Using calculus to derive the volume of a pyramid

[e^(i Pi)+1=0]

Homework Statement

Use calculus to derive the volume of a pyramid

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.

 

[HallsofIvy]

Homework Statement

Use calculus to derive the volume of a pyramid

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

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}

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.

 

[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.

 

[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.

 

[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.

 

[e^(i Pi)+1=0]

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.

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

 

[e^(i Pi)+1=0]

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.

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:

 

[haruspex]

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:

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.

 

[vela]

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.

 

[e^(i Pi)+1=0]

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.

Of course! Thank you