I posted this question to stackexchange and after 24 hours it's got five votes and no answers, so let's see if mathoverflow can say more than that.

Consider two propositions in geometry:

  • Circumscribe a right circular cylinder about a sphere. The surface area of the cylinder between any two planes orthogonal to the cylinder's axis equals the surface area of the sphere between those two planes. (Archimedes showed this.)
  • Let a curve running from the south pole to the north pole on a perfectly spherical earth meet every meridian of longitude at the same angle $\gamma$. Then the length of the curve is proportional to $\sec\gamma$. (In particular, it's infinite of $\gamma$ is a right angle, and one could even go beyond a right angle and consider it an oriented length, so that the oriented length is negative if the curve goes from the north pole to the south pole. In that case, the infinite length would be the $\infty$ that's at both ends of the real line rather than $+\infty$ or $-\infty$, so the length depends continuously on $\gamma$.)

Both propositions admit the same kind of proof: For the first, consider what happens when the distance between the two planes is infinitesimal; then show that if it works then, then it also works for larger-than-infinitesimal distances. For the second, show that the length of an infinitesimal increment of the curve is $\sec\gamma$ times the latitude-component of the distance, just by drawing an infinitesimal right triangle.

You might think that in both cases one is finding an integral, but it's an integral of a constant: just the constant itself times the length of the interval over which one integrates. So it's as if you don't need to know about integrals and you certainly don't need to find any antiderivatives, but the method of proof doesn't work for larger-than-infinitesimal quantities.

One would like to say things like "energy is force times distance; it is only when the force is not constant that one must use integrals and multiply the force at an instant by the infinitesimal distance and then integrate". But if proving the force is constant itself requires an infinitesimal viewpoint, then despite the triviality of computing the integral by multiplying two numbers, it seems necessary to the logic of the argument to view it as an integral.


  • What other instances besides these two are known? (I suspect there are many in geometry.)
  • What else of interest can be said about this phenomenon?
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    $\begingroup$ Archimedes theorem generalises to the Duistermaat-Heckman theorem, en.wikipedia.org/wiki/Duistermaat%E2%80%93Heckman_formula, which in turn has many further generalisations. But they remain "nontrivial trivial integrals" in the sense that the integral is only computed via infinitesimal manipulations of cohomology classes rather than by locating primitives or computing Riemann sums. (I discussed this theorem recently at terrytao.wordpress.com/2013/02/08/… .) $\endgroup$
    – Terry Tao
    Commented Feb 19, 2013 at 18:31
  • $\begingroup$ Of related interest: mathoverflow.net/questions/33129/… $\endgroup$ Commented Feb 20, 2013 at 14:19
  • $\begingroup$ Are you asking for examples where first-order approximations are sufficient to get interesting geometric results? $\endgroup$
    – S. Carnahan
    Commented Feb 21, 2013 at 14:05
  • $\begingroup$ I'm not sure I understand the question. These are examples where arguments using infinitesimals are a natural approach. So use infinitesimals. Is the motivation for the question a feeling that infinitesimals are naughty? This may be relevant: mathoverflow.net/questions/16312/… . Re "it seems necessary to the logic of the argument to view it as an integral," Archimedes didn't know about integrals, except in the sense that the method of exhaustion was very similar to the notion of an integral. $\endgroup$
    – user21349
    Commented Feb 21, 2013 at 16:32
  • $\begingroup$ Math.SE copy of the question: Nontrivial trivial integrals. $\endgroup$ Commented Apr 18, 2017 at 6:25

3 Answers 3


Your first example (Archimedes) makes use of $\cos x/\cos x$. There are many further possible functions like $f(x)/f(x)$ or $f(x) + C - f(x)$ leading to trivial integrals. Cavalieri’s principle comes to mind: The slices of sphere and conus inscribed into a cylinder being equal to the slice of the cylinder.

The most important physical application is probably the constancy of energy $E = T + V$ in conservative fields as the sum of potential energy $V$ and kinetic enery $T$ such that the integral of action is simply $Et$

One example, the energy of the harmonic oscillator, $E_0(\sin^2 x + \cos^2 x)$, appears frequently in exams for freshmen in the form $(\sin^2 x + \cos^2 x)^n$ - a trivial integrand for those who know.


Let me phrase the proof for the first statement in modern language.

