Find the area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$.

$2(\pi - \frac{8}{3})$ square units

The correct answer is Option (3) → $2(\pi - \frac{8}{3})$ square units

We have,

$y^2 = 2x \quad \dots(i)$

and

$x^2 + y^2 = 4x \quad \dots(ii)$

On solving Eqs. (i) and (ii), we get

$x^2 + 2x = 4x$

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

$\Rightarrow x(x - 2) = 0$

$\Rightarrow x = 0, 2$

Also, $x^2 + y^2 = 4x \Rightarrow x^2 - 4x + y^2 = 0$

$\Rightarrow x^2 - 4x + (2)^2 - (2)^2 + y^2 = 0$

$\Rightarrow (x - 2)^2 + y^2 = 2^2 \text{ which is equation of a circle with centre (2, 0) and radius 2 units.}$

$∴$ Required area $= 2 \cdot \int\limits_{0}^{2} [\sqrt{2^2 - (x - 2)^2} - \sqrt{2x}] \, dx$

$= 2 \left[ \left[ \frac{x - 2}{2} \sqrt{2^2 - (x - 2)^2} + \frac{2^2}{2} \sin^{-1} \left( \frac{x - 2}{2} \right) \right] - \left[ \sqrt{2} \cdot \frac{x^{3/2}}{3/2} \right] \right]_{0}^{2}$$

$= 2 \left[ \left( 0 + 0 - 1 \cdot 0 + 2 \cdot \frac{\pi}{2} \right) - \frac{2\sqrt{2}}{3} (2^{3/2} - 0) \right]$

$= \frac{4\pi}{2} - \frac{8 \cdot 2}{3} = 2\pi - \frac{16}{3} = 2 \left( \pi - \frac{8}{3} \right) \text{ sq. units}$