Answer
$\frac{2}{3}$
Work Step by Step
$\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{\sqrt {1-y^{2}}}$$y$$sinx$$dz$$dy$$dx$$=\int_{0}^{\pi}\int_{0}^{1}$$y$$sinx$$z|_{0}^\sqrt {1-y^{2}}dydx$
$=\int_{0}^{\pi}\int_{0}^{1}y\sqrt {1-y^{2}}sinxdydx$
Suppose $u=1-y^{2}$, $du=-2ydy$
and $ydy=\frac{-du}{2}$
when $y=0$ , $u=1$ and when $y=1$ $u=0$
$=\int_{0}^{\pi}\int_{0}^{1}\sqrt {u}(\frac{-du}{2})sinxdudx$
$=\frac{1}{2}\int_{0}^{\pi}\frac{2}{3}u^{3/2}|_{0}^{1}sinxdx$
$=\frac{1}{3}\int_{0}^{\pi}sinxdx$
$=\frac{1}{3}[-cosx]|_{0}^{\pi}$
$=\frac{1}{3}(1-(-1))$
$=\frac{2}{3}$
