The area enclosed by the curves $\mathrm{y}=\sqrt{4-{\mathrm{x}}^2},\mathrm{y}\ge \sqrt{2}\sin \left(\frac{{\displaystyle \mathrm{x\pi }}}{{\displaystyle 2\sqrt{2}}}\right)$ and X-axis is divided by Y-axis in the ratio

Given curves are $\mathrm{y}=\sqrt{4-{\mathrm{x}}^2},\mathrm{y}\ge \sin \left(\frac{{\displaystyle \mathrm{x\pi }}}{{\displaystyle 2\sqrt{2}}}\right)$
Which intersect at $\mathrm{x}=\sqrt{2}$ (by inspection)
Area of the left of y-axis is ${\mathrm{A}}_1=\mathrm{\pi }=\frac{1}{4}\mathrm{th}$ area of circle 

Area to the right of y-axis
${\mathrm{A}}_2=\underset{0}{\overset{\sqrt{2}}{\int }}\left(\sqrt{4-{\mathrm{x}}^2}-\sqrt{2}\sin \frac{\mathrm{x\pi }}{2\sqrt{2}}\right)\mathrm{dx}$
$={\left(\frac{{\displaystyle \mathrm{x}\sqrt{4-{\mathrm{x}}^2}}}{{\displaystyle 2}}+\frac{{\displaystyle 4}}{{\displaystyle 2}}{\sin }^{-1}\frac{{\displaystyle \mathrm{x}}}{{\displaystyle 2}}\right)}_0^{\sqrt{2}}+{\frac{4}{\mathrm{\pi }}\cos \frac{\mathrm{x\pi }}{2\sqrt{2}}|}_0^{\sqrt{2}}$
$\left[1+2.\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\right]+\frac{4}{\mathrm{\pi }}(0-1)=1+\frac{\mathrm{\pi }}{2}-=\frac{4}{\mathrm{\pi }}=\frac{2\mathrm{\pi }-{\mathrm{\pi }}^2-8}{2\mathrm{\pi }}$
$\Rightarrow \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{\mathrm{\pi }}{\left(\frac{2\mathrm{\pi }+{\mathrm{\pi }}^2-8}{2\mathrm{\pi }}\right)}=\frac{2{\mathrm{\pi }}^2}{2\mathrm{\pi }+{\mathrm{\pi }}^2-8}$