Drexel dragonThe Math ForumDonate to the Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Topic: Area and volume of solid of revolution
Replies: 14   Last Post: Nov 28, 2005 11:13 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 11
Registered: 11/26/05
Re: Area and volume of solid of revolution
Posted: Nov 27, 2005 8:21 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

>>>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.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum 1994-2015. All Rights Reserved.