Question

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

• A. $\frac{1}{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{1}{6}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (B)

The required area is
$A=2\displaystyle \int _{0}^{1/\left(\sqrt{2}\right)}\left(\sqrt{1 - x^{2}} - x\right)dx$
$\Rightarrow A=2\left[\frac{x \sqrt{1 - x^{2}}}{2} + \frac{1}{2} sin^{- 1} x - \frac{x^{2}}{2}\right]\frac{1 / \sqrt{2}}{0}$
$=2\left[\frac{1}{2 \sqrt{2}} \frac{1}{\sqrt{2}} + \frac{1}{2} \frac{\pi }{4} - \frac{1}{4}\right]$
$=\frac{\pi }{4}$ sq. units