Answer
$\dfrac{5 \pi}{48}$
Work Step by Step
Here, we have the area function from the point $x=0$ to $x=1$ is given as: $A_1(x)=\dfrac{\pi (\sqrt x)^2}{8}=\dfrac{\pi x}{8}$ and the area function from the point $x=1$ to $x=2$ is given as: $A_2(x)=\dfrac{\pi}{8} (2-x)$
Thus, the total area is:
$A(x)=\dfrac{\pi}{8} \int_0^1 x \ dx+\dfrac{\pi}{8} \int_1^2 (2-x)^2 \ dx$
or, $=\dfrac{\pi}{8} [\dfrac{x^2}{2}]_0^1-\dfrac{\pi}{8}[\dfrac{(2-x)^3}{3}]_1^2$
or, $=\dfrac{\pi}{8} \times \dfrac{1}{2}+\dfrac{\pi}{8} \times \dfrac{1}{3}$
or, $=\dfrac{5 \pi}{48}$
