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