Answer
$\dfrac{2}{3}$
Work Step by Step
Need to integrate the given integral first with respect to $y$, then $z$, and then $x$
$ \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} z \sin x dy dz dx=[-\cos x]_{0}^{\pi} \int_{0}^{1} 2y|_{0}^{\sqrt{1-z^2}}dz$
or, $=(2) \int_{0}^{1} z\sqrt{1-z^2} dz$
plug $a=1-z^2 \implies dz=(-1/2z) du$
or, $= (2)(-1/2)\int_{-1}^0 \sqrt a da$
or, $= (2) (\dfrac{1}{2})[\dfrac{2a^{3/2}}{3}]_0^1$
or, $=\dfrac{2}{3}$
