The area (in square units) of the region bounded by the curves $3y^2 = ax$, $y = a$, $a > 0$ and $y$-axis is:

The curves are $3y^2 = ax$, $y = a$, $a > 0$, and the $y$-axis.

The parabola $3y^2 = ax$ can be rewritten as:

$x = \frac{3y^2}{a}$

This parabola opens to the right and passes through the origin $(0, 0)$.

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The bounded region has:

Left boundary: $y$-axis where $x = 0$

Right boundary: the parabola where $x = \frac{3y^2}{a}$

Bottom: $y = 0$

Top: the line $y = a$

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At any height $y$, the width of the region is:

$\frac{3y^2}{a} - 0 = \frac{3y^2}{a}$

The area is:

$\text{Area} = \int_0^a \frac{3y^2}{a} \, dy$

$= \frac{3}{a} \int_0^a y^2 \, dy$

$= \frac{3}{a} \left[\frac{y^3}{3}\right]_0^a$

$= \frac{3}{a} \left(\frac{a^3}{3} - 0\right)$

$= \frac{3}{a} \cdot \frac{a^3}{3}$

$= \frac{a^3}{a}$

$= a^2$

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Therefore, the area of the region is $a^2$ square units.