Answer
$A=2-\frac{\pi}{2}$
Work Step by Step
From the given sketch, $1-\cos(y) \leq \sin(y)$ for $0\leq y \leq\frac{\pi}{2} $
So the area is:
$$A=\int_{0}^{\frac{\pi}{2}}(\sin (y)-(1-\cos(y)))dy$$
$$A=\int_{0}^{\frac{\pi}{2}}(\sin (y)-1+\cos(y))dy$$
Using the $FTC I$ it follows:
$$[-y+\sin(y)-\cos(y)]_{0}^{\frac{\pi}{2}}=-\frac{\pi}{2}+\sin(\frac{\pi}{2})-\cos(\frac{\pi}{2})-(-0+\sin(0)-\cos(0))=-\frac{\pi}{2}+1+1=2-\frac{\pi}{2}$$
