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Using integrals to prove volume of a pyramid

  • #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.
 

Answers and Replies

  • #2
201
0
I think you made a mistake with the last integral
 
  • #3
201
0
Yep, you made a mistake
actually if the second term is b^2, you get the answer you need
 

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