Question

The area (in sq. units) bounded by $x^2+y^2=1$ and the curve $y^2\ge x^2,$ 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
$A=2{\int }_0^{1/\left(\sqrt{2}\right)}\left(\sqrt{1-x^2}-x\right)dx$
$\Rightarrow A=2\left[\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}si{n}^{-1}x-\frac{x^2}{2}\right]\frac{1/\sqrt{2}}{0}$
$=2\left[\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{1}{2}\frac{\pi }{4}-\frac{1}{4}\right]$
$=\frac{\pi }{4}$ sq. units