Answer:
[tex]8 \pi[/tex]
Step-by-step explanation:
The computation of the volume of water remaining is shown below:
The volume of the cone is
[tex]\pi r^2 \frac{h}{3}\\\\\pi \times 3^2 \times \frac{8}{3} \\\\\frac{72}{3} \pi \\\\24 \pi[/tex]
Now since it is two-third so the remaining is one -third
[tex]= 24 \pi \times \frac{1}{3}[/tex]
[tex]= 8 \pi[/tex]
hence, the volume of water remaining is [tex]8 \pi[/tex]