Draw a rough sketch of the region $\left\{ {(x,y):{y^2} \le 6ax} \right.$ and $\left. {{x^2} + {y^2} \le 16{a^2}} \right\}$. Also find the area of the region sketched using method of integration.

${y^2} \le 6ax,$ which represents the region interior to parabola ${y^2} = 6ax$ towards focus.

And ${x^2} + {y^2} \le 16{a^2}$, which represents the region interior to circle${x^2} + {y^2} = 16{a^2}$.

Solving circle and parabola,
we get

${x^2} + 6ax = 16{a^2}$

$\Rightarrow$ ${x^2} + 6ax - 16{a^2} = 0$

$\Rightarrow$ $(x - 2a)(x + 8a) = 0$

$\Rightarrow$ $x = 2a$

(as $x = - 8a$ is not possible)

Putting $x = 2a$ in parabola, we get
The graph of functions are as shown in the given figure.

From the figure, area of the shaded region

$A = 2\left[ {\int_0^{2a} {\sqrt {6ax} } dx + \int_{2a}^{4a} {\sqrt {{{(4a)}^2} - {x^2}} } dx} \right]$

$= 2\left[ {\sqrt {6a} \left( {\frac{2}{3}{x^{3/2}}} \right)_0^{2a} + \left( {\frac{x}{2}\sqrt {{{(4a)}^2} - {x^2}} + \frac{{{{(4a)}^2}}}{2}{{\sin }^{ - 1}}\frac{x}{{4a}}} \right)_{2a}^{4a}} \right]$

$= 2\left[ {\sqrt {6a} \frac{2}{3}{{(2a)}^{3/2}} + 8{a^2} \cdot \frac{\pi }{2} - \frac{{2a}}{2}\sqrt {16{a^2} - 4{a^2}} - 8{a^2} \cdot \frac{\pi }{6}} \right]$

$= 2\left[ {\sqrt {6a} \frac{2}{3} \cdot 2\sqrt 2 {a^{3/2}} + 4\pi {a^2} - a \cdot 2\sqrt 3 a - \frac{{4{a^2}}}{3}\pi } \right]$

$= 2\left[ {\frac{8}{3}\sqrt 3 {a^2} + 4\pi {a^2} - 2\sqrt 3 {a^2} - \frac{{4{a^2}\pi }}{3}} \right]$

$= 2\left[ {\frac{2}{3}\sqrt 3 {a^2} + \frac{{8{a^2}\pi }}{3}} \right] = \frac{4}{3}{a^2}[\sqrt 3 + 4\pi ]$