Answer
$$V = 4$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = 4 - x - y \cr
& {\text{Let the region be}} \cr
& R = \left\{ {\left( {x,y} \right):x \leqslant y \leqslant 2,{\text{ }}0 \leqslant x \leqslant 2} \right\} \cr
& V = \iint\limits_R zdA = \int_0^2 {\int_x^2 {\left( {4 - x - y} \right)} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& V = \int_0^2 {\left[ {4y - xy - \frac{{{y^2}}}{2}} \right]_x^2} dx \cr
& V = \int_0^2 {\left[ {\left( {4\left( 2 \right) - x\left( 2 \right) - \frac{{{{\left( 2 \right)}^2}}}{2}} \right) - \left( {4\left( x \right) - x\left( x \right) - \frac{{{{\left( x \right)}^2}}}{2}} \right)} \right]} dx \cr
& V = \int_0^2 {\left[ {\left( {6 - 2x} \right) - \left( {4x - {x^2} - \frac{{{x^2}}}{2}} \right)} \right]} dx \cr
& V = \int_0^2 {\left( {6 - 2x - 4x + \frac{3}{2}{x^2}} \right)} dx \cr
& V = \int_0^2 {\left( {6 - 6x + \frac{3}{2}{x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = \left[ {6x - 3{x^2} + \frac{1}{2}{x^3}} \right]_0^2 \cr
& V = \left[ {6\left( 2 \right) - 3{{\left( 2 \right)}^2} + \frac{1}{2}{{\left( 2 \right)}^3}} \right] - \left[ 0 \right] \cr
& V = 4 \cr} $$
