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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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