Answer
$$V = 4$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = {e^{ - \frac{{x + y}}{2}}} \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant y \leqslant \infty ,{\text{ }}0 \leqslant x \leqslant \infty } \right\} \cr
& V = \iint\limits_R zdA = \int_0^\infty {\int_0^\infty {{e^{ - \frac{{x + y}}{2}}}} } dydx \cr
& V = \int_0^\infty {\left[ {\int_0^\infty {{e^{ - \frac{x}{2}}}{e^{ - \frac{y}{2}}}} dy} \right]} dx \cr
& V = \int_0^\infty {{e^{ - \frac{x}{2}}}\left[ {\int_0^\infty {{e^{ - \frac{y}{2}}}} dy} \right]} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate }}\int_0^\infty {{e^{ - \frac{y}{2}}}} dy{\text{ with respect to }}y \cr
& \int_0^\infty {{e^{ - \frac{y}{2}}}} dy = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - \frac{y}{2}}}} dy \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ { - 2{e^{ - \frac{y}{2}}}} \right]_0^b \cr
& = - 2\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - \frac{b}{2}}} - {e^0}} \right] \cr
& = - 2\left[ {{e^{ - \infty }} - {e^0}} \right] \cr
& = - 2\left( {0 - 1} \right) \cr
& = 2 \cr
& {\text{Substitute the previous result in }}\left( {\bf{1}} \right) \cr
& V = \int_0^\infty {{e^{ - \frac{x}{2}}}\left[ 2 \right]} dx{\text{ }}\left( {\bf{1}} \right) \cr
& V = 2\int_0^\infty {{e^{ - \frac{x}{2}}}} dx \cr
& {\text{By symmetry }}\int_0^\infty {{e^{ - \frac{x}{2}}}} dx = \int_0^\infty {{e^{ - \frac{y}{2}}}} dy = 2,{\text{ then}} \cr
& V = 2\left( 2 \right) \cr
& V = 4 \cr} $$
