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 x-axis is divided by 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. $\frac{2{\pi }^2}{{\pi }^2+2\pi -8}$

Answer 1

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

$y=\sqrt{4-x^2}\dots$ (given) $\cdots \left(i\right)$

$\Rightarrow y^2-4x^2$

$\Rightarrow x^2+y^2=4$

$y\ge \sqrt{2}\sin \left(\frac{\pi x}{2\sqrt{2}}\right)$ (given) ... (ii)

To find the period,

$=\frac{2\mathrm{\pi /}}{\mathrm{\pi /}}\times 2\sqrt{2}$

$=4\sqrt{2}\Rightarrow$ period.

$=\frac{4\sqrt{2}}{2}=2\sqrt{2}$

$=2\times 1.732$

$\left[3.4\right]$

To find the intersection point using (i) and (ii)

$x=\sqrt{2}$

Area of circle $=\pi r^2=\pi \times y=4\pi$

Area of circle $=\pi =A_1$.. (iii)

from (-2,0) to (0,0)

$A_2={\int }_0^{\sqrt{2}}\sqrt{4-x^2}-\sqrt{2}\sin \left(\frac{\pi }{2\sqrt{2}}x\right)dx$

$\Rightarrow {\int }_0^{\sqrt{2}}\sqrt{4-x^2}dx-\sqrt{2}{\int }_0^{\sqrt{2}}\sin \left[\frac{\pi x}{2\sqrt{2}}\right]dx$

$\left[\begin{array}{ccc}\therefore & \sqrt{a^2-x^2}& dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\end{array}\right]$

$={{[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^{\sqrt{2}}-\frac{\sqrt{2}}{\left(\frac{\pi }{2\sqrt{2}}\right)}[-\cos \left(\frac{\pi x}{2\sqrt{2}}\right)]}_0^{\sqrt{2}}$

$=[\frac{2}{2}\pi \sqrt{2}+\frac{4}{2}{\sin }^{-1}\left[\frac{1}{\sqrt{2}}\right]]+\frac{2\times 2}{\pi }{\left|\cos \left(\frac{\pi x}{2\sqrt{2}}\right)\right|}_0$

$=[1+\frac{4}{2}{\sin }^{-1}\left[\frac{1}{\sqrt{2}}\right]]+\frac{4}{\pi }[\cos (\frac{\pi }{2\sqrt{2}}\times \sqrt{2})-\cos 0]$

$\Rightarrow [1+\pi /2-4/\pi ]$

$\Rightarrow \frac{2\pi +{\pi }^2-8}{2\pi }$.. (iv)

Using equation (iii) and (iv).

$\frac{A_1}{A_2}=\frac{\pi \times 2\pi }{2\pi +{\pi }^2-8}=\frac{2{\pi }^2}{2\pi +{\pi }^2-8}$ Answer (D)