Answer
$$\dfrac{320}{3}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^4_0 \int^8_z \int^{8-z}_z \space dx \space dy \space dz \\= \int^4_0 \int^8_z(8-2z) \space dy \space dz \\= \int^4_0 (8-2z)\times (8-z) \space dz \\= \int^4_0 (64-24z+2z^2)\space dz \\= [64z-12z^2+\dfrac{2}{3}z^3]^4_0 \\=\dfrac{320}{3}$$
