compute line integral using Greens theorem

Question

The question is: $$\int_\gamma \frac{(x^2+y^2-2)\ dx+(4y-x^2-y^2-2)\ dy}{x^2+y^2-2x-2y+2}$$ $$\gamma:y=2\sin\frac{\pi x}{2} \quad \text{from}\ (2,0)\ \text{to}\ (0,0)$$ Here how i have tried to solve it: I thought that due to the singularity at $(1,1)$ it would be best to use Greens theorem with one circle around the singularity call it $\sigma$ and the line segment between $(0,0)$ to $(2,0)$ and call it $\varphi$ and then $\int_\gamma=-\int_\sigma-\int_\varphi$ but the problem is that the first integral (circle around the point$(1,1)$) is very complicated with $x=1+Rcost$ and $y=1+Rsint$. Do any of you guys have any suggestion or am i on the right track at all?

Answer 1

Hint:

$\frac{\partial}{\partial y} \bigg(\frac{x^2+y^2-2} {(x-1)^2 + (y-1)^2}\bigg) - \frac{\partial}{\partial x} \bigg(\frac{4y - x^2 - y^2 - 2} {(x-1)^2 + (y-1)^2}\bigg) = 0, (x,y) \ne (1,1)$

The vector field is conservative in a domain that does not contain point $(1,1)$.

Point $(1,1)$ is below path $\lambda: y = 2 \sin \frac{\pi x}{2}$. So if we choose a path above $\lambda$, we will have a domain combined with $\lambda$, that does not contain point $(1,1)$. That will allow us to take advantage of path independence.

We can choose circular path $(x-1)^2 + (y-1)^2 = 2$ in anti-clockwise direction from $(2,0)$ to $(0,0)$. We can parametrize it as

$x = 1 + \sqrt 2 \cos t, y = 1 + \sqrt2 \sin t, -\frac{\pi}{4} \leq t \leq \frac{5 \pi}{4}$.

$dx = - \sqrt2 \sin t \ dt, dy = \sqrt2 \cos t \ dt$

The denominator of the vector field reduces to $2$.

Can you take it from here?