Answer
$\dfrac{1}{8}$
Work Step by Step
The point of intersection can be computed as: $\dfrac{y}{2}=y\sqrt {1-y^2} \\ -y^2=\dfrac{1}{4}-1 \\ y = \pm \dfrac{\sqrt 3}{2}$
The area $(A)$ can be calculated as:
$A=\int_{0}^{\frac{\sqrt 3}{2}} [y\sqrt {1-y^2}-\dfrac{y}{2}] \ dy +\int_{\frac{\sqrt 3}{2}}^1 [\dfrac{y}{2}-y\sqrt {1-y^2}] \ dy \\= [-\dfrac{(1-y^2)^{3/2}}{3}-\dfrac{y^2}{4}]_{0}^{\frac{\sqrt 3}{2}} +[\dfrac{(1-y^2)^{3/2}}{3}+\dfrac{y^2}{4}]_{\frac{\sqrt 3}{2}}^1\\=\dfrac{-1}{24}-\dfrac{3}{16}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{3}{16}-\dfrac{1}{24} \\=\dfrac{1}{8} $
