Answer
$$V = \pi $$
Work Step by Step
$$\eqalign{
& z = \sin y \cr
& {\text{From the graph we have the region }}D \cr
& D = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \pi ,{\text{ 0}} \leqslant y \leqslant x} \right\} \cr
& {\text{We can represent the volume as}} \cr
& V = \int_0^\pi {\int_0^x {\sin y} dy} dx \cr
& V = \int_0^\pi {\left[ {\int_0^x {\sin y} dy} \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& V = - \int_0^\pi {\left[ {\cos y} \right]_0^x} dx \cr
& {\text{Evaluate}} \cr
& V = - \int_0^\pi {\left( {\cos x - \cos 0} \right)} dx \cr
& V = - \int_0^\pi {\left( {\cos x - 1} \right)} dx \cr
& V = \int_0^\pi {\left( {1 - \cos x} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& V = \left[ {x - \sin x} \right]_0^\pi \cr
& V = \left[ {\pi - \sin \pi } \right] - \left[ {0 - \sin 0} \right] \cr
& V = \pi \cr} $$
