The area bounded by the curve $y^2=4ax$, $y-$ axis between $y=-a$ and $y=a$ is

The given curves are $y^2=4ax$

To find the area of the region bouned by the curve $y^2=4ax$, $Y-$ axis and the lines $y=a,y=-a$, draw the rough sketch as shown below

The required area is the area of the shaded region

We know that the area of the  region bounded by the curve $x=f\left(y\right)$, $y=a,y=b$ and $Y-$axis is ${\int }_a^{b}f\left(y\right)dy$

Hence, the area of the shaded region is ${\int }_{-a}^a\frac{y^2}{4a}dy$

$\begin{array}{rcl}{\int }_{-a}^a\frac{y^2}{4a}dy& =& 2\left(\frac{1}{4a}\right){\int }_0^a{y}^{2}dy\\ & =& \frac{1}{2a}{\left[\frac{y^3}{3}\right]}_0^a\\ & =& \frac{1}{2a}\left(\frac{a^3}{3}\right)\\ & =& \frac{a^2}{6}\end{array}$

Therefore, the area of the rquired region is $\boxed{\frac{a^2}{6}}$