Given curves $${x^2} = {y \over 4}$$ and $${x^2} = 9y$$ are the parabolas whose equations can be written as $$y = 4{x^2}$$ and $$y = {1 \over 9}{x^2}.$$

Also, given $$y=2.$$

Now, shaded portion shows the required area which is symmetric.



$$\therefore$$ Area $$ = 2\int\limits_0^2 {\left( {\sqrt {9y} - \sqrt {{y \over 4}} } \right)} dy$$

Area $$ = 2\int\limits_0^2 {\left( {3\sqrt y - {{\sqrt y } \over 2}} \right)} dy$$

$$ = 2\left[ {{2 \over 3} \times 3.{y^{{3 \over 2}}} - {1 \over 2} \times {2 \over 3}.{y^{{3 \over 2}}}} \right]_0^2$$

$$ = 2\left[ {2{y^{{3 \over 2}}} - {1 \over 3}{y^{{3 \over 2}}}} \right] = \left. {2 \times {5 \over 3}{y^{{3 \over 2}}}} \right|_0^2$$

$$ = 2.{5 \over 3}2\sqrt 2 = {{20\sqrt 2 } \over 3}$$