JEE Main 2016 — Area Under Curves Question with Solution

$y^2\ge 2x$ (Area outside the parabola)

$x^2+y^2\le 4x$ (Area inside the circle)


Finding point of intersection of the curves, we get, 

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

$\Rightarrow x^2+2x=4x$

$\Rightarrow x^2=2x$

$\Rightarrow x=0, x=2$

If $x=0,$then $y=0$ and if $x=2$, then $y=\pm 2.$

Coordinates of $\mathrm{A}\left(0,\mathrm{ }0\right), B\left(2, 2\right)$

As $x \ge 0 ,y \ge 0$ only area above $x$-axis would be considered.

Hence, area

$=\left[\underset{0}{\overset{2}{\int }}\sqrt{4x-x^2} dx- \sqrt{2} \underset{0}{\overset{2}{\int }}\sqrt{x} dx\right]$

$=\left[\underset{0}{\overset{2}{\int }}\sqrt{4-{\left(x-2\right)}^2} dx- \sqrt{2} \underset{0}{\overset{2}{\int }}\sqrt{x} dx\right]$

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

$=\left[0-2{\sin }^{-1}\left(-1\right)\right]- \sqrt{2} .\frac{2}{3}2\sqrt{2}=$$\left[\pi -\frac{8}{3}\right] \mathrm{sq}.\mathrm{units}$