Part 4/5:
We start by considering the smaller cone formed by the water level. The radius at this height is unknown, so we denote it as ( r_{small} ). Using similar triangles, we can establish a relationship between the dimensions of the large cone and those of the smaller cone.
The triangles formed by the entire cone and the water-filled cone share the angle at the apex. Therefore, we can write:
[ \frac{r_{large}}{h_{large}} = \frac{r_{small}}{h_{small}} ]
Substituting the known values:
[ \frac{2}{6} = \frac{r_{small}}{3} ]
From this proportion, we can rearrange to find that:
[ r_{small} = \frac{2}{6} \times 3 = 1 ]
Having determined the radius of the smaller cone, we can now calculate the volume:
[ V = \frac{1}{3} \pi (1)^2 (3) = \pi ]