Calculus (Minimizing Volume)

Date: 5/13/96 at 21:40:26
From: Anonymous
Subject: Calculus (minimizing volume)


The arch y=sin x,0 less than or = to x less than or = to pi, is 
revolved around the line y=c to generate a solid. Find the value of c 
that minimizes the volume.  

What is the minimum volume?   I've tried integrating pi*integral from 
a to b of (sin x - c)^2 + 2pi*integral from 0 to a of (sin x - c)^2 
where a and b are the two points where y=c crosses y=sin x, but I 
don't get the right answer.

Date: 10/17/96 at 8:53:25
From: Doctor Jerry
Subject: Re: Calculus (minimizing volume)

When you say the arch, I'm assuming you mean the region A that lies 
between the sine curve  and the x-axis, for 0<=x<=pi.  It is this 
region that is revolved about the line y=c, right?  This is what I'll 
assume.

From what you have said, it's okay that the axis of revolution falls 
within A.

I can make several comments.  First, if c>1/2 and we concentrate on 
the curves between a and b, the volume generated by revolving the 
rectangular region below y=c will, as it were, wipe out the volume 
generated by the smaller region above y=c. When c<1/2, the situation 
appear to be more complicated.  Again restricting x to [a,b], the top 
part will generate some volume, but the rectangular region below y=c 
generates volume not otherwise generated, near the line y=c.

My other comment is that you appear to be using the washer method to 
calculate volumes.  You must distinguish between pi*r2^2-pi*r1^2 and 
pi*(r2-r1)^2.  I think it is the former you want to use, the 
difference between two solid disks with the same center.

Interesting problem.  I hope I've understood it properly and have 
given you some help. 

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/