Ok so what we do is how much greater=wasted space=cylindervolume-conevolume
volume of a cinlinder=hpir^2 volume of a cone=(1/3)hpir^2 we can alread sybsitute something (3/3)hpir^2-(1/3)hpir^2=(2/3)hpir^2 so wasted space=(2/3)hpir^2
diameter=4 d=2r d/2=r 4/2=2=r
and 6=height so 2/3(6)(3.14)(2^2) 4(3.14)(4) 16(3.14) 50.24 in^3 wasted
