Why is the quantity of a cone one third of the quantity of a cyndrical tube?

The quantity of a cone with elevation $h$ and also distance $r$ is $\frac{1}{3} \pi r^2 h$, which is specifically one 3rd the quantity of the tiniest cyndrical tube that it fits within.

This can be confirmed conveniently by taking into consideration a cone as a solid of revolution, yet I would love to recognize if it can be confirmed or at the very least aesthetic shown without making use of calculus.

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2019-05-07 01:25:52
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Answers: 3

You can make use of Pappus's centroid theory as in my answer here, yet it does not give much understanding.

If as opposed to a cyndrical tube and also a cone, you take into consideration a dice and also a square - based pyramid where the "leading" vertex of the pyramid (the one contrary the square base) is changed to be straight over one vertex of the base, you can fit 3 such pyramids with each other to create the full dice. (I've seen this as physical toy/puzzle with 3 pyramidal items and also a cubic container.) This might offer some understanding right into the 1/ 3 "sharp point regulation" (for sharp points with comparable, linearly - relevant cross - areas) that Katie Banks reviewed in her comment.

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2019-05-09 01:07:52
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One can reduce a dice right into 3 pyramids with square bases - - so for such pyramids the quantity is without a doubt 1/ 3 hS. And afterwards one makes use of Cavalieri's principle to confirm that the quantity of any kind of cone is 1/ 3 hS.

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2019-05-09 01:07:38
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An aesthetic demo for the instance of a pyramid with a square base. As Grigory states, Cavalieri's principle can be made use of to get the formula for the quantity of a cone. We simply require the base of the square pyramid to have side size $ r\sqrt\pi$ . Such a pyramid has quantity $\frac13 \cdot h \cdot \pi \cdot r^2. $

Then the location of the base is plainly the very same. The cross - sectional location at range a from the optimal is a straightforward issue of comparable triangulars : The distance of the cone's sample will certainly be $a/h \times r$ . The side size of the square pyramid's sample will certainly be $\frac ah \cdot r\sqrt\pi.$
Once once more, we see that the locations have to be equivalent. So by Cavalieri's concept, the cone and also square pyramid have to have the very same quantity : $ \frac13\cdot h \cdot \pi \cdot r^2$

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2019-05-08 21:34:16
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