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It's well known that a cylinder with height $h$ and the base radius $R$ will have a volume of

$$V = \pi R^2h$$

I'm trying to derive an intuitive understanding of it. I have two ways, one seems to be "working" (as in - it yields the correct result) and another one is failing, see below.

Method 1

If we agree that the area of a circle is known as $\pi R^2$ then we can think of a volume as a sum of all "areas of circles along the heights". So if we imagine that there is such an "object" that has a height of "one point" then its volume is equal to the area of the circle it has as a base. We then see that there will be exactly $h$ points in the height of the cylinder, so there will be exactly $h$ such "objects". Hence we can just multiply a "volume" of one "object" by $h$ which will yield the correct cylinder volume. I believe this method is just a "naive" version of a more rigorous calculus approach with defining the limit for $dx$ where $x$ will be along the height of the cylinder

Method 2

Now let's look at the cylinder as a rectangle rotated along one of its sides. If we have $h$ as the cylinder height and $R$ as a radius, then it's an $h \times R$ rectangle rotated by $2\pi$ around the side with length $h$. So the volume is the "sum" of all the "objects" with the height of "one point" and the base a rectangle. There are exactly $2\pi R$ points along the circumference of the circle that the rectangle will follow. Therefore we will have $2\pi R$ of these "objects". So we just need to multiple the "volume" of one such "object" by the number of these "objects" to get the cylinder volume. The "volume" of this "object" will be just the area of a rectangle, since it has a height of "one point". To conclude, we have $2\pi$ of these "objects", thus the volume must be $V = 2\pi R \times V_{\text{object}} = 2\pi R \times h R = 2\pi R^2h$.

But method 2 yields incorrect result, it's exactly $2$ times more than it should be. Why is that? I.e. why is that the point-based construction worked in the first case but failed in the second case? And how to make it work properly? What about rotation in general (i.e. not only a cylinder)?

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  • $\begingroup$ You may also use Cavalieri's principle which states that the volume of two solids lying between same set of two parallel planes is same provided the area of cross sections of both solids cut by any parallel plane is same. $\endgroup$ – Paramanand Singh Feb 20 at 13:02
  • $\begingroup$ I have discussed few applications of Cavalieri's principle in this blog post. $\endgroup$ – Paramanand Singh Feb 20 at 13:04
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The problem with the second approach is that although the outer side of the rectangle travels a distance of $2\pi R$ during your rotation, the inner side travels a distance of $0$, so you're not creating as much volume as you would if you extruded the rectangle "straight out". In fact, since the distance traveled by a point on the rectangle is a linear function of its horizontal position, the average distance traveled is the just the average of the inner and outer values: $\pi R$. Multiplying this average by the area of the rectangle gives you the correct volume.

What you've discovered is that it's not safe to think of "one point" as a well-defined positive distance. The cylinder is indeed the disjoint union of a bunch of one-point-thick sheets as you've described, but in decomposing it like this you've lost information about the "relative sizes" of the different "zero" thicknesses at different places in the sheet.

Instead of these sheets, you should consider the thin wedge of the cylinder that corresponds to a very small change $d\theta$ in the rotation angle. When viewed from the top, this wedge looks like a thin triangle with base $R$ and height $R\times d\theta$, and therefore area $R^2d\theta/2$. As we sweep $\theta$ from $0$ to $2\pi$, the $d\theta$s add up to $2\pi$, so the wedge volumes $\frac{R^2h}2d\theta$ add up to $\pi R^2h$.

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    $\begingroup$ Ah, this makes a lot of sense! It actually explains why decomposing by method 1 naively "worked" while method 2 failed. If I read this right, key thing here is "information about the structure" that is lost on a level of an individual slice if we don't do work to preserve that information. $\endgroup$ – Alma Do Feb 20 at 12:20
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    $\begingroup$ Yeah! This is related to the Borel-Kolmogorov paradox, which shows how things can go wrong when we naively condition on an event of probability 0. The resolution involves "preserving the structure" with a $\sigma$-algebra that describes how we've sliced up a space to arrive at the probability-0 event. en.wikipedia.org/wiki/Borel%E2%80%93Kolmogorov_paradox $\endgroup$ – Karl Feb 20 at 19:12
  • $\begingroup$ And yeah, "preserving information on the level of an individual slice" is what we're doing in calculus when manipulating expressions involving differentials like $d\theta$. In particular, we're only preserving first-order information, i.e. a linear approximation of the relationships among the "small changes" in the different variables. The insight of calculus is that this is all we need. $\endgroup$ – Karl Feb 20 at 19:24
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Well done for trying to get this intuition. The insight comes from calculus. It’s because in your 2nd method, you put too much material near the center.

