Surface integral through a cube. I am doing some fluid dynamics work and the question asks: 

Consider the mass flow vector:
$\rho \vec{u} = (4x^2y,xyz,yz^2)$
Compute the net mass outflow through the cube formed by the planes x=0, x=1, y=0, y=1, z=0, z=1

So I figure that in order to find the net mass outflow I compute the surface integral of the mass flow normal to each plane and add them all up. That is:
$\iint_S (\rho \vec{u} \cdot \hat{n}) dA$
To make the work easier I use the divergence theorem, to replace the surface integral with a volume integral and a much shorter integration:
Volume Integral:
$ \rightarrow \iint_S (\rho \vec{u} \cdot \hat{n}) dA =
\iiint_V (\nabla \cdot \rho \vec{u})dV$
$ \rightarrow \iiint_V(\nabla \cdot (4x^2y,xyz,yz^2))dV$
$ \rightarrow \int \limits_0^1 \int \limits_0^1 \int \limits_0^1(8xy+xz+2yz)dxdy dz$
$ \rightarrow\int \limits_0^1\int \limits_0^1(4y +\frac{z}{2}+2yz) dy dz$
$ \rightarrow\int \limits_0^1(2 +\frac{z}{2}+z) dz$
$ \rightarrow \iiint_V (\nabla \cdot \rho \vec{u})dV = \frac{11}{4}$
I am pretty sure this part is done correctly, however just to check I do the question the long way as well; obtaining 6 surface integrals: 
Surface Integral:
$\hat{n} = (1,0,0) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (4x^2y) dy dz = \int \limits_0^1 2x^2dz = 2x^2 $
$\hat{n} = (-1,0,0) \; \;  \int \limits_0^1 \int \limits_0^1 (-4x^2y) dy dz = \int \limits_0^1 -2x^2dz = -2x^2 $
$\hat{n} = (0,1,0) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (xyz) dx dz = \int \limits_0^1 \frac{yz}{2}dz = \frac{y}{4} $
$\hat{n} = (0,-1,0) \; \; \int \limits_0^1 \int \limits_0^1 (-xyz) dx dz = \int \limits_0^1 \frac{-yz}{2}dz = \frac{-y}{4} $
$\hat{n} = (0,0,1) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (yz^2) dy dx =  \frac{z^2}{2} $
$\hat{n} = (0,0,-1) \; \; \int \limits_0^1 \int \limits_0^1 (-yz^2) dy dx =  \frac{-z^2}{2} $
Summing all of these results in zero which contradicts the answer I obtained earlier of $\frac{11}{4}$. What am i doing wrong can someone please explain, I am sure it is some trivial mistake but I have checked it over and am not sure. Thanks to all for your time!
 A: I have determined my mistake I have to input the values of the planes to get my final answer :
Surface Integral:
$\hat{n} = (1,0,0) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (4x^2y) dy dz = \int \limits_0^1 2x^2dz = 2x^2 = \;2  \rightarrow at \; x = 1  $
$\hat{n} = (-1,0,0) \; \;  \int \limits_0^1 \int \limits_0^1 (-4x^2y) dy dz = \int \limits_0^1 -2x^2dz = -2x^2 = \;0  \rightarrow at \; x = 0 $
$\hat{n} = (0,1,0) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (xyz) dx dz = \int \limits_0^1 \frac{yz}{2}dz = \frac{y}{4} = \frac{1}{4} \;  \rightarrow at \; y = 1 $
$\hat{n} = (0,-1,0) \; \; \int \limits_0^1 \int \limits_0^1 (-xyz) dx dz = \int \limits_0^1 \frac{-yz}{2}dz = \frac{-y}{4} = \;0  \rightarrow at \; y = 0  $
$\hat{n} = (0,0,1) \; \; \; \; \; \int \limits_0^1 \int \limits_0^1 (yz^2) dy dx =  \frac{z^2}{2} = \; \frac{1}{2}  \rightarrow at \; z = 1 $
$\hat{n} = (0,0,-1) \; \; \int \limits_0^1 \int \limits_0^1 (-yz^2) dy dx =  \frac{-z^2}{2} = \; 0 \rightarrow at \; z = 0  $
Summing gets you $\frac{11}{4}$ as required,
