Mathematics - Area Under Curves Question with Solution | TestHub

Left of the y-axis the region is a quarter circle of radius 2: area $\pi$. On the right the curves meet at $x=\sqrt{2}$, so the area is ${\int }_0^{\sqrt{2}}\left(\sqrt{4-x^2}-\sqrt{2}\sin \frac{x\pi }{2\sqrt{2}}\right)dx=\left(1+\frac{\pi }{2}\right)-\frac{4}{\pi }$. Ratio left : right $=\pi :\left(1+\frac{\pi }{2}-\frac{4}{\pi }\right)=$ $2{\pi }^2:\left(2\pi +{\pi }^2-8\right)$.