Question

The area of the smaller portion between curves $x^2+y^2=8$ and $y^2=2x$ is

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

Answer 1

Answer: (C)

$2\pi +\frac{4}{3}$

Given curves
$x^2+y^2=8$ ....(1)
Center is (0,0) and radius $r=2\sqrt{2}$
$y^2=2x$ ....(2)
Solving (1) and (2), we get
$x^2+2x=8$
$\Rightarrow x=2,-4$
At $x=2\Rightarrow y=2,-2$
Hence, the point of intersection of the curves is $(2,2)$ and $(2,-2)$
Now, required area $=2\left(areaOPA\right)$
$=2\left[ar\right(OPB)+ar(PBA\left)\right]$
$=2{\int }_0^2\sqrt{2x}dx+2{\int }_2^{2\sqrt{2}}\sqrt{8-x^2}dx$
$=2\sqrt{2}[\frac{x^{3/2}}{3/2}{]}_0^2+2[\frac{x}{2}\sqrt{8-x^2}+4{\sin }^{-1}\frac{x}{2\sqrt{2}}{]}_2^{2\sqrt{2}}$
$=\frac{4\sqrt{2}}{3}\left(2\sqrt{2}\right)+2(0+2\pi -2-\pi )$
Required Area $=(2\pi +\frac{4}{3})$ sq.units