Suppose that the sphere is

$$\Sigma:=\lbrace x^2+y^2+z^2=1\;\rbrace, $$

and the cylinder is

$$ C=\lbrace x^2+y^2=1\rbrace. $$

For $a,b\in (-1,1)$, $a<b$, set

$$ \Sigma_{a,b}: =\Sigma\cap \lbrace a\leq z\leq b\rbrace,\;\; C_{a,b} := C\cap \lbrace a\leq z\leq b\rbrace. $$

We are supposed to prove that $\DeclareMathOperator{\area}{area}$

$$\area(\Sigma_{a,b})=\area(C_{a,b}). $$

Denote by $r,\theta,z$ the cylindrical coordinates in $\mathbb{R}^3$. we have a natural diffeomorphism $f: C_{a,b}\to \Sigma_{a,b}$ wich associates to a point $P\in C_{a,b}$ with cylindrical coordinates $(r=1, \theta,z)$ a point $\hat{P}$ on $\Sigma$ with cylindrical coordinates $(r=\sqrt{1-z}^2, \theta, z)$. In other words, if we coordinatize $\Sigma_{a,b}$ using the cylindrical coordinates $(\theta,z)$ and we coordinatize $C_{a,b}$ using the cylindrical coordinates $(\theta,z)$, then with these choices of coordinates the map $f$ is described as the identity map.

The area form on the cylinder is $dA_C=d\theta\wedge dz$, while the area density on the the sphere is

$$dA_\Sigma|= f^*dA_\Sigma. $$

The punch line is that $\area(\Sigma_{a,b})=\area(C_{a,b})$ because there exists an area preserving diffeomorphism between these two surfaces.

The second statement has a similar proof and the common principle behind these two statements is the same: if $f: (M_0,g_0)\to (M_1,g_1)$ is a finite-to-one covering map between Riemann manifolds with constant Jacobian, $J_f(x)=c$, $\forall x\in M_0$, then

$$ c\operatorname{vol}\;(M_0, g_0)= (\deg f)\operatorname{vol}\;(M_1, g_1). $$

  • 1
    $\begingroup$ If what I'd been asking for was proofs, then this answer would do it, but the question was something else. $\endgroup$ Commented Feb 19, 2013 at 17:39
  • 1
    $\begingroup$ The above proof is essentially Archimede's. Any instance when you have maps with constant Jacobians will lead to formulas like this. I've encountered similar situations when working with Grassmannians. $\endgroup$ Commented Feb 19, 2013 at 18:26
  • $\begingroup$ So the question is: when would it be the case that you need a strictly local argument to show that the Jacobian or whatever sort of derivative it is is constant? $\endgroup$ Commented Feb 19, 2013 at 20:41
  • 2
    $\begingroup$ The Jacobian is a local quantity. $\endgroup$ Commented Feb 20, 2013 at 0:31

In a static universe with homogeneous density of stars, the solid angle covered by the stars of a shell with radius $r$ and thickness $dr$ is $C r^{-2} r^2 \, dr $. The solid angle covered by all stars of the sphere with radius $r$ is $Cr$. (Olber's paradox)

  • $\begingroup$ So far I don't understand this. Do you mean a sphere with us at the center? I would think the solid angle "covered" by it would be the whole sphere. $\endgroup$ Commented Feb 23, 2013 at 23:57
  • $\begingroup$ (I do understand, however, that the mass of stars within the shell would be $Cr^2\,dr$ and the inverse-square law would make the intensity of light reaching us from them proportional to $1/r^2$. $\endgroup$ Commented Feb 23, 2013 at 23:59
  • $\begingroup$ Yes, I mean the sphere around us as center. The solid angle covered by the cross section $A$ of a star within a spherical shell at distance $r$ is $A/r^2$. So, given a constant density of stars, every shell contributes the same solid angle. Of course there is overlapping, But in an infinite universe the whole solid angle should be covered by stars. $\endgroup$
    – user112109
    Commented Feb 24, 2013 at 10:13
  • $\begingroup$ So you meant the solid angle covered by one star, and then summed over all the stars in that shell. What I thought at first, when you said "covered by the stars in that shell", is that you meant the whole volume of the shell, within which those stars are found. Saying "a star"---singular---makes it clear that that's not what was meant. $\endgroup$ Commented Feb 26, 2013 at 15:05

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