Answer
$$\dfrac{2}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^1_0 \int^{1-x}_0 \int^{2-2z}_0 \space dy \space dz \space dx\\=\int^1_0 \int^{1-x}_0(2-2z) \space dz \space dx\\=\int^1_0 [2z-z^2]^{1-x}_0 \space dx\\=\int^1_0 (1-x^2)dx \\=[x-\dfrac{x^3}{3}]^1_0 \\=(1-0)-(\dfrac{1}{3}-0)\\=\dfrac{2}{3}$$
