Answer
$\dfrac{9 \pi-8}{3}$
Work Step by Step
Consider $A= \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} (3-y) dy dx$
or, $= \int_{0}^{2} [3y-\dfrac{y^2}{2}]_{0}^{\sqrt{4-x^2}} dx$
or, $= \int_{0}^{2} [3 \sqrt {4-x^2} -\dfrac{4-x^2}{2}]dx$
or, $= [\dfrac{3\sqrt {4-x^2}x}{2}-2x +\dfrac{x^2}{3}+6 \sin^{-1} (\dfrac{x}{2})] _0^2 $
Thus, we have $I=6( \dfrac{\pi}{2})-4+\dfrac{8}{6}=\dfrac{9 \pi-8}{3}$
