- #1

- 18

- 1

[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.