Question

The area of the smaller portion enclosed by the curves $x^2+y^2=9$ and $y^2=8x$ is

• A. $\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)$
• B. $2(\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right))$
• C. $2[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}+\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)]$
• D. $[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}+\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)]$
  

Answer 1

The correct option is B $2(\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right))$



$x^2+y^2=9$
$x^2+8x-9=0$
$x=\frac{-8\pm \sqrt{64+36}}{2}$
$x=\frac{-8\pm 10}{2}=-9,1$
$x=1$
Area enclosed =$2[{\int }_0^{1}2\sqrt{2x}dx+{\int }_1^3\sqrt{9-x^2}dx]=2[2\sqrt{2}{\int }_0^{1}2\sqrt{x}dx+{\int }_1^3\sqrt{9-x^2}dx]$
On simplifying we get
$=2[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}si{n}^{-1}\left(\frac{1}{3}\right)]$