Thomas' Calculus 13th Edition

$\frac{20}{3}$
V=$\int^4_0 \int^{\sqrt{4-x}}_0 \int^{2-y}_0$ dz dy dx =$\int^4_0 \int^{\sqrt{4-x}}_0 (2-y)$ dy dx =$\int^4_[2\sqrt{4-x}-(\frac{4-x}{2})]dx$ =$[-\frac{4}{3}(4-x)^{3/2}+\frac{1}{4}(4-x^2)]^4_0$ =$\frac{4}{3}(4)^{3/2}-\frac{1}{4}(16)$ =$\frac{32}{3}-4$ =$\frac{20}{3}$