Simpsons formula and the volume of a pyramid frustum

[jorgenhansen501]

Simpsons formula and the volume of a pyramid frustum

Normally I consider myself quite adept in mathematics, but I simply lack the right idea and/or the mathematical creativity to solve this assignment:

"Prove that the Simpsons formula V = 1/6 * h * (Ab + 4Am + At) can be used to calculate the volume of a pyramid frustrum."

V = Volume of the pyramid frustrum
h = height of the pyramid frustrum
Ab = Area of the "bottom" of the pyramid frustrum
Am = Area of a horisontal slice of the pyramid at 1/2*h
At = Area of the top of the pyramid frustrum.

---

My (fruitless) attempts at a solution:

The normal formula for the volume of a pyramid frustrum is V = 1/3 * h * (Ab + At + SQR(Ab*At))

Initially I thought the assignment could be quite easily solved by viewing the pyramid frustrum as a part of a (imaginary) large pyramid (0) consisting of the pyramid frustrum(1) and a smaller normal pyramid(2) on top of the frustrum:

1/6 * h * (Ab0 + 4Am0 + At0) = ( 1/6 * h * (Ab1 + 4Am1 + At1)) + ( 1/6 * h * (Ab2 + 4Am2 + At2))

In an earlier assignment I proved that Ab + 4*Am + At = 2*Ab holds true for pyramids and cones. Am for the big and the smal pyramid can therefore be written as: (Ab - At) / 4

Substituting this into the earlier equation gives:

(1/6) * h0 * (Ab0 + 4*((Ab0-At0)/4) + At0) = ( (1/6) * h1 * (Ab1 + 4Am1 + At1)) + ( (1/6) * h2 * (Ab2 + 4*((Ab2-At2)/4) + At2))

We should even remember that the top of the pyramid frustrum (At1) is the same as the bottom of the top pyramid (Ab2) and that At2 will equal 0:

(1/6) * h0 * (Ab0 + 4*((Ab0-At0)/4) + At0) = ( (1/6) * h1 * (Ab1 + 4Am1 + Ab2)) + ( (1/6) * h2 * (Ab2 + 4*(Ab2/4))

As the bottom of the pyramid frustrum equals the bottom of the imaginary larger pyramid Ab0 will be equal to Ab1 and that At0 will be zero (as it is the very top of the imaginary pyramid)

(1/6) * h0 * (Ab1 + 4*(Ab1/4)) = ( (1/6) * h1 * (Ab1 + 4Am1 + Ab2)) + ( (1/6) * h2 * (Ab2 + 4*(Ab2/4))

However, after playing around with this for at couple of hours I seem to be getting nowhere.

I would really appreciate if anybody is able to offer some insight or just a hint at the solution!

Jorgen Hansen

 

[Dr.Peterson]

Simpsons formula and the volume of a pyramid frustum

Normally I consider myself quite adept in mathematics, but I simply lack the right idea and/or the mathematical creativity to solve this assignment:

"Prove that the Simpsons formula V = 1/6 * h * (Ab + 4Am + At) can be used to calculate the volume of a pyramid frustrum."

V = Volume of the pyramid frustrum
h = height of the pyramid frustrum
Ab = Area of the "bottom" of the pyramid frustrum
Am = Area of a horisontal slice of the pyramid at 1/2*h
At = Area of the top of the pyramid frustrum.

---

My (fruitless) attempts at a solution:

The normal formula for the volume of a pyramid frustrum is V = 1/3 * h * (Ab + At + SQR(Ab*At))

Initially I thought the assignment could be quite easily solved by viewing the pyramid frustrum as a part of a (imaginary) large pyramid (0) consisting of the pyramid frustrum(1) and a smaller normal pyramid(2) on top of the frustrum:

1/6 * h * (Ab0 + 4Am0 + At0) = ( 1/6 * h * (Ab1 + 4Am1 + At1)) + ( 1/6 * h * (Ab2 + 4Am2 + At2))

In an earlier assignment I proved that Ab + 4*Am + At = 2*Ab holds true for pyramids and cones. Am for the big and the smal pyramid can therefore be written as: (Ab - At) / 4

Substituting this into the earlier equation gives:

(1/6) * h0 * (Ab0 + 4*((Ab0-At0)/4) + At0) = ( (1/6) * h1 * (Ab1 + 4Am1 + At1)) + ( (1/6) * h2 * (Ab2 + 4*((Ab2-At2)/4) + At2))

We should even remember that the top of the pyramid frustrum (At1) is the same as the bottom of the top pyramid (Ab2) and that At2 will equal 0:

(1/6) * h0 * (Ab0 + 4*((Ab0-At0)/4) + At0) = ( (1/6) * h1 * (Ab1 + 4Am1 + Ab2)) + ( (1/6) * h2 * (Ab2 + 4*(Ab2/4))

As the bottom of the pyramid frustrum equals the bottom of the imaginary larger pyramid Ab0 will be equal to Ab1 and that At0 will be zero (as it is the very top of the imaginary pyramid)

(1/6) * h0 * (Ab1 + 4*(Ab1/4)) = ( (1/6) * h1 * (Ab1 + 4Am1 + Ab2)) + ( (1/6) * h2 * (Ab2 + 4*(Ab2/4))

However, after playing around with this for at couple of hours I seem to be getting nowhere.

I would really appreciate if anybody is able to offer some insight or just a hint at the solution!

Jorgen Hansen

You mentioned a previous assignment, which suggests that this problem fits into your context. Can you tell us what facts you know that might be of use, in addition to what you have mentioned? Also, what course is this part of, and what topics have been covered? I think this formula applies more generally, and can be proved by calculus; is that allowed, or are you using only synthetic geometry, or what?

I myself have in the past proved the formula V = h/3 * (Ab + sqrt(Ab*At) + At); I imagine this could be used to prove your formula, once you prove a relationship between Am and Ab and At. (For this one, I used the completed cone as you suggest.)