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.
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 $T$.