20
$\begingroup$

Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.

A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^\circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_\text{rect} = R \cdot h$ multiplied by the circumference of the rotation circle $C_\text{circ} = 2\pi R$: \begin{align} V = A_\text{rect} \cdot C_\text{circ} = 2 \pi R^2 \cdot h \end{align}

Of course, the right volume of a cylinder with radius $R$ and height $h$ is \begin{align} V = A_\text{circ} \cdot h = \pi R^2 \cdot h \end{align} where $A_\text{circ} = \pi R^2$ is the area of the base circle of the cylinder.

Question: Where is the error in my previous argument based on infinitesimals?

$\endgroup$
  • 5
    $\begingroup$ rotated for 360 but why? for me it looks like 180 $\endgroup$ – Flame239 Oct 1 '18 at 9:00
  • 3
    $\begingroup$ @Flame239 : The base has radius $R$, not diameter $D=2R$. $\endgroup$ – Mefitico Oct 1 '18 at 16:05
  • 1
    $\begingroup$ Your link says to use parallel planes. Parallel planes, not planes intersecting along the axis of the cylinder. That's the error. $\endgroup$ – David K Oct 1 '18 at 17:51
  • 2
    $\begingroup$ What you're looking for in this case is the Pappus theorem; a conceptually similar approach applicable to revolution solids. $\endgroup$ – Euro Micelli Oct 1 '18 at 23:37
  • 2
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Paracosmiste Oct 2 '18 at 12:48
31
$\begingroup$

The rotation contributes to the volume by $2\pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.

Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 \leq r \leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2\pi r$, so its contribution to the volume of the cylinder is $dV = h \cdot dr \cdot 2 \pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e. \begin{align} V = \int dV = 2 \pi h \int_0^R r dr = 2 \pi h \left[\frac{r^2}{2} \right]^R_0 = \pi R^2 h. \end{align}

$\endgroup$
53
$\begingroup$

Where is the error in my previous argument based on infinitesimals?

The error is here:

Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for 360°.

If you approximate the cylinder with areas of a finite thickness you can see that these "areas of such rectangles of ... thickness" are not cuboids but triangular prisms.

The volume of a triangular prism however is not $A_\text{rect}\cdot l$ but only $\frac 1 2 A_\text{rect}\cdot l$.

Therefore you have to calculate: $V=\frac 1 2 A_\text{rect}\cdot C_\text{circ}$

... which leads to the correct volume of a cylinder.

$\endgroup$
  • 5
    $\begingroup$ Being a more visual person, I put together a quick diagram in Geogebra that I think represents what you're explaining (when viewed from the top of the cylinder) to help sanity-check my understanding: drive.google.com/open?id=1Lg8WgFvREt-W_h_5PjwcxLIYR1SZGYb5 If this is correct, it may be a useful addition for other visual thinkers to an already excellent answer. $\endgroup$ – Morgen Oct 1 '18 at 15:35
9
$\begingroup$

You are assuming that all points follow a circular trajectory of circumference $2\pi R$. But this is wrong, $2\pi r$ goes from $0$ to $2\pi R$ linearly, and its average value over the rectangle is $\pi R$.

$\endgroup$
8
$\begingroup$

This happens for the same reason, for which the following happens:

If you apply your method to derive the formula for circle area, you get $r 2 r \pi = 2r^2\pi$, even though the correct formula is $r^2\pi$.

The thing is, Cavalieri's method works intuitively when all points of the "sweeping" object travel the same distance (most obvious when they are translated). With rotation, however, each point travel distance depends on its distance from the axis of rotation.

Consider a point $t \in [0, r]$. Travel distance $D(t)$ of $t$, under one full rotation is $D(t) = 2 t \pi$. Its path is the circle of radius $t$, not $r$ (unless $t = r$). My intuitive belief is that correct formula can be thus expressed as integral $$\int\limits_0^rD(t)dt = \pi\int\limits_0^r 2 t dt = \pi(r^2 - 0) = \pi r^2$$

It's easy to see that similar reasoning can be applied to the cylinder case. Admittedly, this kind of brings your problem more into integral calculus, rather than Cavalieri's principle, which primarily targets straight line sweeping paths anyway.

$\endgroup$
5
$\begingroup$

The elementary volumes that you get by rotating the base rectangle are not parallelepipeds, they are prisms with a triangular basis (like when opening a book ..): thus half of the parallelepiped $B \times H \times d\alpha$ that you considered.

$\endgroup$
5
$\begingroup$

One other way to see the error is that at the point that you multiply $A_{rect}$ with a curve length, this length should not be the circumference of the cylinder (what is that circumference in the general application of Cavallieri?), but should be the length of the curve that the center of mass of the slices traces. As that is a circle at half the radius, you get the correct volume $$V=A_{rect}\cdot C_{center}=Rh⋅2\pi\frac{R}2=\pi R^2 h.$$

In general this approach requires that no intersections of the slices occur, that is, the extend of the slices should be smaller than the curvature radius of the center-of-mass curve.

$\endgroup$
1
$\begingroup$

Well, first of all, the infinitesimals don't match up. You have rectangles of thickness of, say, $dw$, and to get from one to the next, you're rotating through an angle of $d\theta$. The second problem is that these volumes are not distinct. If you take a slab with height $h$ and base $R$ with thickness $dw$ and rotate it by an angle of $d \theta$, you will end up with a volume that intersects the volume that it occupied before the rotation. Because of this "double counting" (note that not all points are double-counted; but on average you're counting each point twice), you calculate twice the volume you should.

You can see similar issue when calculating the area of a circle by sweeping a rectangle with length $R$ and thickness $dw$ through the circle. If you rotate it in small enough angles that every point on the circumference is covered by a rectangle, then points in the interior will be covered by multiple rectangles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.