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MathGroup Archive 1997

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Re: mathematica problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7606] Re: [mg7540] mathematica problem
  • From: Hugh Walker <hwalker at uh.edu>
  • Date: Thu, 19 Jun 1997 15:53:23 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

 Stephanie Gill <sgill at winnie.fit.edu> asks

<<******** Consider the region enclosed by y=sin^-1x, y=0, and x=1.  Find
the volume of the solid generated by revolving the region about the x-axis
using
a)  disks;
b)  cylindrical shells.

How to do with Mathematica? *********>>
*****************

Greetings Stephanie:

First you must set up the integral corresponding to what you want to do.
The volume you want has the shape of a "flared cone" symmetric about the
x-axis. Case a) results from slicing this "cone" perpendicular to the
x-axis. A slice at x of thickness dx is a disk of volume dV = Pi y(x)^2 dx;
the volume contained in the "cone" for 0 < x < 1 is

			V = PiIntegrate[y[x]^2,{x,0,1}]

In our case, y[x] = ArcSin[x], so the above integral is evaluated by the
mathematica command

			Vdisk = Pi Integrate[ArcSin[x]^2,{x,0,1}]

For b) one take cylindrical donuts about the x-axis of radius y[x], height
(1-x), and thickness dy. The volume of such a donut is dV = 2 Pi
y(1-x(y))dy. The volume calculated by this formulation, using x[y] = Sin[y]
and 0 < y < Pi/2, is

			Vcyl = 2 Pi Integrate[ y Sin[y] ,{y,0,Pi/2})]

There are, of course, sill other ways to find the same volume. For example,
use the disks as in a) , but write everthing in sight in terms of y instead
of x. This leads to the mathematica command

			Vother = Pi Integrate[y^2 Cos[y],{y,0,Pi/2}]

I hope you find my suggestions useful.



Hugh Walker
hwalker at uh.edu
FAX:	713-729-5434
Phone:	713-729-3093




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