Answer
$$\dfrac{1}{3}$$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$
We can express the region $D$ as follows:
$$D=\left\{ (x, y) | 0 \leq x \leq 1, \ x \leq y \leq 2-x \right\}
$$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\=\int_{0}^{1} \int_{x}^{2-x} x \ dy \ dx \\ =\int_{0}^{1} [x(2-x) -x(x) ] \ dx \\= \int_{0}^{1} 2x-x^2 dx\\=[x^2-\dfrac{2x^3}{3}]_0^1 \\= 1-\dfrac{2}{3}\\=\dfrac{1}{3}$$
