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