Answer
$2-\dfrac{\pi}{2}$
Work Step by Step
Our aim is to integrate the integral as follows:
$\int^2_{\sqrt{2}} \int^y_{\sqrt{4-y^2}} \space dy \space dx = \int^{\pi/2}_{\pi/4} \int^{2 /\sin \theta}_{2}r \space dr \space d\theta $
or, $=\int^{\pi/2}_{\pi/4}(2 \csc^2 (\theta)-2) \space d\theta $
or, $=[-2 \cot (\theta)-\dfrac{\theta}{2}]^{\pi/2}_{\pi/4}$
or, $=2-\dfrac{\pi}{2}$
