Answer
$1-\dfrac{2}{3}\\=\dfrac{1}{3}$
Work Step by Step
The volume under the surface is given by :$z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $\ Volume ; V=\iint_{D} f(x,y) \space dA$
Our aim is to calculate the volume of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | 0 \leq x \leq 1, \ x \leq y \leq 2-x \right\}
$
Now, $V =\iint_{D} f(x,y) \ dA \\=\int_{0}^{1} \int_{x}^{2-x} x \ dy \ dx \\ =\int_{0}^{1} [x(2-x) -x(x) ] \ dx \\= \int_{0}^{1} 2x-x^2 \space dx\\=[x^2]_0^1 -\dfrac{2}{3} \times [x^3]_0^1 \\= 1-\dfrac{2}{3}\\=\dfrac{1}{3}$
