Given curves are
$$
\begin{aligned}
& y=a x^2 \\
& x=a y^2
\end{aligned}
$$
and
Put the value of $y$ by Eq. (ii) in Eq. (i), we get


$$
\begin{aligned}
& & \frac{\sqrt{x}}{\sqrt{a}}=a x^2 \Rightarrow \frac{\sqrt{x}}{\sqrt{a}}-a x^2 & =0 \\
\Rightarrow & & \sqrt{x}\left(\frac{1}{\sqrt{a}}-a x^{3 / 2}\right) & =0 \\
\Rightarrow & & \sqrt{x}\left(1-a^{3 / 2} x^{3 / 2}\right) & =0 \Rightarrow x=0, \frac{1}{a} \\
& \text { When, } & x=0, y & =0 \\
& \text { and } & x=\frac{1}{a}, y & =\frac{1}{a}
\end{aligned}
$$
When, and
$$
\begin{aligned}
& x=0, y=0 \\
& x=\frac{1}{a}, y=\frac{1}{a}
\end{aligned}
$$
Since, shaded part lies in first quadrant, hence points of intersection of curves $y=a x^2$ and $x=a y^2$ are $(0,0)$ and $\left(\frac{1}{a}, \frac{1}{a}\right)$
$\therefore$ Required area
$$
\begin{aligned}
3 & =\int_0^{\frac{1}{a}} \frac{\sqrt{x}}{\sqrt{a}} d x-\int_0^{\frac{1}{a}} a x^2 d x \\
\Rightarrow \quad 3 & =\frac{1}{\sqrt{a}} \int_0^{\frac{1}{a}} \sqrt{x} d x-a \int_0^{\frac{1}{a}} x^2 d x \\
\Rightarrow \quad 3 & =\frac{1}{\sqrt{a}} \frac{2}{3}\left[x^{3 / 2}\right]_0^{\frac{1}{a}}-a\left[\frac{x^3}{3}\right]_0^{\frac{1}{a}}
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow & 3 & =\frac{2}{3 \sqrt{a}}\left[\left(\frac{1}{a}\right)^{3 / 2}\right]-\frac{a}{3}\left[\left(\frac{1}{a}\right)^3\right] \\
\Rightarrow & 3 & =\frac{2}{3 \sqrt{a}} \times \frac{1}{a \sqrt{a}}-\frac{a}{3} \times \frac{1}{a^3} \\
\Rightarrow & 3 & =\frac{2}{3 a^2}-\frac{1}{3 a^2} \\
\Rightarrow & 3 & =\frac{2-1}{3 a^2} \Rightarrow 9 a^2=1 \\
\Rightarrow & a^2 & =\frac{1}{9} \Rightarrow a=\frac{1}{3}
\end{aligned}
$$
[since, shaded region lies in first quadrant] Hence, value of $a=\frac{1}{3}$.