Question

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

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

Answer 1

The correct option is B

$2(\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}{\text{sin}}^{-1}\left(\frac{1}{3}\right))$

$x^2+y^2=9$

$x^2+8x-9=0$

$x=\frac{-8\pm \sqrt{64+36}}{2}$

$x=\frac{-8\pm 10}{2}-9,1$

$x=1$

Area closed $=2[{\int }_0^{1}2\sqrt{2}\sqrt{x}dx+{\int }_1^3\sqrt{9-x^2}dx]=2[2\sqrt{2}{\int }_0^1\sqrt{x}dx+{\int }_1^3\sqrt{9-x^2}dx]$

On simplifying we get

$=2[\frac{\sqrt{2}}{3}+\frac{9\pi }{4}-\frac{9}{2}{\text{sin}}^{-1}\left(\frac{1}{3}\right)]$