Question

The area enclosed by the curve $y=\sqrt{4-x^2},y\ge \sqrt{2}\sin \left(\frac{x\pi }{2\sqrt{2}}\right)$ and the x-axis 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 $\frac{2{\pi }^2}{2\pi +{\pi }^2-8}$

$y=\sqrt{4-x^2},y=\sqrt{2}\sin \left(\frac{x\pi }{2\sqrt{2}}\right)$
intersect at $x=\sqrt{2}$
Area to the left of y-axis is $\pi$
Area to the right of y-axis
$=\underset{0}{\overset{\sqrt{2}}{\int }}(\sqrt{4-x^2}-\sqrt{2}\sin \frac{x\pi }{2\sqrt{2}})dx$
$={(\frac{x\sqrt{4-x^2}}{2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2})}_0^{\sqrt{2}}+{\left(\frac{4}{\pi }\cos \frac{x\pi }{2\sqrt{2}}\right)}_0^{\sqrt{2}}$
$=(1+2\times \frac{\pi }{4})+\frac{\pi }{4}(0-1)$$=1+\frac{\pi }{2}-\frac{4}{\pi }$
$=\frac{2\pi +{\pi }^2-8}{2\pi }\text{sq. units}$

$\therefore \text{Ratio}=\frac{2{\pi }^2}{2\pi +{\pi }^2-8}$