## Thomas' Calculus 13th Edition

$\frac{8\pi}{3}$
V=$4\int^{\pi/2}_0 \int^1_0 \int^{4-4r^2}_{r^4-1}dz$ r dr $d\theta$ =$4\int^{\pi/2}_0 \int^1_0 (5r-4r^3-r^5)drd\theta$ =$4\int^{\pi/2}_0 (\frac{5}{2}-1-\frac{1}{6})d\theta$ =$4\int^{\pi/2}_0d\theta$ =$\frac{8\pi}{3}$