Question

The area (in sq. units) of the smaller region bounded by the curves $x^2+y^2=5$ and $y^2=4x$ is

• A. $\frac{\pi }{4}+\frac{1}{6}$
• B. $\frac{\pi }{4}-\frac{1}{6}$
• C. $\frac{1}{3}+\frac{5\pi }{4}-\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{5}}$
• D. $\frac{2}{3}+\frac{5\pi }{2}-5{\sin }^{-1}\frac{1}{\sqrt{5}}$

Answer 1

The correct option is D $\frac{2}{3}+\frac{5\pi }{2}-5{\sin }^{-1}\frac{1}{\sqrt{5}}$

$x^2+y^2=5$ and $y^2=4x$
The intersection point is $x=1$


Required area $=2\left(\underset{0}{\overset{1}{\int }}2\sqrt{x}dx+\underset{1}{\overset{\sqrt{5}}{\int }}\sqrt{5-x^2}dx\right)$
$=2{\left[\frac{4}{3}{x}^{3/2}\right]}_0^1+2{[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}]}_1^{\sqrt{5}}$
$=\frac{8}{3}+2(0-1+\frac{5\pi }{4}-\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{5}})$
$=\frac{2}{3}+\frac{5\pi }{2}-5{\sin }^{-1}\frac{1}{\sqrt{5}}$ sq. units