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jorgenhansen501

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

"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.

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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 botton 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