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,
Let $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ be a vector field on a region $D \subset \mathbb{R}^2$. As far as I am aware, there is no single concept that mathematicians have designated as "the" integral of $\mathbf{F}$ on $D$. As such, we will have to define it ourselves.
I think the most reasonable definition would be
$$\iint_D \mathbf{F}\,dA = \left( \iint_D P\,dA, \iint_D Q\,dA \right)$$
but this results in a vector, and the OP wants a scalar. To that end, the following integrals are all very natural, and could be taken to be the definition of "the integral" of $\mathbf{F}$ on $D$:
\begin{align*}
\iint_D \text{div}(\mathbf{F})\,dA & = \iint_D (P_x + Q_y)\,dA \tag{1} \\
\iint_D \text{curl}(\mathbf{F}) \cdot \mathbf{k}\,dA & = \iint_D (Q_x - P_y)\,dA \tag{2} \\
\iint_D \Vert \mathbf{F}\Vert\,dA & = \iint_D \sqrt{P(x,y)^2 + Q(x,y)^2}\,dA \tag{3} \\
\left\Vert\iint_D \mathbf{F}\,dA\right\Vert & = \sqrt{ \left( \iint_D P(x,y)\,dA \right)^2 + \left( \iint_D Q(x,y)\,dA \right)^2 } \tag{4}
\end{align*}
Which one of these we designate as "the" integral of $\mathbf{F}$ on $D$ is a matter of taste, or perhaps a matter of what we want our integral to capture.
As far as I can tell, the OP wants the integral of $\mathbf{F}$ on $D$ to measure how "much" of $\mathbf{F}$ there is on $D$. Here is how to do this: First partition $D$ into sub-rectangles $P_1, \ldots, P_n$ of areas $\Delta A_i$, and choose sample points $(x_i, y_i) \in P_i$. Second, define a scalar quantity called $$\text{Much}(\mathbf{F})(x,y)$$
that measures how "much" of $\mathbf{F}$ there is at $(x,y)$. Finally, form the Riemann sum
$$\sum_i \text{Much}(\mathbf{F})(x_i,y_i)\,\Delta A_i$$
so that the limit as the partition gets finer and finer is your integral:
$$\iint_D \text{Much}(\mathbf{F})\,dA = \lim \sum_i \text{Much}(\mathbf{F})(x_i,y_i)\,\Delta A_i.$$
In order for this to work, you need to know (at the very least) what you want $\text{Much}(\mathbf{F})$ to mean for, say, a constant vector field like $\mathbf{F}(x,y) = (1,0)$ or $\mathbf{F}(x,y) = (3,4)$.
Heuristically, it follows that
$$\text{Much}(\mathbf{F})(x,y) = \lim_{\text{Area}(D) \to 0} \frac{ \iint_D \text{Much}(\mathbf{F})\,dA }{\text{Area}(D)}$$
where the regions $D$ contain the point $(x,y)$, and are shrinking reasonably nicely to the point $(x,y)$ in the limit.
Notice that if we take $\text{Much}(\mathbf{F})$ to mean $\text{div}(\mathbf{F})$, then we get
$$\text{div}(\mathbf{F})(x,y) = \lim_{\text{Area}(D) \to 0} \frac{ \iint_D \text{div}(\mathbf{F})\,dA }{\text{Area}(D)} = \lim_{\text{Area}(D) \to 0} \frac{\oint_{\partial D} \mathbf{F} \cdot \mathbf{n}\,ds}{\text{Area}(D)}$$
by using a form of Green's Theorem (or Stokes' Theorem) in the last equality.
Best Answer
Here's the beautiful result you need to use. Denote by $\phi_t$ the flow of your vector field $\vec v$ — i.e., $\dfrac d{dt} \phi_t(\vec x) = \vec v(\phi_t(\vec x))$. If you start with a region $\Omega$ (the unit cube, in your case), let $\Omega_t = \phi_t(\Omega)$ be the region obtained by flowing for time $t$. Then the rate at which the volume of $\Omega_t$ is changing at time $t$ is $\displaystyle\int_{\Omega_t} \text{div}\,\vec v\,dV$. (This is a variant of the divergence theorem, as the flux of $\vec v$ across the boundary of $\Omega_t$ tells you the rate at which the volume of $\Omega_t$ is changing.)
In your first example, $\text{div}\,\vec v = 0$, so the volume remains constant. In your second example, $\text{div}\,\vec v = 1$, so you must solve the ODE $\dfrac{d\,\text{vol}}{dt} = \text{vol}$.