Question

If the area of the curve enclosed by $y|y|\pm x|x|=1$ and $y=|x|$ is $A,$ then $\frac{\pi }{A}$ is
(exclude infinite area regions enclosed between the curves)

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is C $4$

$y|y|\pm x|x|=1$ in the $1^{st}$ and $2^{nd}$ quadrant is
$y^2+x^2=1$ and $y^2-x^2=1$ respectively.
In $3^{rd}$ and $4^{th}$ quadrant, it is $y^2+x^2=-1$ and
$x^2-y^2=1$ respectively


The point of intersection of the curves $x^2+y^2=1$and $y=x$ is $x=\frac{1}{\sqrt{2}}$
$\therefore \text{Area}=2\left(\underset{0}{\overset{1/\sqrt{2}}{\int }}\sqrt{1-x^2}-xdx\right)$
$=2[{(\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}{\sin }^{-1}x)}_0^{1/\sqrt{2}}-{\left(\frac{x^2}{2}\right)}_0^{1/\sqrt{2}}]$
$=\frac{\pi }{4}$
$\Rightarrow \frac{\pi }{A}=4$