Using integrals to prove volume of a pyramid

[TromboneNerd]

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)

$$0\leq z\leq -\frac{h}{c}x+h$$

$$0\leq y\leq -\frac{c}{h}z+c$$

$$0\leq x\leq c$$

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.

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

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

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.

 

[Roni1985]

I think you made a mistake with the last integral

 

[Roni1985]

Yep, you made a mistake
actually if the second term is b^2, you get the answer you need