bounds for surface integral of a plane I need to calculate the next surface integral, but i'm having troubles with the bounds;
$$\iint -2 \, dS,$$
where $S$ is the part of the plane $x+2y+z=2$ that is cut off in the first octant.
My function is $\vec{\rm F}\left(x,y,z\right)=\left(y - x,x - z,x - y\right)$. I found the $-2$ with Stokes' theorem and i'm sure that part is correct. Now I still have to prove that the surface integral is $1$. Can someone please help me ?.
 A: 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.
