$F(x,y)=(x^2-2xy,y^2-x^2+e^{y^2})$ is a vector field. $ɣ_{1}:$ it is the oriented segment that unites $A=(1,1)$ with $B=(-1,1)$ ($A → B$) $ɣ_{2}:$ parabola section that unites $B$ with $A$ ($B → A$) $ɣ_{3}:$ circumference $(x-2)^2+(y-1)^2=1$ traveled from point A clockwise. calculate:
$a)$ $$\int_{ɣ_{1}} \vec{F}d \vec{r}$$
$b)$ $$\int_{ɣ_{2}} \vec{F}d \vec{r}$$
$c)$ $$\int_{c} \vec{F}d \vec{r}$$ $c=ɣ_{1}Uɣ_{2}Uɣ_{3}$
I haven't problem with $a)$ and $b)$. For the first I parameterized the curve as $\vec{r}(t)=(1-2t,1)$ with $tϵ[0,1]$ then $ $$\int_{ɣ_{1}} \vec{F}d \vec{r}$$ $=$ $ $$\int_{0}^{1} \vec{F}(\vec{r}(t))\vec{r}'(t) dt$$$=-7$
Regarding to $b)$ since $\vec{F}$ is a conservative field because $rot \vec{F} = \vec{0}$ then $$\int_{ɣ_{1-}} \vec{F}d \vec{r}$$ $=$ $$\int_{ɣ_{2+}} \vec{F}d \vec{r}$$ $=-7$
but I don't know how to do $c)$. I already appreciate all help and correction.
Parametrize $\gamma_3$ as $$\gamma_3(t)=\big(2+\cos(t),1+\sin(t)\big)\text{ for }0\leq t\leq2\pi$$ and do as $(a)$.