In two dimensions, there is a (are many) straightforward explanation(s) of the fact that the perimeter (i.e. circumference) and area of a circle relate to the radius by $2\pi r$ and $\pi r^2$ respectively. One argument proceeds by approximating these quantities using regular cyclic polygons (equilateral, equiangular, on the circle of radius $r$), noting that such a polygon with $n$ sides can be decomposed into $n$ isosceles triangles with peak angle $\frac{2\pi}{n}$, base length $~2r\sin\frac{\pi}{n}$, and altitude $~r \cos \frac{\pi}{n}$ . Then, associating the circle with the limiting such polygon, we have, $$ P = \lim_{n\to\infty} n \cdot \text{base length } = \lim_{n\to\infty}2r \cdot \pi \frac{n}{\pi} \sin \frac{\pi}{n} = 2\pi r ~~, $$ and similarly, (via trig identity) $$ A = \lim_{n\to\infty} n\left(\frac{1}{2} \text{ base } \times \text{ altitude }\right) = \lim_{n\to\infty}\frac{r^2\cdot 2\pi}{2} \frac{n}{2\pi} \sin \frac{2\pi}{n} = \pi r^2 ~~. $$ Question:

Could someone offer intuition, formulas, and/or solutions for performing a similarly flavored construction for the surface area and volume of a sphere?

Images and the spatial reasoning involved are crucial here, as there are only so many platonic solids, so I am not seeing immediately the pattern in which the tetrahedra (analogous to the 2D triangles) will be arranged for arbitrarily large numbers of faces. Thus far my best result has been a mostly-rigorous construction relying on this formula (I can write up this proof on request). What I'd like to get out of this is a better understanding of how the solid angle of a vertex in a polyhedron relates to the edge-edge and dihedral angles involved, and perhaps a "dimension-free" notion for the ideas used in this problem to eliminate the need to translate between solid (2 degrees of freedom) and planar (1 degree) angles.

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    $\begingroup$ The problem is that in $3$ dimensions the analogous regular polyhedra (generalizing the regular $n$-gons) simply do not exist (there are none with more faces than the icosahedron), so one has to sacrifice some symmetry and the corresponding calculations will be harder. I cannot imagine a polyhedral argument which is easier than Archimedes' argument using Cavalieri's principle. $\endgroup$ Aug 1, 2012 at 2:39
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    $\begingroup$ As for solid angles, this is a very different question. There should be a discrete version of Gauss-Bonnet (en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem) relevant here. $\endgroup$ Aug 1, 2012 at 2:42
  • $\begingroup$ @QiaochuYuan I'm not entirely sure how updating new answers works, but below are my 'correspondingly harder calculations'. :) $\endgroup$ Aug 13, 2012 at 9:38

1 Answer 1


Alright, I've come up with a proof in what I think is the right flavor.

Take a sphere with radius $r$, and consider the upper hemisphere. For each $n$, we will construct a solid out of stacks of pyramidal frustums with regular $n$-gon bases. The stack will be formed by placing $n$ of the $n$-gons perpendicular to the vertical axis of symmetry of the sphere, centered on this axis, inscribed in the appropriate circular slice of the sphere, at the heights $\frac{0}{n}r, \frac{1}{n}r, \ldots,\frac{n-1}{n}r $ . Fixing some $n$, we denote by $r_\ell$ the radius of the circle which the regular $n$-gon is inscribed in at height $\frac{\ell}{n}r$ . Geometric considerations yield $r_\ell = \frac{r}{n}\sqrt{n^2-\ell^2}$ .

As noted in the question, the area of this polygonal base will be $\frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n}$ for each $\ell$ . I am not sure why (formally speaking) it is reasonable to assume, but it appears visually (and appealing to the 2D case) that the sum of the volumes of these frustums should approach the volume of the hemisphere.

So, for each $\ell = 1,2,\ldots,n-1$, the term $V_\ell$ we seek is $\frac{1}{3}B_1 h_1 - \frac{1}{3}B_2 h_2 $, the volume of some pyramid minus its top. Using similarity of triangles and everything introduced above, we can deduce that $$ B_1 = \frac{n}{2}r_{\ell-1}^2 \sin\frac{2\pi}{n}~,~B_2 = \frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n} ~,~h_1 = \frac{r}{n}\frac{r_{\ell-1}}{r_{\ell-1}-r_{\ell}}~,~h_2=\frac{r}{n}\frac{r_{\ell}}{r_{\ell-1}-r_{\ell}} ~~. $$ So, our expression for $V_\ell$ is $$ \frac{r}{6} \sin\frac{2\pi}{n} \left\{ \frac{r_{\ell-1}^3}{r_{\ell-1}-r_{\ell}} - \frac{r_{\ell}^3}{r_{\ell-1}-r_{\ell}} \right\} = \frac{\pi r}{3n} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ r_{\ell-1}^2 + r_\ell^2 + r_{\ell-1}r_\ell \right\} $$ $$ = \frac{\pi r^3}{3n^3} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ (n^2 - (\ell-1)^2) + (n^2-\ell^2) + \sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)} \right\} ~~. $$ So, we consider $ \lim\limits_{n\to\infty} \sum_{\ell=1}^{n-1} V_\ell$ . The second factor involving sine goes to 1, and we notice that each of the three terms in the sum is quadratic in $\ell$, and so the sum over them should intuitively have magnitude $n^3$. Hence, we pass the $\frac{1}{n^3}$ into the sum and evaluate each sum and limit individually, obtaining 2/3, 2/3, and 2/3 respectively (the first two are straightforward, while the third comes from the analysis in this answer).

Thus, we arrive at $\frac{\pi r^3}{3} (2/3+2/3+2/3) = \frac{2}{3}\pi r^3$ as the volume of a hemisphere, as desired.

So was this too excessive or perhaps worth it? I'll leave that to all of you. :)


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