we know the base of the cone has a diameter of 6, thus its radius must be half that or 3.
[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=8 \end{cases}\implies \begin{array}{llll} V=\cfrac{\pi (3)^2(8)}{3}\implies V=24\pi \\\\\\ V\approx 75.4~in^2 \end{array}[/tex]
