The respective equations for the parabola and the circle are:
${\mathit{y}}^2$ = 4x ... (1)
$4{\mathit{x}}^2+4{\mathit{y}}^2$ = 9 ... (2)
or ${\mathit{x}}^2+{\mathit{y}}^2$ = ${\left(\frac{3}{2}\right)}^2$
Equation (1) is a parabola with vertex (0, 0) which opens to the right and equation (2) is a circle with centre (0, 0) and radius $\frac{3}{2}$
From equations (1) and (2), we get:
$4{\mathit{x}}^2$ + 4 (4x) = 9
$4{\mathit{x}}^2$ + 16x - 9 = 0
$4{\mathit{x}}^2$ + 18x - 2x - 9 = 0
2x (2x + 9) - 1 (2x + 9) = 0
(2x + 9) (2x - 1) = 0
x = - $\frac{9}{2},\frac{1}{2}$
For x = $-\frac{9}{2}$ , ${\mathit{y}}^2$ = 4 $\left(-\frac{9}{2}\right)$ , which is not possible, hence x = $\frac{1}{2}$
Therefore, the given curves intersect at x = $\frac{1}{2}$

Required area of the region bound by the two curves
= 2 $\underset{0}{\overset{\frac{1}{2}}{\int }}2\sqrt{\mathit{x}}\mathit{d}\mathit{x}$ + $2\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\int }}\sqrt{\frac{9}{4}-{\mathit{x}}^2}\mathit{d}\mathit{x}$
= 4 ${\left|\frac{2}{3}{\mathit{x}}^{\frac{3}{2}}\right|}_0^{\frac{1}{2}}$ + 2
= $\frac{8}{3}{\left(\frac{1}{8}\right)}^{\frac{1}{2}}$ + 2
= $\frac{8}{3}\left(\frac{1}{2\sqrt{2}}\right)$ + $\frac{9}{4}\left(\frac{\mathit{\pi }}{2}\right)$ - $\frac{\sqrt{2}}{2}-\frac{9}{4}\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{1}{3}\right)$
= $\frac{2\sqrt{2}}{3}+\frac{9\mathit{\pi }}{8}$ - $\frac{\sqrt{2}}{2}-\frac{9}{4}\mathit{s}\mathit{i}{\mathit{n}}^{-1}\left(\frac{1}{3}\right)$