The volume [tex]V[/tex] of a cone is given by the formula [tex]V=\frac{1}{3}A_b\cdot h[/tex] , where [tex]A_b[/tex] is the area of the base and [tex]h[/tex] is the height of the cone.
The cone in question here is a right circular base cone. The volume of a right circular base cone with radius [tex]r[/tex] is,
[tex]V=\frac{1}{3} \pi r^2h\\[/tex].
In this problem, the diameter is given instead of the radius. If the diameter is 8 units, the radius of the cone is then 4 units because it is half of the radius.
The volume of this cone is therefore,
[tex]V=\frac{1}{3} \pi r^2=\frac{1}{3}\pi (4)^2\times 6 =\frac{\pi} {3}\times 16\times 6= 32\pi .[/tex]
The volume of this cone is [tex]32\pi units^3[/tex]
