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