Date: Nov 27, 2005 8:21 PM
Author: RCA
Subject: Re: Area and volume of solid of revolution
>>>Take a thin slice of this sphere perpendicular to the

>>>xy-plane 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*(1-x^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*(1-x^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*(1-x^2)* ds.