Volume Using Cross Sections

Date: 05/28/2001 at 00:56:34
From: Purvi
Subject: Volume using cross-sections

I need to find the volume of the region between  y = |x| and 
y = -|x|+6 with cross sections that are equilateral triangles and 
perpendicular to the x-axis. 

The integral I came up with is S from -3 to 3 of (-2|x|+6)*(.5|x|)dx 
but I'm not so sure about it. I might have to change my graph a little 
because the volume of the region must be no less than 64 cubic inches 
and no more than 512 cubic inches. I'm trying to make a pyramid out of 
this because with this project we get graded on creativity and 
complexity. I would really appreciate it if you are able to help me.

Date: 05/28/2001 at 17:45:32
From: Doctor Jaffee
Subject: Re: Volume using cross-sections

Hi Purvi,

I have a few suggestions for you that I think will help your project.  

First, let's talk geometry; specifically equilateral triangles.  
Suppose you have an equilateral triangle whose sides each have length 
's'.  If you draw the altitude from one of the vertices, you split the 
triangle into two 30-60-90 triangles. You recall that the shorter leg 
is half as long as the hypotenuse and the longer leg is sqrt(3) times 
as long as the shorter leg. So, using this information you should be 
able to write a formula for the area in terms of 's'.

Now, if you look at the region bounded by y = 6 - |x| and y = |x|, 
you can see that it is symmetrical to the y-axis. In the first 
quadrant your two functions can be expressed simply as y = 6 - x and 
y = x, and as you figured out, they intersect at (3,3).

So, my suggestion is that you let 0 and 3 be the limits of 
integration, and then double your answer. The integrand will be the 
area of the equilateral triangle where s = 6 - 2x.

Give it a try and if you want to check your solution with me, write 
back. If you are having difficulties, let me know and show me what you 
have done so far, and I'll try to help you some more.

Good luck. 

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/