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Using integrals to prove volume of a pyramid

  1. Mar 30, 2010 #1
    the first question i have is if i set it up the iteration right. The I've simplified to one quarter of the pyramid, so the value c is actually 1/2 the base of the whole pyramid. and h is the height of the pyramid. (the pinnacle of the pyramid is along the z axis)

    [tex]0\leq z\leq -\frac{h}{c}x+h[/tex]

    [tex]0\leq y\leq -\frac{c}{h}z+c[/tex]

    [tex]0\leq x\leq c[/tex]

    If all that is correct, then ill show my integration work and you can tell me where i went wrong. I don't know how to place lower and upper limits on integrals using Latex, so just use the limits from above.

    [tex]V=4\int \int \int dydxdz[/tex]
    in respect to y
    [tex]=4\int \int (-\frac{c}{h} z+c)dzdx[/tex]
    in respect to z
    [tex]=4\int [-\frac{c}{2h} (-/frac{h}{c} x+h)^{2} +c(-\frac{h}{c} x+h)]dx[/tex]
    finally in respect to x
    [tex]=4(-\frac{c^{2} h}{6} +ch)[/tex]

    [tex]=-\frac{2c^{2} h}{3} +ch)[/tex]
    substitute in .5b for c
    [tex]=-\frac{b^{2} h}{6} +\frac{bh}{2}[/tex]

    this is obviously not anything close to the formula for the volume of a pyramid, and is completely non sensical. according to this, only very small values of b and h will produce a positive volume. where did i go wrong? I skipped quite a few steps for the sake of typing space, so if your not sure where i got this or that from ill show you the more detailed work where the problem might lie.
  2. jcsd
  3. Mar 30, 2010 #2
    I think you made a mistake with the last integral
  4. Mar 30, 2010 #3
    Yep, you made a mistake
    actually if the second term is b^2, you get the answer you need
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