<s>[INST] Divergence theorem approach The question is 
I am unable to solve this question. I am confused whether to use the left or right side of the divergence theorem and what region is being asked.
Divergence theorem:-
$\iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\,dV= \iint {\displaystyle \scriptstyle S}  {\displaystyle (\mathbf {F} \cdot \mathbf {n} )\,dS.} $
 [/INST] The Divergence theorem fundamentally states that the divergence is a measure of how much vector field flux is coming out of a point, so we can add all of the divergence within a closed region to get the total flux leaving the boundary of the region:

The solid region being considered can be described as:
\begin{align*}
R &= \left\{ (x,y,z) \in \mathbb{R}^3 \mid 0 \leq x \leq 6,\, 0 \leq y \leq 3,\, 0 \leq z \leq 3-\frac{1}{2}x \right\} \\
&= \left\{ (x,y,z) \in \mathbb{R}^3 \mid 0 \leq x \leq 6-2z,\, 0 \leq y \leq 3,\, 0 \leq z \leq 3 \right\}
\end{align*}
displayed below:

We are asked to find the flux coming out of $\partial R$, the boundary of $R$, which consists of 5 faces, of the vector field:
$$ \mathbf{F}(x,y,z) = \langle 2xy, yz^2, xz \rangle $$
This is tedious since the boundary consists of 5 faces for which we have to break down the flux surface integral into 5 surface integrals, each with its own parametrization. Hence, we calculate the divergence of the vector field:
\begin{align*}
\nabla \cdot \mathbf{F} &= \frac{\partial}{\partial x} 2xy + \frac{\partial}{\partial y} yz^2 + \frac{\partial}{\partial z} xz \\
&= 2y + z^2 + x
\end{align*}
and apply the Divergence Theorem:
\begin{align*}
\iint_{\partial R} \mathbf{F} \cdot d\mathbf{s} &= \iiint_R (\nabla \cdot \mathbf{F}) \, dV \\
&= \iiint_R (x + 2y + z^2) \, dV \\
&= \int_0^3 \int_0^3 \int_0^{6-2z} (x + 2y + z^2) \, dx \, dz \, dy \\
&= \int_0^3 \int_0^3 \left[\frac{(6-2z)^2}{2} + (6-2z)(2y + z^2) \right] \, dz \, dy \\
&= \int_0^3 \int_0^3 (18 + 12 y - 12 z - 4 y z + 8 z^2 - 2 z^3) \, dz \, dy \\
&= \int_0^3 \left[(18 + 12 y)z - (6 + 2y)z^2 + 8 \frac{z^3}{3} - \frac{z^4}{2} \right]_0^3 \, dy \\
&= \int_0^3 \left[3(18 + 12 y) - 9(6 + 2y) + 8(9) - \frac{81}{2} \right] \, dy \\
&= \int_0^3 (\frac{63}{2} + 18y) \, dy \\
&= \left[\frac{63}{2}y + 9y^2 \right]_0^3 \\
&= \frac{63(3)}{2} + 9(9) \\
&= \frac{351}{2}
\end{align*}
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