If V₁, V₂ and V₃ be the volumes of a right circular cone. A sphere and a right circular cylinder having the same radius and same height then

$$\eqalign{ & {\text{Volume of cone }}{V_1} = \frac{1}{3}\pi {r^2}h \cr & {\text{Volume of sphere }}{V_2} = \frac{4}{3}\pi {r^3} \cr & = \frac{2}{3}\pi {r^2}\left( {2r} \right)\,\,\,\,\,\left[ {\therefore h = 2r} \right] \cr & = \frac{2}{3}\pi {r^2}h \cr & {\text{Volume of cylinder }}{V_3} = \pi {r^2}h \cr & {\text{Cone}}\left( {\pi {r^2}h} \right) = 3{V_1} \cr & {\text{Sphere}}\left( {\pi {r^2}h} \right) = \frac{3}{2}{V_2} \cr & {\text{Cylinder}}\left( {\pi {r^2}h} \right) = {V_3} \cr & 3{V_1} = \frac{3}{2}{V_2} = {V_3} \cr & 6{V_1} = 3{V_2} = 2{V_3} \cr} $$