Frustum of a Pyramid with a Rectangular Base

Date: 02/20/2002 at 12:42:03
From: Brad Nelson
Subject: Frustum of a pyramid with rectangular base

Back in 2000 you gave a solution you derived using calculus to 
someone wanting to know how to figure the volume of a rectangular 
based frustum of a pyramid see: 

   Volume of a Frustum-Like Structure
   http://mathforum.org/dr.math/problems/cunningham.5.12.00.html   

   V = a1b1 + a2b2 + (a1b2 + a2b1)/2 x h/3

I am an engineer with a water treatment agency and need to figure the 
amount of water per foot of elevation in our reservoirs that happen to 
have the same shape as previously described. To satisfy my curiosity, 
could you please send me a copy of your derivation? I have had up 
through differential equations in college and am a bit rusty, so lay 
it on me.

Thanks in advance.

Date: 02/20/2002 at 14:58:30
From: Doctor Peterson
Subject: Re: Frustum of a pyramid with rectangular base

Hi, Brad.

You'll be interested in this later page, where I answered a question 
like yours about the volume contained by such a shape up to a given 
depth:

   Volume of a Trapezoidal Solid
   http://mathforum.org/dr.math/problems/greg.11.15.00.html   

It will also be of interest to you that the shape under discussion is 
a special case of a more general shape I wasn't aware of at the time, 
called the prismoid, or, even more generally, a prismatoid:

   MathWorld - Eric Weisstein:

   http://mathworld.wolfram.com/Prismoid.html   
   http://mathworld.wolfram.com/Prismatoid.html   

The latter page gives the same formula as in my reference above, for 
the volume in this much more general case, where you just have two 
parallel polygonal bases joined to one another by straight edges:

    V = h/6 (A1 + 4M + A2)

where h is the altitude, A1 and A2 are the areas of the bases, and 
M is the area of the cross section halfway between.

Now, my calculus derivation for our special case was based on the fact 
that the length and width are changing linearly with height. So the 
dimensions of the rectangular cross-section at height x will be

    a = a1 + (a2-a1) * x/h
    b = b1 + (b2-b1) * x/h

(To check these, see what they are at x=0 and x=h.)

Then I integrated the product ab with respect to x, from 0 to h. It's 
a pretty easy integration.

I later rederived the same formula by dissecting the shape into 
several prisms and pyramids. That takes a little more visualization 
and calculation, but no calculus.

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

Date: 02/21/2002 at 10:08:06
From: Brad Nelson
Subject: Frustum of a pyramid with rectangular base

Thank you for the information. I was able to integrate it and come up 
with the same answer. My only other question is what principle or 
theorem you used for the a and b of the rectangular cross-section at 
any height x?

   a = a1 + (a2-a1) * x/h
   b = b1 + (b2-b1) * x/h

Once again, thanks for your prompt response to my inquiry.

Date: 02/21/2002 at 10:21:06
From: Doctor Peterson
Subject: Re: Frustum of a pyramid with rectangular base

Hi, Brad.

You can just draw a side view:

                a1
          +------------+ -------------
         /            / \       |x  |
        /            /   \      |   |
       +------------+-----+ ------  |h
      /            /       \        |
     /     a1     /  a2-a1  \       |
    +------------+-----------+ -------
                a2

You can see that the width at distance x from the top is 
a1 + (a2-a1)x/h, using similar triangles. (I've got this upside-down 
from the original labeling, but that doesn't affect the math.)

Is that what you wanted?

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

Date: 02/21/2002 at 11:30:39
From: Brad Nelson
Subject: Frustum of a pyramid with rectangular base

Aha, I see the light. I guessed you were probably using similar 
triangles but was having a hard time visualizing it for some reason.
Thank you for your time and patience with a math-rusty engineer. You 
have been most helpful.