The curl of your vector field is
$$\nabla \times \vec{F} = (0,-1,0).$$
Therefore, by Stokes's theorem we know that
$$\int\limits_{C} \vec{F} \cdot d \vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot d \vec{S}.$$
Since this is a plane the normal is $(1,2,1)$. Plugging these yields
$$\iint\limits_{S} (\nabla \times \vec{F}) \cdot d \vec{S} = \iint\limits_{D} (-2) \, dA = -2 \iint\limits_{D} \, dA.$$
I'm denoting by the $D$ the area in the plane. This means that circulation of $\vec{F}$ is $-2$ times the area in the plane projected by the surface.
Whenever we integrate over surfaces the basic idea is to parametrize it by a region in $\mathbb{R}^2$. When we perform the integration, the idea is to "pull back" the integration on the surface to integration in $\mathbb{R}^2$, which is what we know.
This is a slanted plane, but if you managed to view it from the $z$ axis you would not be able to distinguish it from a triangle in the $xy$ plane. That is the projected area. It is the region in the plane bound by
$$
\begin{cases}
x \geq 0, \\
y \geq 0, \\
x+2y \leq 2.
\end{cases}
$$
The last inequality was found setting $z=0$ in the plane equation, giving the boundary. The area of this triangle is simple: it has $2$ units as base and $1$ as height, therefore
$$\int\limits_{C} \vec{F} \cdot d \vec{r} = -2 \iint\limits_{D} \, dA = -2 \cdot \left( \frac{1}{2} \cdot 2 \cdot 1 \right) = -1.$$
Minus sign is due to orientation. I believe we used the counterclockwise orientation all along.
Best Answer
$\displaystyle \iint_{S}\frac{1}{\sqrt{1+4z}} \ dS$
Surface $S$ is $z = x^2+y^2$, bound above by plane $z = 3 - 2y$
Now we know that,
$\displaystyle \iint_{S}\frac{1}{\sqrt{1+4z}} \ dS = \iint_{E}\frac{1}{\sqrt{1+4z}} \sqrt{1+ \bigg(\frac{\partial z}{\partial x}\bigg)^2 + \bigg(\frac{\partial z}{\partial y}\bigg)^2} \ \ dA$
where $E$ is the projection of the surface in $XY$ plane and $dA$ is the area element of the projection.
This indeed simplifies to $\displaystyle \iint_{E} 1 \cdot dA$
Now to find projection of the surface in $XY$ plane,
$z = x^2+y^2 = 3 - 2y \implies x^2+(y+1)^2 = 4$
We substitute $x = r \cos\theta, y = -1 + r\sin\theta$ and we have $0 \leq r \leq 2, 0 \leq \theta \leq 2\pi$
So the integral becomes $\displaystyle \int_0^{2\pi} \int_0^2 r \ dr \ d\theta = 4 \pi$