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Using calculus to derive the volume of a pyramid

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Use calculus to derive the volume of a pyramid

    3. The attempt at a solution

    There's probably a simpler way to go about this, but I wanted a challenge. I decided to calculate 1/4 of the pyramid in the first octant and then multiply my final answer by 4.

    First we have a pyramid of height h, length L and width w, (measuring from the origin) so the corners are at (0,0,h), (L,0,0) and (0,w,0).

    Calculating the plane that defines this quarter of the pyramid yields

    [itex]whx+lhy+lw(z-h)=0[/itex] so [itex]z=h-\frac{hx}{l}-\frac{hy}{w}[/itex]

    We integrate this over dy from 0 to [itex]y=\frac{-wx}{l}+w[/itex] which is the diagonal line the pyramid would make in octant one when looking straight down and then over dx from 0 to L.

    Multiply this answer by 4 and we get [itex]\frac{2hwl}{3}[/itex], so it comes out double for some reason? :cry:

    Thanks for looking.
     
    Last edited: Jun 21, 2013
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  3. Jun 21, 2013 #2

    HallsofIvy

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    What do you mean by this? What plane? You can, in fact, use a number of different planes. Planes parallel to the three coordinate planes are the obvious ones.
    Taking [tex]z= z_0[/tex] we would, by "similar triangles" have [tex]\frac{x}{l}= \frac{y}{w}= \frac{z_0}{h}[/tex] so that we integrate with respect to z from 0 to h and, for each z, with respect to x from 0 to [tex]\frac{lz}{h}[/tex] and with respect to y from 0 to [tex]\frac{wz}{h}[/tex]

     
  4. Jun 21, 2013 #3
    "What plane?"

    As I thought I made clear, the plane defining the surface of the pyramid in the first octant.
     
  5. Jun 21, 2013 #4
    With respect, I think you are making this unnecessarily difficult, and you are calculating only for a symmetrical pyramid. Imagine any pointy thing with straight edges between base and top, with a base area 'A' in the xy plane, and height h. Then consider a parallel plane thickness dz, at height z. Calculate the volume of the slice, and then integrate between o and h.
     
  6. Jun 21, 2013 #5

    haruspex

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    Didn't follow the logic in the OP completely, but I notice the absence of a step where the area of a triangle is involved. That's where the 1/2 should come from.
     
  7. Jun 21, 2013 #6
    What made you think I'm calculating only for a symmetrical pyramid?
     
  8. Jun 21, 2013 #7
    That would be nice, but I'm fairly sure that's not it as the double integral accounts for the area of the triangle. I did it twice and got the same answer so I'm fairly certain it's not an algebra mistake. Hopefully this will make things clearer:

    G5GwXw0.jpg
     
  9. Jun 22, 2013 #8

    haruspex

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    The integrand is only valid for z=0. If you think I've misunderstood, pls write out in more detail how you're carving up the pyramid for the purpose of integration.
     
  10. Jun 23, 2013 #9

    vela

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    You're not solving the problem you think you are. You found the volume of a pyramid where the diagonals of the base are length 2L and 2w. The area of the base is 4(wL/2) = 2wL, and to get the volume, you'd multiply the area of the base by h/3, which yields the same result as what you got.
     
    Last edited: Jun 23, 2013
  11. Jun 24, 2013 #10

    Of course! Thank you
     
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