## Thomas' Calculus 13th Edition

$\frac{3\pi-4}{6}$
$V=\int^{\pi/2}_0 \int^{\cos\theta}_0 \int^{3\sqrt{1-r^2}}_0dz$ r dr $d\theta$ =$\int^{\pi/2}_0 \int^{\cos\theta}_0 3r\sqrt{1-r^2}drd\theta$ =$\int^{\pi/2}_0 [-(1-r^2)^{3/2}]^{cos\theta}_0 d\theta$ =$\int^{\pi/2}_0[-(1-cos^2\theta)^{3/2}+1]d\theta$ =$\int^{\pi/2}_0 (1-sin^3\theta) d\theta$ =$[\theta+\frac{sin^2\theta\cos\theta}{3}]^{\pi/2}_0-\frac{2}{3}\int^{\pi/2}_0 \sin\theta d\theta$ =$\frac{\pi}{2}+\frac{2}{3}[cos\theta]^{\pi/2}_0$ =$\frac{\pi}{2}-\frac{2}{3}$ =$\frac{3\pi-4}{6}$