Answer
$$\int_{0}^{\pi / 2} \int_{0}^{y / 2} \int_{0}^{1 / y} \sin y d z d x d y=\frac{1}{2}$$
Work Step by Step
Given $$\int_{0}^{\pi / 2} \int_{0}^{y / 2} \int_{0}^{1 / y} \sin y \ d z d x d y$$
So, we have
\begin{align}
I&=\int_{0}^{\pi / 2} \int_{0}^{y / 2} \int_{0}^{1 / y} \sin y \ \ d z d x d y\\
&=\int_{0}^{\pi / 2} \int_{0}^{y / 2} (z)_0^{1/y} \sin y \ d x d y\\
&=\int_{0}^{\pi / 2} \int_{0}^{y / 2} \sin y(\frac{1}{y} -0)d x d y\\
&=\int_{0}^{\pi / 2} \int_{0}^{y / 2} \frac{\sin y}{y} d x d y\\
&=\int_{0}^{\pi / 2} [x]_{0}^{y / 2} \frac{\sin y}{y} d y\\
&=\int_{0}^{\pi / 2} (\frac{y}{2}-0)\frac{\sin y}{y} d y\\
&=\frac{1}{2} \int_{0}^{\pi / 2} \sin y d y\\
&=\left[-\frac{1}{2} \cos y\right]_{0}^{\pi / 2}\\
&=-\frac{1}{2} \left[ \cos \frac{ \pi}{2}- \cos 0 \right] \\
&=\frac{1}{2}
\end{align}
