The area (in sq . units ) of the region (x, y): y ^ 2 ≥ 2x and x ^...

${y}^{2} \ge 2x$ (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 ${y}^{2} = 2x$

$⇒ {x}^{2} + 2x = 4x$

$⇒ {x}^{2} = 2x$

$⇒ x = 0, x = 2$

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

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

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

Hence, area

$= \left[\mathop{∫}_{0}^{2}\sqrt{4x - {x}^{2}}dx - \sqrt{2}\mathop{∫}_{0}^{2}\sqrt{x}dx\right]$

$= \left[\mathop{∫}_{0}^{2}\sqrt{4 - {\left(x - 2\right)}^{2}}dx - \sqrt{2}\mathop{∫}_{0}^{2}\sqrt{x}dx\right]$

$= {\left[\left(\dfrac{x - 2}{2}\right)\sqrt{4x - {x}^{2}} + \dfrac{4}{2} . \sin^{- 1}\left(\dfrac{x - 2}{2}\right) - \sqrt{2}\left(\dfrac{x\dfrac{3}{2}}{\dfrac{3}{2}}\right)\right]}_{0}^{2}$

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