Answer
$$V = 4$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = 4 - {y^2} \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):x \leqslant y \leqslant 2,{\text{ }}0 \leqslant x \leqslant 2} \right\} \cr
& V = \iint\limits_R zdA = \int_0^2 {\int_x^2 {\left( {4 - {y^2}} \right)} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& V = \int_0^2 {\left[ {4y - \frac{{{y^3}}}{3}} \right]_x^2} dx \cr
& V = \int_0^2 {\left[ {\left( {4\left( 2 \right) - \frac{{{{\left( 2 \right)}^3}}}{3}} \right) - \left( {4\left( x \right) - \frac{{{{\left( x \right)}^3}}}{3}} \right)} \right]} dx \cr
& V = \int_0^2 {\left[ {\frac{{16}}{3} - \left( {4x - \frac{{{x^3}}}{3}} \right)} \right]} dx \cr
& V = \int_0^2 {\left( {\frac{{16}}{3} - 4x + \frac{{{x^3}}}{3}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = \left[ {\frac{{16}}{3}x - 2{x^2} + \frac{{{x^4}}}{{12}}} \right]_0^2 \cr
& V = \left[ {\frac{{16}}{3}\left( 2 \right) - 2{{\left( 2 \right)}^2} + \frac{{{{\left( 2 \right)}^4}}}{{12}}} \right] - \left[ 0 \right] \cr
& V = 4 \cr} $$
