= Solution
Use the prism bounds read from the PDF diagram and the <divergence> from part (ii):
$$
\begin{aligned}
\int_V\nabla\cdot\mathbf F\,dV
&=\int_0^2\int_0^{4-2x}\int_0^1(2y+4x-8-2z)\,dz\,dy\,dx\\
&=\int_0^2\int_0^{4-2x}(2y+4x-9)\,dy\,dx.
\end{aligned}
$$
Set $A=4-2x$ in the inner evaluation. Since $4x-9=-2A-1$, the inner integral is $A^2-(2A+1)A=-A^2-A$. Therefore
$$
\boxed{\int_V\nabla\cdot\mathbf F\,dV
=-\int_0^2[(4-2x)^2+(4-2x)]\,dx
=-\frac{32}{3}-4=-\frac{44}{3}}.
$$
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