Flux of vector field across surface S The question I am working on is: 
Let S be the part of the plane $1\!x + 3\!y + z = 3$ which lies in the first octant, oriented upwards. Find the flux of the vector field $\mathbf{F} = 3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$ across the surface S.
My work so far:
$\int F dS$ 
Parametrization: $T: (x, y, 3-3y-x)$
$T_x  (1, 0, -1)$
$T_y  (0,1,-3)$
$T_x \times T_y (-1,3,1)$
This is the correct orientation (going upwards).
$\int \int (3,2,3) \cdot (-1,3,1)dxdy = -3+6+3 = 6$
$\int\int6dxdy$ with $0 \leq x \leq 3-3y$ and $0 \leq y \leq 1/3$
This isn't yielding me the correct answer. I think my bounds are incorrect but I'm a bit stumped on how to find the correct ones. 
 A: The normal is $n = (1,3,1)$, there was the error.
The projected area onto the $x$-$y$-plane is $A_p = 3/2$.
The flux is $\Phi=(F\cdot n)\,A_p =  ((1,3,1) \cdot (3,2,3))\, 3/2 = 18$.

(Large Version)
A: Let the parametric equation of the surface be
$${\bf{r}} = {\bf{r}}(u,v)$$
Then the flux across the surface is give by
$$\eqalign{
  & \varphi  = \int\limits_S {{\bf{F}}({\bf{x}}(u,v)).d{\bf{a}}}   \cr 
  & d{\bf{a}} = \frac{{\partial {\bf{r}}}}{{\partial u}} \times \frac{{\partial {\bf{r}}}}{{\partial v}}dudv \cr} $$
Now, consider the following
$$\eqalign{
  & {\bf{r}} = x{\bf{i}} + y{\bf{j}} + \left( {3 - x - 3y} \right){\bf{k}}  \cr 
  & \frac{{\partial {\bf{r}}}}{{\partial x}} = {\bf{i}} - {\bf{k}},\,\,\,\,\,\frac{{\partial {\bf{r}}}}{{\partial y}} = {\bf{j}} - 3{\bf{k}}  \cr 
  & \frac{{\partial {\bf{r}}}}{{\partial x}} \times \frac{{\partial {\bf{r}}}}{{\partial y}} = \left( {{\bf{i}} - {\bf{k}}} \right) \times \left( {{\bf{j}} - 3{\bf{k}}} \right) = {\bf{i}} + 3{\bf{j}} + {\bf{k}}  \cr 
  & d{\bf{a}} = \frac{{\partial {\bf{r}}}}{{\partial x}} \times \frac{{\partial {\bf{r}}}}{{\partial y}}dxdy = \left( {{\bf{i}} + 3{\bf{j}} + {\bf{k}}} \right)dxdy  \cr 
  & {\bf{F}}({\bf{x}}) = 3{\bf{i}} + 2{\bf{j}} + 3{\bf{k}}  \cr 
  & {\bf{F}}({\bf{x}}).d{\bf{a}} = 3 + 6 + 3 = 12\,dxdy\cr} $$
and hence
$$\eqalign{
  & \varphi  = \int_{y = 0}^1 {\left( {\int_{x = 0}^{3 - 3y} {12dx} } \right)dy}  = \int_{y = 0}^1 {12(3 - 3y)dy}  = 36\int_{y = 0}^1 {(1 - y)dy}   \cr 
  & \,\,\,\,\, = 36\left. {\left( {y - \frac{{{y^2}}}{2}} \right)} \right|_0^1 = 18 \cr} $$

Your Mistakes
1) The normal was wrong.
2) The bounds was wrong.
