since the diameter of the cone is 2 inches, then its radius is half that, namely 1 inch, thus
[tex]\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=1\\ h=6 \end{cases}\implies V=\cfrac{\pi (1)^2(6)}{3}\implies V=2\pi \\\\[-0.35em] ~\dotfill\\\\ \cfrac{vanilla}{chocolate}\qquad 2:1\qquad \cfrac{2}{1}\qquad \qquad \cfrac{2\cdot \frac{2\pi }{2+1}}{1\cdot \frac{2\pi }{2+1}}\implies \cfrac{~~\frac{4\pi }{3}~~}{\frac{2\pi }{3}} \\\\[-0.35em] ~\dotfill\\\\ vanilla\implies \cfrac{4\pi }{3}\qquad \approx\qquad 4.19~in^3[/tex]
