Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {{y^3}\sin x{y^2}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 2,\,\,\,\,0 \leqslant y \leqslant \sqrt {\pi /2} } \right\} \cr
& {\text{Convert to an iterated integral substituting the region }}R \cr
& = \int_0^{\sqrt {\pi /2} } {\int_0^2 {{y^3}\sin x{y^2}} } dxdy \cr
& = \int_0^{\sqrt {\pi /2} } {\left[ {\int_0^2 {{y^3}\sin x{y^2}} dx} \right]} dy \cr
& {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr
& = y\int_0^2 {{y^2}\sin x{y^2}} dx \cr
& {\text{integrating}} \cr
& = - y\left( {\cos x{y^2}} \right)_0^2 \cr
& {\text{evaluating the limits for the variable }}x \cr
& = - y\left( {\cos \left( 2 \right){y^2} - \cos \left( 0 \right){y^2}} \right) \cr
& = - y\cos 2{y^2} + y \cr
& \cr
& \int_0^{\sqrt {\pi /2} } {\left[ {\int_0^2 {{y^3}\sin x{y^2}} dx} \right]} dy = \int_0^{\sqrt {\pi /2} } {\left( {y - y\cos 2{y^2}} \right)} dy \cr
& {\text{integrating}}{\text{, }} \cr
& = \left( {\frac{{{y^2}}}{2} - \frac{1}{4}\sin 2{y^2}} \right)_0^{\sqrt {\pi /2} } \cr
& {\text{evaluate}} \cr
& = \left( {\frac{{{{\left( {\sqrt {\pi /2} } \right)}^2}}}{2} - \frac{1}{4}\sin 2{{\left( {\sqrt {\pi /2} } \right)}^2}} \right) - \left( {\frac{{{{\left( 0 \right)}^2}}}{2} - \frac{1}{4}\sin 2{{\left( 0 \right)}^2}} \right) \cr
& = \frac{{\pi /2}}{2} - \frac{1}{4}\sin \left( \pi \right) - 0 \cr
& = \frac{\pi }{4} \cr} $$
