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}+4 x-2 x-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]
$=2[\int\limits_{0}^{2} \sqrt{2x} \,dx+\int\limits_{2}^{2 \sqrt{2}} \sqrt{8-x^{2}} d x]$
$=2\left[\sqrt{2}\left(x^{3 / 2}. \frac{2}{3}\right)_{0}^{2}+\left(\frac{x}{2} \sqrt{8-x^{2}}\right.\right.$
$+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}})_{2}^{2 \sqrt{2}}]$
$=2\left[\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}\right)+4 \times \frac{\pi}{2}-2-4 \times \frac{\pi}{4}\right]$
$=2\left[\frac{2 \sqrt{2}}{3}. 2 \sqrt{2}+2 \pi-2-\pi\right]$
$=2\left[\frac{8}{3}-2+\pi\right]=2\left(\frac{2}{3}+\pi\right)$
$=2 \pi+\frac{4}{3}$