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

We have, ${y^2} = 2x$ and ${x^2} + {y^2} = 4x$
${y^2} = 2x$ is parabola opening to the right of positive direction x-axis

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

$\Rightarrow$ ${(x - 2)^2} + {y^2} = 4,$ which is circle having centre at (2,0) and radius'2' Solving the curves we get,

${x^2} + 2x = 4x$

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

$\Rightarrow$ $x = 0,2$

When $x = 0,y = 0$ and when $x = 2,y \pm 2$

Thus points of intersection are (0,0),(2,2) and (2,-2)

The graph is as shown in the given figure.

From the figure, area of shaded region,
$= 2 \cdot \int_0^2 {\left[ {\sqrt {{2^2} - {{(x - 2)}^2}} - \sqrt {2x} } \right]dx}$

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

$= 2\left[ {\left( {0 + 0 - 0 + 2 \cdot \frac{\pi }{2}} \right) - \frac{{2\sqrt 2 }}{3}\left( {{2^{3/2}} - 0} \right)} \right] = 2\pi - \frac{{16}}{3}$ sq units