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MathGroup Archive 1997

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Re: mathematica problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7592] Re: [mg7540] mathematica problem
  • From: Allan Hayes <hay at haystack.demon.co.uk>
  • Date: Thu, 19 Jun 1997 03:13:55 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 13 Jun 1997
Stephanie Gill <sgill at winnie.fit.edu>
in [mg7540] mathematica problem
wrote

> 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.
>
>I do not understand how to do this on Mathematica, can you help?

Stephanie:

Here is the graph of  y = ArcSin[x]   (that is y = sin^-1x) from x  
= 0 to x = 1.

In[1]:=
Plot[ArcSin[x],{x,0,1},
Epilog->
	{{PointSize[.02],Point[{.7, ArcSin[.7]}],
	 Text[{"x","y"},{.7, ArcSin[.7]},{-1,1}],
	  Line[{{1,0},{1, ArcSin[1]}}]
	 }}
]

You can look at the curve as as given by either
(1)  {x,y} such that y = ArcSin[x]  for x in [0,1]                  
or
(2)  {x,y} such that x = Sin[y]     for y in [0,ArcSin[1]]            

Using (1) we get the volume from disks:

In[2]:=
	Integrate[Pi*ArcSin[x]^2, {x, 0, 1}]
Out[2]=
	1/4*Pi*(-8 + Pi^2)

Using (2) we get the volume from shells:

In[3]:=
	Integrate[2*Pi*y*(1 - Sin[y]), {y, 0, ArcSin[1]}]
Out[3]=
	1/4*Pi*(-8 + Pi^2)

This will look clearer if you convert the input and output cells  
for the integrations to TraditionalForm (to do this, select the  
cells and use the menu Cell>ConvertTo>TraditionalForm).

If you are interested in writing an fuller explanation of what is  
happening then Mathematica can produce graphics to show the disks  
and shells - please get in touch if you have any further questions  
about this.

Allan Hayes
hay at haystack.demon.co.uk
http://www.haystack.demon.co.uk/training.html
voice:+44 (0)116 2714198
fax: +44 (0)116 2718642
12 Copse Close, Leicester, LE2 4FB, UK



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