## Thomas' Calculus 13th Edition

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