Answer
$$V = 4$$
Work Step by Step
$$\eqalign{
& z = 2 - 4x \cr
& {\text{Let }}z = 0 \cr
& 2 - 4x = 0 \cr
& 4x = 2 \cr
& x = \frac{1}{2},{\text{ So the region }}D{\text{ is:}} \cr
& D = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \frac{1}{2},{\text{ 0}} \leqslant y \leqslant 8} \right\} \cr
& {\text{We can represent the volume as}} \cr
& V = \int_0^8 {\int_0^{1/2} {\left( {2 - 4x} \right)} dx} dy \cr
& {\text{Integrate with respect to }}x \cr
& V = \int_0^8 {\left[ {2x - 2{x^2}} \right]_0^{1/2}} dy \cr
& V = \int_0^8 {\left[ {2\left( {\frac{1}{2}} \right) - 2{{\left( {\frac{1}{2}} \right)}^2}} \right]} dy \cr
& V = \int_0^8 {\frac{1}{2}} dy \cr
& V = \frac{1}{2}\left( {8 - 0} \right) \cr
& V = 4 \cr} $$
