Answer
$$V = 32$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = 6 - 2y \cr
& {\text{Let the region be}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant 2,{\text{ }}0 \leqslant x \leqslant 4} \right\} \cr
& V = \iint\limits_R zdA = \int_0^2 {\int_0^4 {\left( {6 - 2y} \right)} } dxdy \cr
& {\text{Integrate with respect to }}x \cr
& V = \int_0^2 {\left[ {6x - 2xy} \right]_0^4} dy \cr
& V = \int_0^2 {\left[ {\left( {6\left( 4 \right) - 2\left( 4 \right)y} \right) - \left( {6\left( 0 \right) - 2\left( 0 \right)y} \right)} \right]} dy \cr
& V = \int_0^2 {\left( {24 - 8y} \right)} dy \cr
& {\text{Integrate}} \cr
& V = \left[ {24y - 4{y^2}} \right]_0^2 \cr
& V = \left[ {24\left( 2 \right) - 4{{\left( 2 \right)}^2}} \right] - \left[ {24\left( 0 \right) - 4{{\left( 0 \right)}^2}} \right] \cr
& V = 32 \cr} $$
