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Homework Help: Volume of a Frustrum of a Pyramid

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the volume, using only the variables a, b, and h in your answer.

    A frustum of a pyramid with square base of side b, square top of side a, and height h:

    riajvr.gif


    2. Relevant equations

    Area = length*width.

    Length of side S = ??

    3. The attempt at a solution

    I know that generally when finding volume, it's going to be the integral from a[bottom] to b[top] of a cross-section's area. usually that'd just be integral[pi*r^2] a..b for a cylinder.

    For this, obviously, I need to find an equation for the length side S that is a function of h, so S(h) = SL.

    My first thought was (b-a)h, but that wouldn't change as the integral changes, so it's out. I can't think of a dynamic equation to suit the problem, and would appreciate help pushing me in that direction.
     
  2. jcsd
  3. Sep 9, 2010 #2

    Dick

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    Homework Helper

    Let x be the height of your cross section. So x ranges from 0 to h. Let S(x) be the side length as a function of x. So S(0)=b and S(h)=a and S varies linearly in between. Now can you write an explicit form for S(x)?
     
  4. Sep 9, 2010 #3
    Ahh, soo..


    S(x) = b + [(a-b)x]/h


    So it would be the integral from 0..h of ( b + [(a-b)x]/h )^2 dx..

    And the volume is:

    V = [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP105419c6ggha31a9g60d000062bg6e58533dg93g?MSPStoreType=image/gif&s=16&w=117&h=36 [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Sep 9, 2010 #4
    Update - got this problem right, then did another frustrum (of a cone) and got that right as well. Thanks for the pointer, appreciate it! :D

    I love how this parallels the derivative formula in the easy/hard way...(1/3)*h*(a-b)^2 every time, give or take a Pi...lol.
     
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