This one says the area of a ball is the same with the are of cylinder surrounding it.


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    $\begingroup$ On the Sphere and Cylinder is Archimedes' work that contains the proof of this. I think this may be a translated PDF of the work. $\endgroup$ Aug 3, 2020 at 7:08
  • $\begingroup$ In "On the Sphere and Cylinder" mentioned, Archimedes proved the general formula for the lateral area of a cylinder "The surface of any right cylinder excluding the bases is equal to a circle whose radius is a mean proportional between the side of the cylinder and the diameter of its base". [Translated quote from the book "God created the Integers"] $\endgroup$
    – Aravind
    Aug 3, 2020 at 7:15
  • $\begingroup$ The proof method is by contradiction; for the area of a circle, this is by inscribing and circumscribing regular polygons; for the cylinder, this is similar: inscribe/circumscribe such polygons on the base and then extend them to prisms that inscribe/circumscribe the cylinder. $\endgroup$
    – Aravind
    Aug 3, 2020 at 7:21
  • $\begingroup$ Here's a youtube video with a geometric run through of how he derived it. $\endgroup$ Aug 3, 2020 at 17:17

3 Answers 3


If you are looking for something intuitive back then, there is one way. If we take a rope of fixed (but minimal thickness), it can act as an integrable element (like discs of rings used in calculus).

1. Now start wrapping the rope around the sphere. Cut-off the rope at the length where it has covered the whole sphere. This is A1.


2a. Now again take another rope. Wrap it around the cylinder along it's height. Cut off at the length where the whole cylinder gets wrapped. This is B1.

2b. Take another piece of rope. From the centre of the circular base/top of the cylinder, start coiling it along the boundary until the whole area of the circle is covered. Cut at this lenght. This is B2. Do the same for the top/base. This will be B3.

Now when you compare the length of A1 vs the length B1+B2+B3 ,under the errors of physical measurement, they will be of same length. Multiply the length of the rope with it's mean diameter, and you will have nearly equal areas too! (since the area of that elemental strip will be it's length into it's width).

Here's the pic for help in visualization:

enter image description here

It's probably one of the ways early mathematicians could have done it without any proper knowledge of trigonometry or calculus.

  • $\begingroup$ This won't do. Not rigorous. I am sure Archimedes know better $\endgroup$
    – user4951
    Aug 9, 2020 at 9:41

Archimedes showed that the surface area of a sphere is equal to four times the area of the base circle. Since the cylinder of the same height has the same area (by unfolding it), the assertion follows.

His proof involved approximating a sphere by a surface of revolution of a polygon inscribed in the circle, as follows: [Modern notation is in square brackets]

$\hspace{2cm}$enter image description here

Each section of the polygon-surface is a frustum of a cone. It was well-known that the curved surface area of a cone is the base circle times the ratio of the side length to the radius $$[\ =\pi r^2(l/r)=\pi rl\ ]$$ So the relevant yellow part is equal to the difference of two cones, equal to the sloping side length times the mean perimeter $$[\ \pi r_1l_1-\pi r_2l_2=2\pi\frac{r_1+r_2}{2}(l_1-l_2);\ \textrm{note that } r_1/r_2=l_1/l_2].$$

So the problem reduces to adding the mean perimeters of the polygon-surface, since the side-lengths are fixed [$l_1-l_2=BC$]. This sum is the same as adding the vertical diameters at the vertices $$[\pi\sum_i\frac{d_i+d_{i+1}}{2}=\pi\sum_id_i]$$

$\hspace{3cm}$enter image description here

This is where Archimedes' ingenuity comes to the fore: Draw another set of diameters, as in the faint red lines. They are cut by the off-diagonal $AC$ in two similar triangles, each similar to $\triangle ABC$. Each red triangle has a fixed ratio of its hypotenuse/height to its base [$ab:bc=AB:BC$], hence this is also the ratio of their sums. But the sum of the verticals is the required sum [$\sum_i(ab)_i=\sum_id_i$], while that of the bases is $AC$ [$=\sum_i(bc)_i$]. The conclusion is that the sum [$\sum_id_i$] is equal to the diameter times the ratio of the off-diagonal to the side [$AB.AC/BC$].

Combining the two results gives that the area of the polygon-surface is the base circumference times the off-diagonal [$\pi AB.AC$], which is less than four times the base circle area [$\pi AB^2$].

An identical argument with a circumscribed polygon, shows that the polygon-surface has area larger than an $\epsilon$ of the required area. Hence the result. $$[\pi AB.AC\to \pi {AB}^2,\quad \textrm{as } BC\to0.]$$

$$\ \\[1cm]$$ P.S. In modern trigonometric terms, Archimedes' main part of the proof of the sum of diagonals is equivalent to $\sum_{n=1}^N\sin\frac{n\pi}{N}=\cot\frac{\pi}{2N}$.


Put the sphere inside the cylinder and let $\phi\in[0,\pi]$ be the oriented angle of a vector with the oriented axis of symmetry in the construction. Then take two parallel planes (perpendicular to the cylinder) cutting the sphere at angles $\phi$ and $\phi+\delta \phi$, respectively (with $\delta \phi$ being small).

The slice on the sphere between the two cuts is comparable to a 'tilted' strip of width $r\delta \phi$ and of circumference $2\pi r \sin \phi$. Whence its area is of order: $$ \Delta A = ( r \ \delta\phi) \ (2 \pi r \cdot \sin(\phi))$$

Now the slice on the cylinder is a vertical strip of width $r \sin \phi \ \delta \phi$ (the factor $\sin \phi$ coming from projection) and circumference $2\pi r$. Whence of area: $$ \Delta A' = (r\ \sin (\phi) \ \delta \phi) \ (2\pi r),$$ i.e. the two slices have the same area. Applying some limiting argument we obtain the result as well as the following generalization: Take two parallel planes cutting the sphere at angles $\phi_1<\phi_2$ then the areas of the slice on the sphere and the cylinder are the same. This leads to Lamberts equal-area projection which provides maps of the earth respecting areas (but highly distorted near the poles).

  • $\begingroup$ The question is about the methods Archimedes might have used. $\endgroup$ Aug 8, 2020 at 9:09
  • $\begingroup$ @PeterPhipps Yes, I am aware. The above succinct version can easily be made into an Archimedian like argument with polygonal approximations for the two strips. Drawings would clearly help on the comprehension of the projections but I am somewhat short of time. $\endgroup$
    – H. H. Rugh
    Aug 8, 2020 at 10:20
  • $\begingroup$ Archimedes doesn't have trigonometry and my friends don't either $\endgroup$
    – user4951
    Aug 9, 2020 at 2:09
  • $\begingroup$ That being said, this actually look great $\endgroup$
    – user4951
    Aug 12, 2020 at 12:23
  • $\begingroup$ Even though this is not technically correct as an answer, I actually like this answer a lot. I would teach this to my children. $\endgroup$
    – user4951
    Aug 12, 2020 at 13:55

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