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/
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