The area of the region $\mathrm{R}=\left\{(x, y):|x| \leq|y|\right.$ and $x^{2}$ $\left.+y^{2} \leq 1\right\}$ is

• A

  $\frac{3 \pi}{8}$ sq units
• B

  $\frac{5 \pi}{8} \mathrm{sq}$ units
• Correct answer
  C

  $\frac{\pi}{2}$ sq units
• D

  $\frac{\pi}{8}$ sq unit

Solution

Required area $=4$ (Area of the shaded region in first quadrant) $$ =4 \int_{0}^{1 \sqrt{2}}\left(y_{1}-y_{2}\right) d x=4 \int_{0}^{1 \sqrt{2}}\left(\sqrt{1-x^{2}}-x\right) d x $$ $=4\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\frac{x^{2}}{2}\right]_{0}^{1 / \sqrt{2}}$ $=4\left[\frac{1}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{\pi}{4}-\frac{1}{4}\right]$ $=4\left[\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}\right]=\frac{4 \pi}{8}=\frac{\pi}{2}$ sq units

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