# Using calculus to derive the volume of a pyramid

1. Jun 21, 2013

### e^(i Pi)+1=0

1. The problem statement, all variables and given/known data

Use calculus to derive the volume of a pyramid

3. The attempt at a solution

There's probably a simpler way to go about this, but I wanted a challenge. I decided to calculate 1/4 of the pyramid in the first octant and then multiply my final answer by 4.

First we have a pyramid of height h, length L and width w, (measuring from the origin) so the corners are at (0,0,h), (L,0,0) and (0,w,0).

Calculating the plane that defines this quarter of the pyramid yields

$whx+lhy+lw(z-h)=0$ so $z=h-\frac{hx}{l}-\frac{hy}{w}$

We integrate this over dy from 0 to $y=\frac{-wx}{l}+w$ which is the diagonal line the pyramid would make in octant one when looking straight down and then over dx from 0 to L.

Multiply this answer by 4 and we get $\frac{2hwl}{3}$, so it comes out double for some reason?

Thanks for looking.

Last edited: Jun 21, 2013
2. Jun 21, 2013

### HallsofIvy

Staff Emeritus
What do you mean by this? What plane? You can, in fact, use a number of different planes. Planes parallel to the three coordinate planes are the obvious ones.
Taking $$z= z_0$$ we would, by "similar triangles" have $$\frac{x}{l}= \frac{y}{w}= \frac{z_0}{h}$$ so that we integrate with respect to z from 0 to h and, for each z, with respect to x from 0 to $$\frac{lz}{h}$$ and with respect to y from 0 to $$\frac{wz}{h}$$

3. Jun 21, 2013

### e^(i Pi)+1=0

"What plane?"

As I thought I made clear, the plane defining the surface of the pyramid in the first octant.

4. Jun 21, 2013

### pongo38

With respect, I think you are making this unnecessarily difficult, and you are calculating only for a symmetrical pyramid. Imagine any pointy thing with straight edges between base and top, with a base area 'A' in the xy plane, and height h. Then consider a parallel plane thickness dz, at height z. Calculate the volume of the slice, and then integrate between o and h.

5. Jun 21, 2013

### haruspex

Didn't follow the logic in the OP completely, but I notice the absence of a step where the area of a triangle is involved. That's where the 1/2 should come from.

6. Jun 21, 2013

### e^(i Pi)+1=0

What made you think I'm calculating only for a symmetrical pyramid?

7. Jun 21, 2013

### e^(i Pi)+1=0

That would be nice, but I'm fairly sure that's not it as the double integral accounts for the area of the triangle. I did it twice and got the same answer so I'm fairly certain it's not an algebra mistake. Hopefully this will make things clearer:

8. Jun 22, 2013

### haruspex

The integrand is only valid for z=0. If you think I've misunderstood, pls write out in more detail how you're carving up the pyramid for the purpose of integration.

9. Jun 23, 2013

### vela

Staff Emeritus
You're not solving the problem you think you are. You found the volume of a pyramid where the diagonals of the base are length 2L and 2w. The area of the base is 4(wL/2) = 2wL, and to get the volume, you'd multiply the area of the base by h/3, which yields the same result as what you got.

Last edited: Jun 23, 2013
10. Jun 24, 2013

### e^(i Pi)+1=0

Of course! Thank you