$x^2+y^2 \leq 16a^2$
Let us consider $x^2+y^2=16a^2$
This represents a circle with centre at the orgin and radius 4a
The praola is $y^2=6ax$
The required area bounded by the parabola $y^2=6ax$ and the circle $x^2+y^2=16a^2$ is shown by the shades portion.
The points of intersection can be found by solving the two equations:
$x^2+y^2=16a^2$-----(1)
$y^2=6ax$-----(2)
substitute equ(2) in equ(1)
$x^2+6ax=16a^2$
$=>x^2+6ax-16a^2=0$
$=>(x+8a)(x-2a)=0$
Therefore $x=-8a,2a$
If $x=-8a,$ y is imaginary
If $x=2a,y^2=12a^2=>y=\pm 2 \sqrt 3 a$
Therefore The points of intersection $(2a,2 \sqrt 3 a)\;ana\;(2a,-2 \sqrt 3 a)$
Therefore area=2 [area of OAMO+area of AMPA]
$=2\bigg[\int \limits_0^{2a} \sqrt {6ax}\; x+\int \limits_{2a}^{4a} \sqrt {16a^2-x^2}dx\bigg]$
$=2 \sqrt {6a} \bigg[\large\frac{x^{3/2}}{3/2}\bigg]_0^{2a}+2\bigg[\frac{x}{2}\sqrt {16a^2-x^2}+\frac{16a^2}{2} \sin^{-1}(\frac{x}{4a})\bigg]_{2a}^{4a}$
On Applying the limits,
$=2 \sqrt {6a} \times \large\frac{2}{3} \bigg[(2a)^{3/2}-0\bigg]+2\bigg[\frac{16a}{2} \sqrt {16a^2-16a^2}+\frac{16a^2}{2} \sin^{-1}(\frac{4a}{4a})-\frac{4a}{2} \sqrt {16a^2-4a^2}-\frac{16a^2}{2} \sin^{-1}(\frac{2a}{4a})\bigg]$
$=\large\frac{16a^2}{\sqrt 3}+0+16a^2 \sin^{-1}(1)-\frac{4a^2}{2} \sqrt{12}-16a^2 \sin^{-1}(1/2)$
$=\large\frac{16a^2}{\sqrt 3}+16a^2.\frac{\pi}{2}-4 \sqrt 3 a^2 - 16a^2 \frac{\pi}{6}$
$\large\frac{4a^2}{\sqrt 3}+\frac{16a^2 \pi}{3}$
$=\large\frac{4a^2}{3}(4 \pi+\sqrt 3) sq.units.$