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I would like to react to one of the answers on this thread (I don't have enough rep to make a comment):

Use cylinder's formula for frustum (conical frustum)

Where is answered:

Essentially, what you'd need is the average of the areas of the horizontal slices into which the frustrum is cut by planes paralell to its base, not their diameters.

I would like to ask if it is really working, I tried it with a conical frustum of r1 = 4, r2 = 2, h = 10 and a get result of V =100$\pi$ , the result give by formula $\frac1 3\pi h(r^2+R^2+rR)$ is $293.215$

Did I something wrong in the calculation or is the idea of "transforming" a cone frustum into cylinder with base area equal to an average area of two frustum's slices wrong ?

Thank you for your answers.

Community
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Caller
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  • I am sorry if I misunderstood something, but author of question in the link is asking why avg. of radii is not working, not the avg. of the areas. And the last answer in the link is states that he should take avg. of the areas instead of avg. of radii, but it does not work for me, so the matter of my question is whether I am wrong, or the answer in the link is wrong. – Caller Jan 16 '16 at 20:23
  • Oh, I see now, I've rescinded my close vote. – Travis Willse Jan 16 '16 at 20:53

1 Answers1

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While the idea of averaging the areas is correct, the problem here is that it is not a linear average, since the area doesn't change linearly with the height. The correct average is obtained by integration:

$$ \int_0^{h}\pi\left(r_1 - \frac{(r_1-r_2)}{h}r\right)^2 \;\mathrm{d}r. $$

This averages the area function where the radius changes linearly between $r_1$ and $r_2$ with $h$, and integrates to the formula you gave.

Since the answer to the other question didn't specify what kind of average, it is technically not wrong, just misleading.

David Kleiman
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  • This is the usual interpretation of "average" for the quantities described in the answer; the only "misleading" part is that the answer failed to point out that the "averaging" method in the question https://math.stackexchange.com/questions/696120, namely "take the average of the smallest and largest values," does not generally work. – David K Dec 22 '18 at 14:03