ x > 0\}$, the gradient of $f_1(x,y) = \arctan \dfrac{y}{x}$ is the vector field $\left(\dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)$, which is continuous over this region. 
Since the curve $C_1$ lies entirely in this region, we have $\displaystyle\int_{C_1}\dfrac{x\,dy-y\,dx}{x^2+y^2} = f_1(4,4)-f_1(2,-2)$.
Over the simply connected region $\{(x,y) \in \mathbb{R}^2 \ | \ y > 0\}$, the gradient of $f_2(x,y) = \text{arccot} \dfrac{x}{y}$ is the vector field $\left(\dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)$, which is continuous over this region. 
Since the curve $C_2$ lies entirely in this region, we have $\displaystyle\int_{C_2}\dfrac{x\,dy-y\,dx}{x^2+y^2} = f_2(-5,5)-f_2(4,4)$.
The curve $C_3$ is a straight line segment, which is parameterized by $x = t$, $y = -t$, for $t \in [-5,2]$. This is easy enough to evaluate (you should get that the integrand is $0$ on $C_3$). 
Finally, add the three pieces together to get the answer. 
