Originally Posted by

**Alterah** Ok, I am able to use integration to solve for the volume of a pyramid (I know that the formula for Volume is V = (1/3)Bh. Where B = Area of Base. Anyhow, the problem is as follows...

A pyramid is 9m tall and has bases 6m x 6m. Find the volume of the solid formed with a plane horizontal to the base intersects the pyramid at height 3m.

One way to do this is to use either the formula, or integration to find the volume of the pyramid, then use it again to find the volume of the smaller pyramid which is formed by the plane intersecting the pyramid and subtracting the two.

I am wondering if there is a way to set this up with triple integrals. I've got this so far for the volume of a pyramid with double integrals:

$\displaystyle

V = 4\int_{0}^{3\sqrt2}

\int_{0}^{3\sqrt2 -x} (9 - \frac{3\sqrt2}{2}x -

\frac{3\sqrt2}{2}y) dydx$

However, that gives the whole volume, not the portion cut out by the plane. Thanks for help.