Why doesn't the author take the volume of a modern cpu or hard disk into account when
calculating the storage/calculation power of a current laptop? I mean the data storing volume
of a 2.5" hard disk platter is about
(3.14*6cm^2*50um) = 6e-4 dm^2 ~= 1e-3 dm^3. So only 16 orders of magnitude to go... . Same
applies for the volume of a cpu die. Or do I miss something?