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 expressed as: $$\ Volume =\iint_{D} f(x,y) \space dA$$
The region $D$ can be defined as:
$D=\left\{ (x, y) | 0 \leq x \leq 1, \ x \leq y \leq 2-x \right\}
$
Now, $\ V =\int_{0}^{1} \int_{x}^{2-x} x \ dy \ dx \\ =\int_{0}^{1} [x(2-x) -x(x) ] \ dx \\= \int_{0}^{1} 2x-x^2 dx\\=[x^2-\dfrac{2x^3}{3}]_0^1 \\= (1-0)-[\dfrac{2(1)^3-0}{3}]\\=\dfrac{1}{3}$
