$\text{The area of the region }\left\{(x,y):y^2\le 4x\right.\text{ , }\left.4x^2+4y^2\le 9\right\}$

$\text{ Given inequalities are }$

$y^2\le 4x--------\left(1\right)$

$\text{ and }4x^2+4y^2\le 9-----------\left(2\right)$

$\text{ Points satisfying (1) lies interior to parabola }{y}^2=4x\text{ and those }$$\text{ of }\left(2\right)\text{ lies inside circle }4x^2+4y^2=9$

$\text{ Solving the curves we get, }$

$4x^2+16x=9$

$\text{ or }(2x-1)(2x+9)=0$

$\text{ or }x=1/2\text{ (as }x=\frac{-9}{2}\text{ not possible })$

$\text{ Therefore, the points of intersection of both curves are }\left(\frac{1}{2},\sqrt{2}\right)$$\text{ and }\left(\frac{1}{2},-\sqrt{2}\right)\text{ . }$

The graph of these two curves and the common region of the points satisfying both the inequalities is as shown in adjacent figure

From the figure, required area is given by
$A=2{\int }_1^{\frac{1}{2}} 2\sqrt{x}dx+2{\int }_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{2}\sqrt{9-4x^2}dx$

$\text{ Putting }2x=t\text{ , dx=}\frac{dt}{2}\text{in the second integral, we get }$

$\begin{array}{r} \\ \therefore A=2\left[{\int }_0^{\frac{1}{2}} 2\sqrt{x}dx+\frac{1}{4}{\int }_1^3 \sqrt{(3{)}^2-(t{)}^2}dt\right]\\ =2\left[{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{1}{2}}+\frac{1}{4}{\left[\frac{t}{2}\sqrt{9-t^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{t}{3}\right)\right]}_1^3\right)\end{array}$

$\begin{array}{c}=2\left(\frac{2}{3\sqrt{2}}+\frac{1}{4}\left[\left\{0+\frac{9}{2}{\sin }^{-1}\left(1\right)\right\}-\left\{\frac{1}{2}\sqrt{8}+\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\right\}\right]\right)\\ =2\left(\frac{9\pi }{16}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12}\right)\end{array}$