To make everything explicit:
Knowing now that $[0,3]\times[0,2]$ is a rectangle in the $xy$-plane, we also have our limits of integration. We write $x^2yx=x^2y(1+2x+3y)=x^2y+2x^3y+3x^2y^2$ and integrate over the region $\{(x,y)|0\leq x\leq 3, 0 \leq y \leq 2\}$. Thus our integral is
$\displaystyle\int_{x=0}^3\int_{y=0}^2(x^2y+2x^3y+3x^2y^2)dydx$.
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
Here is the solution,
$$ \iint f(x,y,z) dS=\int_{0}^{1}\int_{0}^{1-x} (x+y) \sqrt{1+(\frac{dz}{dx})^{2}+(\frac{dz}{dy})^2}dy\,dx$$
$$ = \sqrt{3} \int_{0}^{1}\int_{0}^{1-x} (x+y)dydx= \frac{\sqrt{3}}{3}. $$
Note that, S is the first octant part of the plane, then you will have the region in xy-plane bounded by the $x$-axis, $y$-axis and the $y=1-x$, $x\geq 0, y\geq 0.$