Evaluating the surface integral $\iint_\Sigma \mathbf{f} \cdot d \mathbf{a}$ where $\mathbf{f}(x,y,z)=(x^2,xy,z)$ 
Evaluate the surface integral $\iint_\Sigma \mathbf{f} \cdot d \mathbf{a}$ where $\mathbf{f}(x,y,z)=(x^2,xy,z)$ and $\Sigma$ is the part of the plane $6x+3y+2z=6$ with $x,y,z\geq 0$.

I changed the function to parametric form, $\mathbf{r} (u,v)=(u^2,uv,3-3u-3v/2)$.
Then $\mathbf{r}_u\times \mathbf{r}_v=(-3v/2+3u,3u,2u^2)$.
So the integral is $\int^2_0\int^1_0(-3u^2v/2+6u^2-3u^3)\,dudv=3/2$.
But the answer turn out to be $15/4$, what's wrong with my solution?

You can choose a parametrization of the plane as
$$\textbf{r}(x,y) = \left( x, y, \frac{6-6x-3y}{2} \right).$$
We find the expression of the field in the surface by finding the composite, yielding
$$\textbf{f}(\textbf{r}(x,y)) = \left( x^2, xy, \frac{6-6x-3y}{2} \right).$$
To find the projected area in the $xy$ plane we make $z=0$, finding
$$\frac{6-6x-3y}{2} = 0 \iff 6x+3y=6 \iff 2x+y=2.$$
The normal vector is computed as $\textbf{r}_x \times \textbf{r}_y = (3, 3/2, 1)$.
We have that the surface integral is
$$\iint\limits_{\Sigma} \textbf{f} \cdot d \textbf{a} = \int_0^1 \int_0^{2-2x} 3x^2 + \frac{3xy}{2} + \frac{6-6x-3y}{2} \, dy \, dx.$$
