Step 1:

The curve $x^2=4y$ is the equation of the parabola with vertex at the origin and axis along y-axis and open upwards.

Let $R_1$ be the region lying inside the parabola $x^2=4y\Rightarrow y=\frac{x^2}{4}$

We have $4x^2+4y^2=9$

$x^2+y^2=\large\frac{9}{4}$$\Rightarrow y^2=\large\frac{9}{4}-x^2$

$y=\sqrt {\large\frac{9}{4}-x^2}$

Let $R_2$ be the region lying inside the circle.

Hence the area of the required region is bounded by the parabola $x^2=4y$ and the circle $x^2+y^2=\large\frac{9}{4}.$

Step 2:

To find the point of intersection let us substitute $x^2=4y$ in the equation of the circle.

$4y+y^2=\large\frac{9}{4}$$\Rightarrow 4y^2+16y-9=0$

On factorising we get,

(2y+9)(2y-1)=0

$\Rightarrow x=\large\frac{-9}{2}$

and $y=\large\frac{1}{2}$.

Hence the points of intersection are ($\sqrt 2,\frac{1}{2})(\sqrt 2,\frac{-1}{2}).$

Step 3:

The required area is $A=\int_0^{\Large\frac{1}{\sqrt 2}}(R_2-R_1)dx$

$\Rightarrow\int_0^{\Large\frac{1}{\sqrt 2}}\large\frac{\sqrt {9-4x^2}}{2}-\int_0^\frac{1}{\sqrt 2}\frac{x^2}{4}$$dx$.

$A=\frac{1}{2}\int_0^{\Large\frac{1}{\sqrt 2}}\sqrt {9-4x^2}-\frac{1}{4}\int_0^\sqrt 2x^2dx.$

On integrating we get,

$\Rightarrow\large\frac{1}{2}.\frac{1}{2}\begin{bmatrix}\frac{2x}{2}\sqrt{9-4x^2}+\frac{9}{2}\sin^{-1}\frac{2x}{3}\end{bmatrix}_0^\sqrt 2-\frac{1}{4}\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^\sqrt 2$

On applying the limits we get,

$\Rightarrow\large\frac{\sqrt 2}{12}+\frac{9}{8}\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.

Step 4:

But the curves are symmetrical about x-axis the required area is

$A=2\bigg[\large\frac{\sqrt 2}{12}+\frac{9}{8}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)\bigg]$

$\;\;\;=\large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\frac{2\sqrt 2}{3}\big)$.

Hence the required area is $\large\frac{\sqrt 2}{6}+\frac{9}{4}$$\sin^{-1}\big(\large\frac{2\sqrt 2}{3}\big)$.