Archimedes derived a formula for the area of a spherical cap.
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so Archimedes says that the curved surface area of a spherical cap is equal to the area of a circle with radius equal to the distance between the vertex at the curved surface and the base of the spherical cap. $$A = \pi(h^2+a^2)$$
I want to know how Archimedes derived this formula. I have searched on the net and only found solutions using integration. Is there a method to do this without using integration?

  • $\begingroup$ According to Morris Kline in Mathematical Thought from Ancient to Modern Times, Volume I, page 109, he used the method of exhaustion. See this for a description of the method (phrased in modern terms) applied to this problem. $\endgroup$ Jun 28, 2013 at 15:05
  • $\begingroup$ In Archimedes codex, or the Palimpset, they go into great detail how he arrived at this. He used a parabolic segment, the area intercepted between a parabola and a straight line crossing through it. I see the formula posted but thought you were looking for the history or thought behind it. $\endgroup$
    – Anon
    Nov 26, 2015 at 1:25

4 Answers 4


Enclose the sphere inside a cylinder of radius $r$ and height $2r$ just touching at a great circle. The projection of the sphere onto the cylinder preserves area.

That is the way Archimedes derived that the area of the sphere is same as lateral surface area of the cylinder which is $= (2 \pi r) (2r)=4\pi r^2$. The projection of the cap on the cylinder has area $(2 \pi r)h$. And since $a^2=r^2-(r-h)^2=2rh-h^2 \Rightarrow 2rh=h^2+a^2$, the area of the cap is $\pi (2rh) = \pi (h^2+a^2).$

Edit: corrected grammar

  • 3
    $\begingroup$ To convince yourself that the projection preserves area, note that although the radius of circle after projection increases(by a factor of $\cos \theta$) the width of the shell decreases(by a factor of $\cos \theta$) as the width is at an angle on the sphere and vertical on the cylinder, where $\theta$ is the angle between the circle and the great circle touching the cylinder. The two effects cancel out. $\endgroup$
    – nsoum
    Jun 28, 2013 at 14:18
  • $\begingroup$ what is the great circle? could you show me a diagram? $\endgroup$
    – udiboy1209
    Jun 28, 2013 at 14:21
  • $\begingroup$ Just imagine a cylinder of radius $r$ and height $2r$, put it on a table and drop the sphere inside it. It just touches the cylinder along a circle. That is the great circle I was talking about. $\endgroup$
    – nsoum
    Jun 28, 2013 at 14:43
  • $\begingroup$ And what is the point and method of projection to the cylindrical surface? It can't be a plate carree on the sides as that is not equal-area, nor can it be a shadow projection from a light source at the center as that is Mercator on the curved part of the cylinder and gnomonic on the top and bottom, neither of which are equal area projections either. $\endgroup$ Apr 1, 2017 at 6:46
  • $\begingroup$ The projection of sphere onto the walls of the cylinder, excluding the caps. $\endgroup$
    – Tosha
    Oct 1, 2020 at 12:25

Because a visual solution was requested, I have tried to explain @nsoum 's very correct answer using a diagram.


Consider a horizontal slice of the cylinder of thickness $z$. Let the length of the arc subtended on the sphere be $w$. From the figure, we can see that we can assume the arc to be small enough to be a straight line segment, if the thickness $z$ is small enough.

$$ \frac{x}{r} = \sin(\alpha) = \frac{z}{w} $$

Also, by using these thicknesses we can calculate area of cylindrical slice and spherical slice:

$$ A_{\textrm{cylinder}} = 2\pi r z $$

$$ A_{\textrm{sphere}} = 2\pi x w $$

But we have proved that $xw = rz$, so $ A_{\textrm{cylinder}} = A_{\textrm{sphere}} $.

This can very easily be extended to equating areas of the spherical cap, by slicing the spherical cap into multiple small slices and adding each one's area. We can say that each slice on the sphere will have area equal to the slice on the cylinder, hence area of the spherical cap becomes $2\pi rh = \pi(h^2+a^2)$

  • $\begingroup$ Thanks for your answer! I assume that, since that sphere appears to be touching the cylinder at four distinct points, the cross-section (diagram) that you're suggesting is vertical. Am I correct? (By "vertical" I mean the cross-section surface is perpendicular to the flat surface of the cylinder, and passes through the flat surface's center) $\endgroup$ Oct 12, 2017 at 11:32
  • $\begingroup$ @GaurangTandon Ahh, yes. Rotate the flat image around the vertical axis (perpendicular to the slice) and you would get the sphere and cylinder. $\endgroup$
    – udiboy1209
    Oct 12, 2017 at 20:25

Made a little visualization of it here. This helps my high school students understand this much better.


  • $\begingroup$ The Desmos demonstration is nice, but it is preferable to craft answers that are as self-contained as possible. This is particularly important on the internet, where links can go dead without warning. Could you, perhaps, include a screenshot or two, as well as some explanation as to how this answers the question? $\endgroup$
    – Xander Henderson
    Apr 25, 2018 at 17:40

See "On the Sphere and the Cylinder":

  • propositions 42 and 43,
  • pages 52 and 53 of Heath's 1897 English edition of Archimedes's works,
  • pages 244 and 245 of the pdf here.
  • $\begingroup$ "Link rots" may occur. It'd be really helpful if you at least copy and paste the content from the links you've posted, with proper attribution. Adding your own insights into that source would be a big plus. Thanks! $\endgroup$ Oct 12, 2017 at 11:24
  • $\begingroup$ @GaurangTandon, the question asks for Archimedes's derivation, and the accepted answer provides those ideas in a modern form, so I don't see much to add here. You're welcome to edit, or to expand on the bibliographical reference for Archimedes's works. $\endgroup$
    – user210229
    Oct 12, 2017 at 12:16

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