Answer
$$V = {\left( {e - 1} \right)^2}$$
Work Step by Step
$$\eqalign{
& z = {e^{x + y}};\,\,\,\,\,\,\,0 \leqslant x \leqslant 1,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr
& {\text{by the equation just given}}{\text{, the volume is}} \cr
& V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr
& {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr
& V = \int_0^1 {\int_0^1 {{e^{x + y}}} dxdy} \cr
& V = \int_0^1 {\left[ {\int_0^1 {{e^{x + y}}} dx} \right]dy} \cr
& {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr
& = \int_0^1 {{e^{x + y}}} dx \cr
& {\text{integrate by using }}\int {{e^u}du} \cr
& = \left[ {{e^{x + y}}} \right]_0^1 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = {e^{1 + y}} - {e^{0 + y}} \cr
& = {e^{1 + y}} - {e^y} \cr
& \cr
& V = \int_0^1 {\left[ {\int_0^1 {{e^{x + y}}} dx} \right]dy} \cr
& V = \int_0^1 {\left( {{e^{1 + y}} - {e^y}} \right)dy} \cr
& {\text{integrating}} \cr
& V = \left[ {{e^{1 + y}} - {e^y}} \right]_0^1 \cr
& {\text{evaluating}} \cr
& V = \left( {{e^{1 + 1}} - {e^1}} \right) - \left( {{e^{1 + 0}} - {e^0}} \right) \cr
& V = {e^2} - e - e + 1 \cr
& V = {e^2} - 2e + 1 \cr
& V = {\left( {e - 1} \right)^2} \cr} $$
