The volume of a normal Cone = π × r² × (h/3)
If the cone were chopped off at "y" then we would remove a smaller cone, whose height would be (h-y) and whose base radius would be r × (h-y)/h
Hence the "removed cone"s volume would = π × (r × (h-y)/h)² × ((h-y)/3)
So the Volume of the truncated cone = π × r² × (h/3) - π × (r × (h-y)/h)² × ((h-y)/3)
The areas of a normal cone are:
Base = π × r²
Surface Area of Side = π × r × s (where s = side length)
The areas of a truncated cone are (h=height of normal cone, y=height of truncated cone):
Base = π × r²
Top = π × (r × (h-y)/h)²
Surface Area of Side = π × r × s × (y/h)
That's a start, but I gotta go ...
]]>