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