设：底圆半径为r,高为h,则圆椎体积V=Pi/3*h*r^2.由内切圆这个条件可得关系式：(h-1)/(h^2+r^2)^0.5=1/r.化简：h=2r^2/(r^2-1),代入：V=2Pi/3*r^4/(r^2-1),求导：V'=2Pi/3*[4r^3(r^2-1)-r^4*2r]/(r^2-1)^2=4Pi/3*r^3*(r^2-2)/(r^2-1)^2,令V'=0得：r=2^0.5.所以：Vmin=8Pi/3.