I need to show that moving the curve to a simply connected region, the integral of the field over the curve will be $0$.
Given $F(x, y) = ((-y)/(x^2+y^2 ), (x/(x^2+y^2 ))$, and $γ$ circle of radius $1$ centered at $(2, 0)$; i.e., $γ ( t ) = (2 + \cos t ,\sin t )$, $0 ≤ t ≤ 2π$. Note that $γ$ is entirely contained in the open half-plane $x > 0$, the half-plane is simply connected, and $F$ is defined everywhere in the half-plane.
How do I show that $$∮F \cdot ds = 2∮_0^{2\pi}(1+2\cos φ)/(5+4\cosφ)\,dφ$$ and subsequently $$∮_0^{\pi}F \cdot ds = π −π = 0 $$ (I know this is true because each path taken is independent)
I tested $∂M/∂y=∂N/∂y$ for conservative field, it checks out. Then I derive the potential functions. However, it seems I need to use the polar coordinates, since I have a circle with points $(3,0), (2,1), (1,0), (2,-1)$ on the $x$-axes and $y$-axes. I just don’t know how to implement what I know.
HELP PLEASE!!