Answer
$$\frac{2}{\pi }$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^1 {\int_0^{\pi /2} {\sin \pi x\cos y\sin 2z} dy} dx} dz \cr
& = \int_0^{\pi /2} {\int_0^1 {\left[ {\int_0^{\pi /2} {\sin \pi x\cos y\sin 2z} } \right]dy} dx} dz \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^{\pi /2} {\int_0^1 {\sin \pi x\sin 2z\left[ {\int_0^{\pi /2} {\cos y} } \right]dy} dx} dz \cr
& = \int_0^{\pi /2} {\int_0^1 {\sin \pi x\sin 2z\left[ {\sin y} \right]_0^{\pi /2}} dx} dz \cr
& = \int_0^{\pi /2} {\int_0^1 {\sin \pi x\sin 2z} dx} dz \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^{\pi /2} {\sin 2z\int_0^1 {\sin \pi x} dx} dz \cr
& = - \int_0^{\pi /2} {\frac{{\sin 2z}}{\pi }\left[ {\cos \pi x} \right]_0^1} dz \cr
& = - \int_0^{\pi /2} {\frac{{\sin 2z}}{\pi }\left( { - 2} \right)} dz \cr
& = \frac{2}{\pi }\int_0^{\pi /2} {\sin 2z} dz \cr
& {\text{Integrate with respect to }}z \cr
& = - \frac{1}{\pi }\left[ {\cos 2z} \right]_0^{\pi /2} \cr
& = - \frac{1}{\pi }\left[ {\cos \pi - \cos 0} \right] \cr
& = \frac{2}{\pi } \cr} $$
