Question

If the area enclosed by the curve $y=\sqrt{x}$ and $x=-\sqrt{y},$ the circle $x^2+y^2=2$ above the x-axis, is $A$ then the value of $\frac{16}{\pi }A$ is

Answer 1

$A_1=\text{Area}ACO$
$A_2=\text{Area}BCO$
$A_3=\text{Area}ACBDA$

$A_1=\underset{0}{\overset{1}{\int }}-\sqrt{y}dy$
$=-{\left[\frac{2}{3}{y}^{3/2}\right]}_0^1$
$=|-\frac{2}{3}|=\frac{2}{3}$

$A_2=\underset{0}{\overset{1}{\int }}{y}^{2}dy$
$=\frac{1}{3}$

From figure, area of the shaded region $A_3$ is,
$A_3=\frac{\pi }{2}-1$

$\therefore A=A_1+A_2+A_3$
$=\frac{\pi }{2}$

$\therefore \frac{16}{\pi }A=\frac{16}{\pi }\times \frac{\pi }{2}=8$