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

Answer: (C)

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

On solving Eqs. (i) and (ii), we get
$\therefore x^2+2x-8=0$
$x^2+4x-2x-8=0$
$x(x+4)-2(x+4)=0$
$(x-2)(x+4)=0$
$\therefore x=2$ and $y=\pm 2$
$\therefore$ Required area
$=2$ [Area of OAP + Area of PAB]
$=2[\underset{0}{\overset{2}{\int }}\sqrt{2x}dx+\underset{2}{\overset{2\sqrt{2}}{\int }}\sqrt{8-x^2}dx]$
$=2[\sqrt{2}{\left(x^{3/2}.\frac{2}{3}\right)}_0^2+(\frac{x}{2}\sqrt{8-x^2}$
$+\frac{8}{2}{\sin }^{-1}\frac{x}{2\sqrt{2}}{)}_2^{2\sqrt{2}}]$
$=2\left[\frac{2\sqrt{2}}{3}\left(2^{3/2}\right)+4\times \frac{\pi }{2}-2-4\times \frac{\pi }{4}\right]$
$=2\left[\frac{2\sqrt{2}}{3}.2\sqrt{2}+2\pi -2-\pi \right]$
$=2\left[\frac{8}{3}-2+\pi \right]=2\left(\frac{2}{3}+\pi \right)$
$=2\pi +\frac{4}{3}$