well then, the volume of the nose cone will just be the sum of the volume of the cylinder below and the cone above.
since the diameter for both is 8, then their radius is half that, or 4.
[tex]\bf \stackrel{\textit{volume of a cone}}{V=\cfrac{\pi r^2 h}{3}}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\cfrac{\pi (4)^2(6)}{3}\implies V=32\pi \\\\\\ \stackrel{\textit{volume of a cylinder}}{V=\pi r^2 h}~~ \begin{cases} r=radius\\ h=height\\ \cline{1-1} r=4\\ h=6 \end{cases}\implies V=\pi (4)^2(6)\implies V=96\pi \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{volume of the nose cone}}{32\pi +96\pi \implies 128\pi }\qquad \approx \qquad 402.12[/tex]
