RCA
Posts:
11
Registered:
11/26/05
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Re: Area and volume of solid of revolution
Posted:
Nov 27, 2005 8:21 PM
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>>>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.
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