# Thread: Volume of Part of a Pyramid

1. ## Volume of Part of a Pyramid

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:

$
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.

2. Most simply, I think, is to orient the pyramid so that the base vertices are on the x and y axes. If we then look only at Octant I, we have the tetrahedron with x- and y-intercepts at $\frac{6}{\sqrt{2}}$. Obviously, the z-intercept is at 9.

Three sides of the tetrahedron are the coordinate planes. The equation of the plane making the fourth side is trivial from the Intercept Form:

$
\frac{x}{\frac{6}{\sqrt{2}}}+
\frac{y}{\frac{6}{\sqrt{2}}} + \frac{z}{9} = 1
$

Solve for z: $z = 9-\frac{3y}{\sqrt{2}}-\frac{3x}{\sqrt{2}}$

Eliminate z and solve for y: $y = 3\sqrt{2}-x$

Now write the integral

$
\int_{0}^{\frac{6}{\sqrt{2}}}\int_{0}^{3\sqrt{2}-x}\int_{0}^{9-\frac{3y}{\sqrt{2}}-\frac{3x}{\sqrt{2}}}dzdydx
$

This should be 1/4 the volume of the pyramid. Only a slight variation should produce the volume of the top point and subtraction the volume of the frustrum.

Note: I solved for z and then y. There are five (5) other ways to proceed.

3. Ok, that's how you would use triple integrals, but that gets you 1/4 the volume of the entire pyramid. In my example we are cutting the pyramid at height 3.... so z = 3 is a plane that intersects the pyramid. I am wanting to know if there's a triple integral to describe that.

4. It's supposed to lead you through a thought process. That pyramid is SO conveniently oriented that it may suggest an easy solution. What happens if you take the larger pyramid and drop it 6m. What would that leave above the xy-plane?

5. Ok, well I've made a graph. I want to make sure we're on the same page here. What is sounds like you are wanting to do is to make the graph shorter, whereas I am wanting an integral only for the portion above the xy-plane and below the plane z = 3, without the need for two integrals using bounds. Though, since we know in this problem z = 3, maybe I should plug it in for z regarding the limits of integration...for a triple integral. Here's what I am using graph wise...

6. 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:

$
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.
If you know "V= (1/3)Bh" you don't need to integrate. The base one side of the entire pyramid is 6m long and the height is 9 m. Slicing the pyramid 3m above the base leaves a smaller pyramid with height 9- 3= 6 m. By "similar triangles, the base, b, of this smaller pyramid is given by b/6= 6/9 so that b= 36/9= 4m. The problem, then, is just to find the volume of a pyramid of height 6m with base a 4m by 4m square.

7. Originally Posted by HallsofIvy
If you know "V= (1/3)Bh" you don't need to integrate. The base one side of the entire pyramid is 6m long and the height is 9 m. Slicing the pyramid 3m above the base leaves a smaller pyramid with height 9- 3= 6 m. By "similar triangles, the base, b, of this smaller pyramid is given by b/6= 6/9 so that b= 36/9= 4m. The problem, then, is just to find the volume of a pyramid of height 6m with base a 4m by 4m square.
I realize that. And the only reason I am trying the other way with integration is for fun really.

8. Originally Posted by Alterah
I realize that. And the only reason I am trying the other way with integration is for fun really.
Okay, enjoy!