Question

The area (in sq.units) of the region $\left\{\right(x,y):y^2\ge 2x$ and $x^2+y^2\le 4x,x\ge 0,y\ge 0$ is

• A. $\pi -\frac{8}{3}$
• B. $\pi -\frac{4\sqrt{2}}{3}$
• C. $\frac{\pi }{2}-\frac{2\sqrt{2}}{3}$
• D. $\pi -\frac{4}{3}$

Answer 1

The correct option is A $\pi -\frac{8}{3}$

Consider $y^2=2x$ and $x^2+y^2=4x$, on solving,

we get $x^2+2x=4x$

$x^2=2x\Rightarrow x=0,2$

The integral lies between $0$ and $2$

$y^2=2x$ $y^2+x^2=4x$

$\Rightarrow y=\sqrt{2x}$ $\Rightarrow y=\sqrt{4x-x^2}$

Area $={\int }_0^2\sqrt{4x-x^2}-\sqrt{2x}dx$

$={\int }_0^2\sqrt{2^2-(2-x{)}^2}-{\int }_0^2\sqrt{2x}dx$

$={[\frac{x-2}{2}\sqrt{4x-x^2}-\frac{4}{2}{\sin }^{-1}\left(\frac{x-2}{2}\right)]}_0^2-{\left[\frac{\sqrt{2}}{3/2}{x}^{3/2}\right]}_0^2$

$=\left|2{\sin }^{-1}\right(-1\left)\right|-\frac{2\sqrt{2}}{3}\cdot 2\sqrt{2}$

$=\pi -\frac{8}{3}$.