Answer
$$A = \sqrt 2 $$
Work Step by Step
$$\eqalign{
& x = \sin y,{\text{ }}x = 0,{\text{ }}y = \pi /4,{\text{ }}y = 3\pi /4 \cr
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{\pi /4}^{3\pi /4} {\left( {\sin y - 0} \right)} dy \cr
& A = \int_{\pi /4}^{3\pi /4} {\sin y} dy \cr
& {\text{Integrate}} \cr
& A = - \left[ {\cos y} \right]_{\pi /4}^{3\pi /4} \cr
& {\text{Evaluate}} \cr
& A = - \left[ {\cos \left( {\frac{{3\pi }}{4}} \right) - \cos \left( {\frac{\pi }{4}} \right)} \right] \cr
& {\text{Simplify}} \cr
& A = - \left( { - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}} \right) \cr
& A = \sqrt 2 \cr} $$
