Surface integral- getting different result using two methods I'm doing my homework and I came to conclusion I'm not sure is right.

I need to find $$\iint_S x dydz+y^2dxdz+z^2dxdy$$ where $S$ is outer side of surface $x=z^2$, and $1\le y \le3$ and $x\le9$.

Now, since for calculating first part of integral ($xdydz$) I need to project surface on plane $x=0$ I will have integral equal $0$. And I get same conclusion for two other parts of integral. Am I right? In the end integral equals $0$?
UPDATE:
But using divergence theorem I get:
$I=\int_1^3 dy \int_{-3}^3 dz \int_{z^2}^9 (1+2y+2z) dx = 360$.
I'm not sure which part I'm doing wrong?
 A: I am not familiar with that way of computing surface integrals. I assume you want to compute the flux of the vector field $\vec{F} = (x,y^2,z^2)$ across that surface.
We can parametrize the surface with
$$\mathbf{r}(u,v) = (u^2, v, u), \quad \begin{cases} -3 \leq u \leq 3, \\ 1 \leq v \leq 3. \end{cases}$$
Computing $\partial \mathbf{r} / \partial u, \partial \mathbf{r} / \partial v$ and taking the cross product we find
$$\vec{N} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = (-1,0,2u).$$
Restricting our vector field to the surface yields
$$\vec{F} (\mathbf{r}(u,v)) = (u^2, v^2, u^2),$$
and the surface integral becomes
$$
\begin{align}
\iint\limits_{\Sigma} \vec{F} \cdot d \vec{S} & = \int_1^3 \int_{-3}^3 (u^2, v^2, u^2) \cdot (-1, 0, 2u) \, du \, dv \\
 & = \int_1^3 \int_{-3}^3 -u^2 +2u^3 \, du \, dv \\
 & = 2 \int_{-3}^3 2u^3 - u^2 \, du \\
 & = 2 - \frac{u^3}{3} \bigg\vert_{-3}^3 \\
 & = 2 \cdot (-2) \cdot 9 = -36.
\end{align}
$$
I used that $2u^3$ is an odd function in a symmetric interval centered at the origin, therefore its integral is zero, leaving just the $-u^2$ to be integrated, an even function. Here's a picture of the surface, courtesy of Mathematica:

The problem of applying the divergence theorem is that it isn't a closed surface, hence we cannot use it.
