## Thomas' Calculus 13th Edition

$\frac{320}{3}$
$V=\int^4_0 \int^8_z \int^{8-z}_z$ dx dy dz =$\int^4_0 \int^8_z(8-2z)dy$ dz =$\int^4_0 (8-2z)(8-z)dz$ =$\int^4_0 (64-24z+2z^2)dz$ =$[64z-12z^2+\frac{2}{3}z^3]^4_0$ $\frac{320}{3}$