Question

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

• A. $\frac{{\pi }^2-8}{{\pi }^2+8}$
  
• B. $\frac{{\pi }^2-4}{{\pi }^2+4}$
  
• C. $\frac{\pi -4}{\pi +4}$
  
• D. $\frac{2{\pi }^2}{2\pi +{\pi }^2-8}$

Answer 1

The correct option is D $\frac{2{\pi }^2}{2\pi +{\pi }^2-8}$

By the graph
Area of the left of y-axis is $\pi$,Area of the right of y-axis = ${\int }_0^{\sqrt{2}}(\sqrt{4-x^2}-\sqrt{2}sin\left(\frac{x\pi }{2\sqrt{2}}\right))dx$