Question

The area of the region bounded by the curves $x^2+y^2=8$ and $y^2=2x$ is

• A. $2\pi +\frac{1}{3}$
• B. $\pi +\frac{1}{3}$
• C. $2\pi +\frac{4}{3}$
• D. $\pi +\frac{4}{3}$

Answer 1

The correct option is C $2\pi +\frac{4}{3}$

Given curves,
$x^2+y^2=8$ .....(i)
and $y^2=2x$ .....(ii)

On solving Eqs. (i) and (ii), we get
$x^2+2x-8=0$
$x^2+4x-2x-8=0$
$x(x-4)-2(x+4)=0$
$(x-2)(x+4)=0$
$\therefore x=2$ and $y=\pm 2$

$\therefore$ Required area
$=2[\text{Area of OAP}+\text{Area of PAB}]$

$=2\left[{\int }_0^2\sqrt{2x}dx+{\int }_2^{2\sqrt{2}}\sqrt{8-x^2}dx\right]$

$=2[\sqrt{2}{(x^{3/2}\cdot \frac{2}{3})}_0^2+{(\frac{x}{2}\sqrt{8-x^2}+\frac{8}{2}{\sin }^{-1}\frac{x}{2\sqrt{2}})}_2^{2\sqrt{2}}]$

$=2\left[\frac{2\sqrt{2}}{3}\right(2^{3/2})+4\times \frac{\pi }{2}-2-4\times \frac{\pi }{4}]$

$=2[\frac{2\sqrt{2}}{3}\cdot 2\sqrt{2}+2\pi -2-\pi ]$

$=2[\frac{8}{3}-2+\pi ]+2(\frac{2}{3}+\pi )=2\pi +\frac{4}{3}$