Answer
$2(\pi -1)$
Work Step by Step
Our aim is to integrate the integral as follows:
$\int^{\pi/2}_0 \int^{2\sqrt{2-\sin2\theta} \space r \space dr \space d\theta} =2\int^{\pi/2}_0(2-\sin2\theta) d\theta $
or, $=(4) (\dfrac{\pi}{2}) -4 \int^{\pi/2}_0 \sin \phi \cos \phi d \phi $
or, $=2\pi -2$
or, $=2(\pi -1)$
