Answer
$1$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^1_0 \int^{2-2x}_0 \int^{3-3x-3y/2}_0 \space dz \space dy \space dx \\= \int^1_0 \int^{2-2x}_0 (3-3x-\dfrac{3y}{2}) \space dy \space dx \\= \int^1_0 [6(1-x)^2-(\dfrac{3}{4})(4) (1-x)^2] dx \\= \int^1_0 3(1-x)^2 dx \\=[-(1-x)^3]^1_0 \\=1$$
