Let $S$ be the surface where $x + 2y + 3z = 0$ and $-1 \leq y \leq 1, 0 \leq z \leq 1$. Compute the surface integral $$\int \int_S (2x,3y-x,1-2y) \cdot \mathbf{\hat{N}}dS$$ where the unit normal vector $\mathbf{\hat{N}}$ points upwards.
My attempt: Using that $f = z = -x/3 -2y/3$ we find that $\mathbf{\hat{N}}$ points in the direction $(-f_x,-f_y,1) = (1/3,2/3,1)$. Further, by looking at the projection of $f$ onto the $xy$-plane, we set $z = 0$ and see that $y = -x/2$, which means $-2 \leq x \leq 2$ when $-1 \leq y \leq 1$as given.
We therefore obtain $\mathbf{F} \cdot \mathbf{\hat{N}}dS = 1dydx$. Further, when integrating over the region in the $xy$-plane given by $y = -x/2$ for $-2 \leq x \leq 2$ and $-1 \leq y \leq 1$ we obtain $0$, since there is as much area above and below the $y$-axis. Hence, I obtain the answer zero for the surface integral, however the correct answer is $6$.
Is anyone able to see where I'm wrong?
A parametrization of the surface is $$ [-1,1]\times [0,1]\ni (y,z)\mapsto \boldsymbol{\phi}(y,z)=\begin{pmatrix}-2y-3z\\y\\z\end{pmatrix}\,. $$ I have shown in this answer that $$ \boldsymbol{\phi}_y\times \boldsymbol{\phi}_z=\begin{pmatrix}-2\\1\\0\end{pmatrix}\times \begin{pmatrix}-3\\0\\1\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix} $$ is the unnormalized normal vector $\mathbf{N}$ to the surface which we use to integrate in this parametrization: \begin{align} &\int_S\mathbf{F}\cdot\mathbf{\hat N}\,dS=\int_{-1}^1\int_0^1\mathbf{F}\cdot\begin{pmatrix}1\\2\\3\end{pmatrix}\,dz\,dy=\int_{-1}^1\int_0^1(2x+6y-2x+3-6y)\,dz\,dy\\ &=\int_{-1}^1\int_0^13\,dz\,dy=6\,. \end{align}
Another parametrization of the surface is $$ [-2y-3,-2y]\times [-1,1]\ni (x,y)\mapsto \boldsymbol{\psi}(x,y)=\begin{pmatrix}x\\y\\\frac{-2y-x}3\end{pmatrix}\,. $$ Note that $(x,y)$ are not from a rectangle anymore. Therefore a slightly sloppy notation.
How to find the bounds on $x$: we want
$$ z=\frac{-2y-x}{3} $$ to be in $[0,1]\,.$ The two inequalities that are equivalent to this are $$ -2y-3\le x\le -2y\,. $$ The normal vector to the surface is now $$ \boldsymbol{\psi}_x\times \boldsymbol{\psi}_y=\begin{pmatrix}1\\0\\-\frac13\end{pmatrix}\times \begin{pmatrix}0\\1\\-\frac23\end{pmatrix}=\frac13\begin{pmatrix}1\\2\\3\end{pmatrix} $$and the integral is \begin{align} &\int_S\mathbf{F}\cdot\mathbf{\hat N}\,dS=\frac13\int_{-1}^1\int_{-2y-3}^{-2y}\mathbf{F}\cdot\begin{pmatrix}1\\2\\3\end{pmatrix}\,dx\,dy=\frac13\int_{-1}^1\int_{-2y-3}^{-2y}3\,dx\,dy=6\,. \end{align}