Where did I got wrong with this surface integral It appears that I don't quite have surface integrals like I thought I did.  The following is a problem from the back of the book (not homework because it wasn't prescribed but I'm working it to learn).  Here's the problem:
Integrate $G(x,y,z) = x + y + z$ over the surface cut from the first octant and the plane $2x + 2y + z = 2$.  This tells me that $S = F(x,y,z) = 2x + 2y + z = 2$ is implicitly given.  Therefore, my integrating formula is $\int \int_S G(x,y,z) d\sigma = \int \int_R G(x,y,z)\frac{\left| \nabla \mathbf{F} \right|}{\left| \nabla \mathbf{F} \cdot \mathbf{p} \right|}$.
For this problem, I have $\nabla \mathbf{F} = 2\mathbf{i} + 2 \mathbf{j} + \mathbf{k}$ and since the Region, $R$, I'm projecting onto is in the $xy$ plane, I'm using $\mathbf{p} = \mathbf{k}$.  Thus, $\nabla \mathbf{F} \cdot \mathbf{p} = 1$.  Also, since this surface is given implicitly, I have $z = 2 - 2x - 2y$.
Therefore, I have setting things up with $0 \le x \le 1$ and $0 \le x \le 1$:
$$
\begin{array}{rcl}
\int \int_R G(x,y,z)\frac{\left| \nabla \mathbf{F} \right|}{\left| \nabla \mathbf{F} \cdot \mathbf{p} \right|} & = & \int_0^1 \int_0^1 x + y + (2 - 2x - 2y)(\frac{\sqrt(3)}{1})dxdy \\
 & = & 3 \int_0^1 \int_0^1 2-2x-2y dxdy \\
 & = & 3
\end{array}
$$
However, my book tells me the answer is 2.
 A: The function to be integrated over the surface $F$ is
$$G(x,y,z)=x+y+z$$
and $F$ is given by 
$$z = 2-2x-2y.$$
A position vector pointing to the surface is
$$r(x,y)=\begin{bmatrix}
x\\
y\\
2-2x-2y
\end{bmatrix}.$$
With this vector the values of $G$ on the surface can be calculated:
$$G(r(x,y))=-x-y+2.$$
The partial derivatives of $r$ with respect to $x$ and $y$ are the following vectors:
$$\frac{\partial r}{\partial x}=\begin{bmatrix}
\ 1\\
\ 0\\
-2
\end{bmatrix}
\text{ and }\ 
\frac{\partial r}{\partial y}=\begin{bmatrix}
\ 0\\
\ 1\\
-2
\end{bmatrix}$$
If we set $z=0$ then the "shadow" of the surface on the $xy$ plane turns out to be
$$T=\{(x,y):0\le x\le 1, 0\le y\le 1-x\}.$$
The calculation of the surface integral of $G$ over $F$, according to  wiki goes like this
$$\iint_FGds=\iint_TG(r(x,y))\left\|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y}\right\|dxdy.$$
First, let's evaluate the vector product
$$\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y}=\begin{bmatrix}
\ 1\\
\ 0\\
-2
\end{bmatrix}\times \begin{bmatrix}
\ 0\\
\ 1\\
-2
\end{bmatrix}= \begin{bmatrix}
\ 2\\
-2\\
\ 1
\end{bmatrix}.$$
Then the absolute value of the vector product:
$$\left\|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y}\right\|=3.$$
Now, the integral
$$\iint_FGds=3\iint_T(-x-y+2)dydx.$$
Oops, here went the first typo...
And finally, considering the definition of $T$ we have
$$\iint_FGds=3\int_0^1\int_0^{1-x}(-x-y+2)dydx=$$
$$=3\int_0^1\left(\frac{3}{2}-2x+\frac{1}{2}x^2\right)dx=2.$$
Apparently the second mistake was the definition of $T$.
