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Given a cylinder's radius and height, $a$ and $z$, and given that $z\ll a$, what is it's volume without using $\pi$?

I was thinking that I could integrate to get the cylinder's circumference, and then divide this by the diameter to get $\pi$, but I haven't tried it yet. Is this correct?

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Just out of curiosity, why do you want to avoid $\pi$? – EuYu Nov 13 '12 at 20:50
You need $\pi$ to get the cylinder's circumference. – Tim Vermeulen Nov 13 '12 at 20:52
It is simply a mind-bender, a challenging question designed to make me think. However, every solution I've tried so far has not worked. – Stan Harvey Nov 13 '12 at 20:53
If the area of the base is $A$, the volume is $zA$. Or if the circumference is $C$ the volume is $aCz/2$ – Ross Millikan Nov 13 '12 at 21:29
You could always take the Archimedes route: place it in a bowl full of water (or a measuring jug) and measure the volume of liquid displaced. – Mark Bennet Nov 13 '12 at 21:51

In order to exactly find the volume, you must use $\pi$, as volume of a cylinder is given by

$$V = \pi a^2 z$$

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@amWhy Haha, thank you! – Argon Nov 27 '12 at 2:50

The area of the base is $\dfrac{\tau a^2}{2}$, so the volume is $\dfrac{\tau a^2 z}{2}$.

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Haha, I know about $\tau$! It does seem like the easy way out, though... – Stan Harvey Nov 13 '12 at 20:55
I think $\tau$ is cheating a bit, because $\tau = 2\pi$! – Argon Nov 13 '12 at 20:56
It reminds me of a cure for hiccups that I was told about as a child: Run three times around the house without thinking of wolves. It is very important not to think of wolves. If you do, the hiccups will continue. Anyway, to get more mathematical, just use the first positive zero of the sine function. No π needed. – Harald Hanche-Olsen Nov 13 '12 at 21:16

Why don't you inscribe the cylinder into a prism whose base is a regular polygon and has $h=z$, and then compute its volume as a function of $n$? The resulting expression does not involve $\pi$ and if you take the limit when $n$ goes to $\infty$ you get the volume of the cylinder. The tricky part is to express the apothema in terms of $n$, so you can check that the limit is actually finite, but it can be done.

To me, this sounds like a typical application of the density of $\mathbb{Q}$ in $\mathbb{R}$.

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Sure you can without pi. Remove the top of the cylinder. Fill it with fine-grained sand. Then pour the sand into a rectangular box, and measure how far it comes up. Of course, it will be an estimate, but not a bad one. (Get an even finer grade sand.) You can do the same by unscrewing the top of a sphere.

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