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 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.

---

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

 

[MarkFL]

Hello, and welcome to MHB, Jorgen! (Wave)

Simpson's Rule is used to approximated a definite integral, so let's begins by setting up the actual integral.

Now, the area \(A\) of a particular cross section will vary as the square of a linear measure of the cross section, such as a side length or diagonal, which we can call \(r\). Hence

\(\displaystyle A=kr^2\)

\(\displaystyle A(0)=kr_0^2=A_b\implies r_0^2=\frac{A_b}{k}\)

\(\displaystyle A(h)=kr_h^2=A_t\implies r_h^2=\frac{A_t}{k}\)

Now, we know that along a vertical axis \(y\) from \(0\) to \(h\), we have:

\(\displaystyle y=\frac{h}{r_h-r_0}(r-r_0)\implies dy=\frac{h}{r_h-r_0}\,dr\)

Hence, the volume \(V\) of the frustum is:

\(\displaystyle V=k\frac{h}{r_h-r_0}\int_{r_0}^{r_h} r^2\,dr\)

Using Simpson's Rule with \(n=2\), we have:

\(\displaystyle V=k\frac{h}{r_h-r_0}\cdot\frac{r_h-r_0}{3\cdot2}\left(r_0^2+4\left(\frac{r_0+r_h}{2}\right)^2+r_h^2\right)=k\frac{h}{3}\left(r_0^2+r_0r_h+r_h^2\right)\)

We know there will be no error because we have a quadratic integrand.

Thus:

\(\displaystyle V= k\frac{h}{3}\left(\frac{A_b+\sqrt{A_bA_t}+A_t}{k}\right)=\frac{h}{3}\left(A_b+\sqrt{A_bA_t}+A_t\right)\)

Does that make sense?

 

[jorgenhansen501]

It makes perfect sense - and I really appreciate that you took the time to write such an elaborate answer.

However, the teacher has explicitly stated that the problem should be solved without the use of integrals... Hence my fruitless attempts shown earlier.

Since the Simpsons formula is about approximating the integral I cannot grasp the idea of using it without integrals, but I guess the teacher want us to develop some kind of higher understanding of the problem - which surely hasn't manifested itself in me yet...

I realize now that I should have stated the teachers requirement of not using integrals in the problem statement. Please accept my apology for that omission.

Again, your effort is truly appreciated

Jorgen Hansen

 

[MarkFL]

It makes perfect sense - and I really appreciate that you took the time to write such an elaborate answer.

However, the teacher has explicitly stated that the problem should be solved without the use of integrals... Hence my fruitless attempts shown earlier.

Since the Simpsons formula is about approximating the integral I cannot grasp the idea of using it without integrals, but I guess the teacher want us to develop some kind of higher understanding of the problem - which surely hasn't manifested itself in me yet...

I realize now that I should have stated the teachers requirement of not using integrals in the problem statement. Please accept my apology for that omission.

Again, your effort is truly appreciated

Jorgen Hansen

I didn't use integration to get the result, but we do need to know what we're approximating and so that's why I set up the integral but didn't actually integrate. The integration has already been done using parabolic arcs in Simpson's Rule which I applied.

 

[MarkFL]

I suppose we could begin by simply applying Simpson's Rule to state:

\(\displaystyle V=\frac{h}{3\cdot2}\left(A_b+4A_m+A_t\right)\)

Now, we need to express \(A_m\) in terms of \(A_b\) and \(A_t\). Suppose:

\(\displaystyle A(r)=kr^2\)

where:

\(\displaystyle A(r_0)=kr_0^2=A_b\)

\(\displaystyle A(r_h)=kr_h^2=A_t\)

And hence:

\(\displaystyle A\left(\frac{r_0+r_h}{2}\right)=k\left(\frac{r_0+r_h}{2}\right)^2=A_m\)

We then find:

\(\displaystyle A_m=\frac{k}{4}(r_0^2+2r_0r_h+r_h^2)=\frac{A_b+2\sqrt{A_bA_t}+A_t}{4}\)

And thus:

\(\displaystyle V=\frac{h}{6}\left(2A_b+2\sqrt{A_bA_t}+2A_t\right)=\frac{h}{3}\left(A_b+\sqrt{A_bA_t}+A_t\right)\)