A conical vessel, whose vertical angle is $30^\circ$ as shown in the diagram, has water poured into it at such a rate that after $t$ seconds, the depth of the water is $x\text{ cm}$ and the radius of the water surface is $r\text{ cm}$.

1. Express the radius $r$ in terms of the depth $x$ using trigonometry.

From the diagram, the vertical angle is $30^\circ$. The right-angled triangle formed by the radius $r$, depth $x$, and the slanted side has an angle of $30^\circ$ at the vertex.
$\tan 30^\circ = \frac{r}{x}$
$r = x \tan 30^\circ = x \left(\frac{1}{\sqrt{3}}\right) = \frac{x}{\sqrt{3}}$

2. Substitute $r$ into the formula for the volume of a cone.

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$.
Here, $h = x$.
$V = \frac{1}{3}\pi \left(\frac{x}{\sqrt{3}}\right)^2 x$

3. Simplify the expression to show the required result.

$V = \frac{1}{3}\pi \left(\frac{x^2}{3}\right) x$
$V = \frac{\pi x^3}{9}$