## Calculus 8th Edition

$\dfrac{2}{3}$
$\int_{0}^{\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} z \sin x dy dz dx=(2) \int_{0}^{1} z\sqrt{1-z^2} dz$ Consider $p=1-z^2; dz=(-\dfrac{1}{2z}) dp$ Then, we have $(2) \int_{0}^{1} z\sqrt{1-z^2} dz= (2)(-\dfrac{1}{2})\int_{-1}^0 p^{1/2} dp \\= [\dfrac{2p^{3/2}}{3}]_0^1 \\Hence, \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} z \sin x dy dz dx=\dfrac{2}{3}$