## Thomas' Calculus 13th Edition

$\frac{3}{2}$
$\int^1_0 \int^{3-3x}_0 \int^{3-3x-y}_0$ dz dy dx =$\int^1_0 \int^{3-3x}_0 (3-3x-y)$ dy dx =$\int^1_0[(3-3x)^2-\frac{1}{2}(3-3x)^2]dx$ =$\frac{9}{2} \int^1_0 (1-x)^2$ dx =$\frac{-3}{2}[(1-x)^3]^1_0$ =$\frac{3}{2}$