A better approximation would be to use triangles (they are pointy at one end, and that point belongs near the center of the cylinder). Triangles’ areas are half that of the rectangle with same height and base, so that’s why you overestimated by a factor of two in the second method.

More rigorously, the area element in polar coordinates is $r \ dr \ d\theta$. You should let me know how much exposure to calculus before I explain further.

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  • $\begingroup$ I think I get the idea of "too much material", but I am not sure I follow the triangles construction, could you elaborate here? And yes, I'm ok with calculus, I just wanted to arrive at the result intuitively and test the limits of what is right there $\endgroup$ – Alma Do Feb 20 at 12:12
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    $\begingroup$ For a visualisation, look at yawnoc's pictures. If you take a thin rectangular prism, copy it, and rotate the copy slightly around one edge, the two prisms will still overlap near the axis of rotation, leading to an overestimate of the volume. However, if you do the same with triangular prisms, they will not overlap. $\endgroup$ – Benjamin Wang Feb 20 at 13:32
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Method 2 fails because rectangular slices overlap each other, especially near the axis of the cylinder. This results in the volume being over-counted.

Instead we need triangular slices (which have half the volume of the rectangular slices).

This is best understood with a diagram:

Left: vertical rectangular slices bunch up and overlap near the centre of the cylinder being approximated. Right: triangular slices do not overlap and fit perfectly.

Source code (Mathematica):

(* Cylinder dimensions *)
cylinderHeight = 2;
cylinderRadius = 1;
(* Slice dimensions *)
numSlices = 16;
sliceAngle = 2 Pi / numSlices;
sliceHalfThickness = cylinderRadius * sliceAngle / 2;
(* Slices *)
rectangularSlice =
  Cuboid[
    {0, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  ];
triangularSlice =
  Prism @ {
    {0, 0, 0},
    {cylinderRadius, -sliceHalfThickness, 0},
    {cylinderRadius, +sliceHalfThickness, 0},
    {0, 0, cylinderHeight},
    {cylinderRadius, -sliceHalfThickness, cylinderHeight},
    {cylinderRadius, +sliceHalfThickness, cylinderHeight}
  };
(* Make diagrams *)
showSlices[slice_] :=
  Graphics3D[
    Table[
      {
        FaceForm[Black],
        EdgeForm @ Directive[Thick, Hue[n / numSlices]],
        Rotate[slice, n * sliceAngle, {0, 0, 1}]
      }
      , {n, numSlices}
    ]
    , Boxed -> False
    , ImageSize -> 240
    , Lighting -> {"Ambient", White}
  ];
Row[
  showSlices /@ {rectangularSlice, triangularSlice}
]
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    $\begingroup$ Very nice. The visual explanation is immediately clear. $\endgroup$ – Peter - Reinstate Monica Feb 20 at 13:13
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For method 2, those "objects" actually have dimensions of $\frac12\times r\times h\times d\theta$ (think triangular prism or cylindrical sector). To calculate the volume, we can sum up the volume of these "objects" along the length of the circumference: $$\int_0^{2\pi r}\frac12rh\,d\theta=\pi r^2h.$$

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I will try and keep the answer simple and certain parts may be more simplified than the real situation.

First reduce the problem to a $2D$ problem as the height $h$ is not your concern. It really boils down to area of a circle being $\pi R^2$ vs. $2\pi R^2$.

a) Draw a rectangle with length of $\pi R$ and width of $2 R$. Now its area is $2 \pi R^2$. If we added height $h$, its volume would be $2\pi R^2h$. Now take a pin of length $R$ (if we added height $h$ to it, we would have a rectangular blade), place it in the rectangle with one end of the pin at the center of the rectangle.

Then rotate the pin around that end. Can you see that as we rotate, it will touch the rectangle at exactly two points (as the rectangle width is $2R$) and at all other points, it leaves space around. This clearly shows that the area of rotation of a pin of length $R$ is less than $2\pi R^2$. Now add height $h$ to it and we see that the solid generated by revolution of a rectangular blade of dimension $R \times h$ has lower volume than $2 \pi R^2 h$.

b) Now if you rotate a pin of length $R$ around one of its end by a very small angle $d\theta$, the arc length is $R d\theta$. Considering the angle is really small, we can approximate the area to a triangle of base $R d\theta$ and height $R$. So, $dA = \frac{1}{2} R \times R d\theta$. Over $2\pi$, this translates to $\pi R^2$.

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