Solution:

Eq. of parabola is $y^{2}=4 x$……(i)
Eq. of circle is $4 x^{2}+4 y^{2}=9 \ldots \ldots$ (ii)

From the diagram, the points of intersection of parabola (i) and circle (ii) are
$A^{\left(\frac{1}{2}, \sqrt{2}\right)} \text { and } B\left(\frac{1}{2},-\sqrt{2}\right)$

The required shaded area OADBO

$=2 \times$ Area $O A D O=2[$ Area OAC + Area CAD]

$=\left[\begin{array}{l}
2\left[\int_{0}^{\frac{1}{2}} 2 \sqrt{x} d x+\int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{\frac{9}{4}-x^{2}} d x\right] \\
\left.=\left[2 \cdot \frac{x^{\frac{3}{2}}}{\frac{1}{2}}\right\}_{0}^{\frac{1}{2}}+\left\{\frac{x \sqrt{\frac{9}{4} x^{2}}}{2}+\frac{9}{2} \sin ^{-1} \frac{x}{3 / 2}\right\}_{\frac{1}{2}}^{\frac{1}{2}}\right\}
\end{array}\right.$

$=\left[\frac{4}{3} \times \frac{1}{2 \sqrt{2}}+\frac{9}{8} \sin ^{-1} 1-\frac{\frac{1}{2} \sqrt{2}}{2}-\frac{9}{8} \sin ^{-1} \frac{1}{3}\right]$
$=\left[\frac{\sqrt{2}}{3}+\frac{9}{8} \cdot \frac{\pi}{2}-\frac{\sqrt{2}}{4}-\frac{9}{8} \sin ^{-1} \frac{1}{3}\right]$
$=\left(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}+\frac{\sqrt{2}}{6}\right) \text { sq. units }$