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Now I'm aware of the proof for the volume of a cone being a third of the volume of the corresponding cylinder and the proof is satisfactory.

Here's an alternate approach that seems to imply that the volume of the cone is actually half the volume of the cylinder. Now it clearly is a wrong result and hence my approach as a whole or some step within it is wrong. I'm not able to spot the error here and would appreciate help. It is as follows,

Imagine a cylinder with a radius r and a height h. Imagine the cylinder to be translucent. Now construct a cone inside the cylinder. Draw an altitude right through the middle of the cone, through the core of the entire structure (cone inside the cylinder). Imagine the structure right in front of your eyes. Consider a 2D vertical cross section of the shape right through the middle. The cone can be divided into two triangles on either side of the altitude with areas equal to (rh)/2. There are two inverted triangles on either side of the cone which would become the parts of the left-over cylinder should the cone be removed. Areas of these triangles are also (rh)/2. This parity of the two inner triangles(inside the cone) and two outer triangles (outside the cone) would be true throughout the cylinder. In other words, an infinite number of such pairs exists about the central altitude line. All such pairs, when added up, make up the solid cylinder. This implies that the volume of the cone and the volume of the left-over shape when the cone is removed are the same, that is, the cone is half the volume of the cylinder. What am I missing?

I apologise for framing such a long and maybe, to some degree, an unintelligible question. I would have provided an animation had I the know-how.

enter image description here

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    $\begingroup$ I've added a depiction of the cross sections you're taking. Feel free to remove if it's not the right interpretation $\endgroup$
    – user170231
    Feb 5, 2023 at 21:57
  • $\begingroup$ @user170231 that is very very helpful. Thank you. What software did you use to render this? $\endgroup$
    – John1085
    Feb 20, 2023 at 3:31
  • $\begingroup$ It was made with Mathematica $\endgroup$
    – user170231
    Feb 20, 2023 at 5:38

5 Answers 5

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The parts of the triangles on the outside contribute more to the volume than the parts on the inside, and the triangles belonging to the cone are on the inside, so they contribute less than half the volume.

The way to see this to imagine not just the cross section, but a whole "slice of cake" (if you imagine the cylinder to be a cake). The cylinder is made up of many such cake slices, each of which is wider on the outside of the cake than at the center. But this is not apparent if you only look at your slices sideways. You have to look at them from above, too.

Such contribution can't be ignored when calculating volumes.

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This is basically expanding on Vercassivelaunos's answer. Upvote their answer too if you upvote this.

Maybe the error is easier to see if instead of a cone you use a smaller cylinder.

You have an outer cylinder of radius r and height h. Inside it, there is a smaller cylinder of radius r/2 and height h. In the cross sections you are considering, the inner cylinder covers half the area of the outer cylinder. And yet, the inner cylinder has a volume of only a quarter of the outer cylinder.

You are essentially cutting a pie into small slices, and ignoring that they are wedge shaped. There is more volume in the thicker end of these wedges, however thin you slice them. If you ignore that, the volume of the outer part is underestimated and the inner part overestimated.

The same goes in the case of the cone. You are overestimating the inner parts relative to the outer parts, leading to an answer that is too high.

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Your argument seems to be that, given that the sections of the cone by planes containing the axis are half of the sections of the cylinder, the volume would have to be half as well.

Under some conditions, having half of the section would imply having half of the volume. That would be the case if, say, you have two solids with equal altitude and their sections by planes, perpendicular to the altitude, are in a ratio $1:2$.

If the sections are radial, that is not the case. There is a theorem by Pappus, according to which if you have a solid obtained by radially rotating a planar shape (as in your case), its volume is equal to the area of the shape times the perimeter of the circle described by its center of mass. The center of mass of your triangle is closer to the axis than the center of mass of the corresponding rectangle. Because of that difference, the ratio of volumes is not $1:2$ but $1:3$.

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The area of the rectangle is $2rh$ so the radius and height contribute equally to the area. The volume of the cylinder is $\pi r^2h$ so the volume is more dependent on the radius than the height. The outside triangles contribute more to the volume because their radii spend more of the height with a greater value than the inside triangle.

You can see that by imagining dividing the cylinder into small horizontal slices. The volume of a slice consists of the inside cylinder whose radius is that of the cone at that height and the volume of the outside of that cylinder. i.e. $$\pi(R^2-r^2)dh+\pi r^2dh$$ Those two pieces contribute the same to the volume when $$R^2-r^2=r^2\implies\frac{R^2}{2}=r^2\implies r=\frac{R}{\sqrt{2}}\approx 0.707R$$ Only the lower $30\%$ of the cone's height contribute more volume to the cylinder than its complement.

That's why they don't contribute equally to the volume of the cylinder even though they do contribute equally to the area of the rectangular cross section.

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Quoting from the question:

Consider a 2D vertical cross section of the shape right through the middle. The cone can be divided into two triangles on either side of the altitude with areas equal to (rh)/2. There are two inverted triangles on either side of the cone which would become the parts of the left-over cylinder should the cone be removed. Areas of these triangles are also (rh)/2. This parity of the two inner triangles(inside the cone) and two outer triangles (outside the cone) would be true throughout the cylinder.

In the image below the 2D cross section as described is the rectangle on the left. We can show that the claim about parity fails by considering a horizontal cross section taken halfway down the cylinder. This appears as two concentric circles. The central circle, representing the cone, has area only one third that of the annulus representing the cylinder outside of the cone. If our argument was correct, these areas would be equal.

This is just a visual demonstration of other answers, showing the importance of distance from the axis of rotation when calculating the volumes.

enter image description here

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