JEE Main 2016 — Mathematics Calculus

${y}^{2}\geq 2x$y2≥2x (Area outside the parabola)

${x}^{2}+{y}^{2}\leq 4x$x2+y2≤4x (Area inside the circle)

Finding point of intersection of the curves, we get,

${x}^{2}+{y}^{2}=4x$x2+y2=4x and ${y}^{2}=2x$y2=2x

$$\Rightarrow {x}^{2}+2x=4x$$⇒x2+2x=4x

$$\Rightarrow {x}^{2}=2x$$⇒x2=2x

$$\Rightarrow x=0, x=2$$⇒x=0,x=2

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

Coordinates of $A(0, 0), B(2, 2)$A(0,0),B(2,2)

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

Hence, area

$$=[\int _{0}^{2}\sqrt{4x-{x}^{2}} dx- \sqrt{2} \int _{0}^{2}\sqrt{x} dx]$$=[∫02​4x−x2​dx−2​∫02​x​dx]

$$=[\int _{0}^{2}\sqrt{4-{(x-2)}^{2}} dx- \sqrt{2} \int _{0}^{2}\sqrt{x} dx]$$=[∫02​4−(x−2)2​dx−2​∫02​x​dx]

$$={[(\frac{x-2}{2}) \sqrt{4x-{x}^{2}}+\frac{4}{2} .{\mathrm{sin}}^{-1}(\frac{x-2}{2})- \sqrt{2} (\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}})]}_{0}^{2}$$=[(2x−2​)4x−x2​+24​.sin−1(2x−2​)−2​(23​x23​​)]02​

$=[0-2{\mathrm{sin}}^{-1}(-1)]- \sqrt{2} .\frac{2}{3}2\sqrt{2}=$=[0−2sin−1(−1)]−2​.32​22​=$[\pi -\frac{8}{3}]\mathrm{sq}.\mathrm{units}$[π−38​]sq.units