The area (in sq. units) bounded by ${\mathrm{x}}^2+{\mathrm{y}}^2=1$ and the curve ${\mathrm{y}}^2\ge {\mathrm{x}}^2,$ above the $\mathrm{x}$-axis is

• A

  $\frac{1}{4}$
• Correct answer
  B

  $\frac{\mathrm{\pi }}{4}$
• C

  $\frac{1}{6}$
• D

  $\frac{\mathrm{\pi }}{6}$

Solution

The required area is
$\mathrm{A}=2{\int }_0^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\left(\sqrt{1-{\mathrm{x}}^2}-\mathrm{x}\right)\mathrm{d}\mathrm{x}$
$\Rightarrow \mathrm{A}=2\left[\frac{\mathrm{x}\sqrt{1-{\mathrm{x}}^2}}{2}+\frac{1}{2}{\sin }^{-1}\mathrm{x}-\frac{{\mathrm{x}}^2}{2}\right]\genfrac{}{}{0pt}{}{1/\sqrt{2}}{0}$
$=2\left[\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{1}{2}\frac{\mathrm{\pi }}{4}-\frac{1}{4}\right]$
$=\frac{\mathrm{\pi }}{4}$ sq. units

Hence, the correct answer is option: 'B'.