# Find the average height of a pyramid

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1. Mar 10, 2015

### Permanence

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

2. Relevant equations
V = (1/3) * A * H [Volume of Pyramid]

3. The attempt at a solution
The first thing I did was to calculate the height of pyramid from the volume formula. I got a perpendicular height of 15. I'm not sure where to go from there.

I'm under the impression that average height has something to do with geometric center. I researched the topic and found something about Pappus Theorem, and that the center of mass is 3/4 of the way from the vertex to the mid point of the base. Using that I come up with an average height of 3.75, but I'm not certain if that number is correct or if it even represents average height.

I found a related problem online, but it involved things that I have not discussed in class and I had trouble following along.

I would really appreciate some guidance. I'd like to know if my answer of 3.75 is correct, and how I would go about solving this using multivariable calculus.

2. Mar 10, 2015

### Dick

Yes, I think it is correct. I get the same thing. I did it by thinking of square cross sections of the pyramid at height x. Like in your picture. If the area of the square at height x is A(x) then the average height is given by $\frac{\int A(x) x dx}{\int A(x) dx}=\frac{\int A(x) x dx}{volume}$. Can you fill in the missing parts like limits on the integral and the form of A(x) in terms of x? Very resourceful solution to the problem, by the way. The average height is the height of the center of mass. Good job.

Last edited: Mar 10, 2015
3. Mar 10, 2015

### Permanence

Hi, thank you for your reply. I understand what you did, but I was hoping you could clarify something for me

A classmate used the following from the book to solve the problem.
He then set it 80/16 to get the answer 5.

That answer is different than both the method you used and the method I used which is leaving me confused. Do you see anything wrong with using that equation?

Thank you

edit:

I thought about it more, and I'm not sure if it's 5 but I see some error in that approach.

Imagine the length of the sides of the base are 1, and the height is 1. You get a volume of 1/3. Solving with that formula you get an average height of 1/3. That is the center for a two dimensional right triangle, but the center of a three dimensional pyramid is 1/4 (according to the slightly related example problem I attached in my original post)

Last edited: Mar 10, 2015
4. Mar 10, 2015

### LCKurtz

I tend to agree with $\bar h = 5$ using the definition $A_b \bar h = V$ where $A_b$ is the area of the base and $V$ is the volume.

5. Mar 10, 2015

### Permanence

This problem has really confused me. I don't understand why different approaches are leading to different answers. I believe my professor is going to want 5, but I see too much conflict to pick one answer.

LCKurtz, can you explain why that method gives an incorrect answer for a base 1, and height 1?

6. Mar 11, 2015

### LCKurtz

No. Offhand I don't see why the answers are different. I will look at it some more tomorrow.

7. Mar 11, 2015

### Dick

Ok, I see what going on here. The answer you are giving is interpreting the 'average' in the question as the average of the height $x$ being $\frac{\int x dA}{A}$. The center of mass answer is interpreting it as being $\frac{\int x dV}{V}$. One's an area average and the other is a volume average. Completely different. But both valid without a context to know which you want.

Last edited: Mar 11, 2015
8. Mar 11, 2015

### LCKurtz

Yeah, after sleeping on it I have another take also. You define the average value of a function $f(x)$ on $[a,b]$ as$$A=\frac 1 {b-a}\int_a^b f(x)~dx$$This is a completely different calculation than calculating the y coordinate $\bar y$ of the center of area enclosed between $f(x)$ and the $x$ axis on that interval.

For example, if $f(x)\equiv 1$ on $[0,1]$, the $y$ value is always $1$ and its average is $1$. But $\bar y=\frac 1 2$.

9. Mar 11, 2015

### Permanence

Thank you for both for your replies.
My professor said that my approach to the problem was incorrect, and that the centroid has nothing to do with the problem. I didn't entirely understand why he dismissed my approach, he said something about not needing to bring the z variable into this.

10. Mar 11, 2015

### LCKurtz

Read my post #8 for a 1 variable example. The centroid does have nothing to do with it. It's the same for two variables.

11. Mar 11, 2015

### Ray Vickson

I think the appropriate definition of "average" depends on context---what you want to do with it. The volume-oriented definition has something to do with centroids (centers of gravity), but other types of averages are also possible. For example, suppose you stand back from the pyramid and just trace the visible outline; what you see looks like a one-variable graph of a "triangular" function, something like $y = 1-a|x|$ for $|x| \leq 1/a$. Now the average value of $y$ would be like the 1-variable average in your thumbnail from the book. This average would be much more relevant than the centroid (the 2-dimensional average) to people who are interested in climbing the pyramid. They would go along surface, so the interior of the pyramid is of little relevance to them.

12. Mar 11, 2015

### Permanence

Thank you again for the replies. Ray your example about the climbing has helped me understand the differences.

13. Mar 11, 2015

### SammyS

Staff Emeritus
Yes. I agree with 5 also.