RCA
Posts:
11
Registered:
11/26/05


Re: Area and volume of solid of revolution
Posted:
Nov 27, 2005 8:21 PM


>>>Take a thin slice of this sphere perpendicular to the >>>xyplane that goes through the point x = a. This thin >>>slice is a disk with radius f(a) and so the area of this >>>disk is pi * [f(a)]^2.
>>>Thus, the volume for our sphere should be the integral >>>of pi * (1  x^2) * dx from 1 to 1. Indeed, if we >>>evaluate this integral we get 4 * pi / 3, which is the >>>volume of a unit sphere.
I agree and understand that we get the right value for volume of a sphere this way. I just dont understand the argument leading up to the result. Why isnt the volume greater than pi*(1x^2)dx  since the curve does slant even in the infinitesimal distance and the volume of the thin disk is in fact greater than pi*(1x^2)dx (since the disk is in fact better approximated by a frustum of a cone with slant height ds than a cylinder with height dx in the general case), and hence is somewhat better represented by pi*(1x^2)* ds.

