Answer
$$V = \frac{3}{8}$$
Work Step by Step
$$\eqalign{
& {\text{The volume of the solid is given by}} \cr
& V = \iint\limits_R zdA,{\text{ }}z = 1 - xy \cr
& {\text{Let the region be }} \cr
& R = \left\{ {\left( {x,y} \right):x \leqslant y \leqslant 1,{\text{ }}0 \leqslant x \leqslant 1} \right\} \cr
& V = \iint\limits_R zdA = \int_0^1 {\int_x^1 {\left( {1 - xy} \right)} } dydx \cr
& {\text{Integrate with respect to }}y \cr
& V = \int_0^1 {\left[ {y - \frac{{x{y^2}}}{2}} \right]_x^1} dx \cr
& V = \int_0^1 {\left[ {\left( {1 - \frac{x}{2}} \right) - \left( {x - \frac{{{x^3}}}{2}} \right)} \right]} dx \cr
& V = \int_0^1 {\left( {1 - \frac{x}{2} - x + \frac{{{x^3}}}{2}} \right)} dx \cr
& V = \int_0^1 {\left( {1 - \frac{{3x}}{2} + \frac{{{x^3}}}{2}} \right)} dx \cr
& {\text{Integrate}} \cr
& V = \left[ {x - \frac{{3{x^2}}}{4} + \frac{{{x^4}}}{8}} \right]_0^1 \cr
& V = \left[ {1 - \frac{{3{{\left( 2 \right)}^2}}}{4} + \frac{{{{\left( 1 \right)}^4}}}{8}} \right] - \left[ 0 \right] \cr
& V = \frac{3}{8} \cr} $$
