## Looking for an interesting problem/riddle involving triple integrals.

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Does anyone know some good problem in real analysis, the solution of which involves triple integrals, and which is suitable for second semester Analysis students?

Thanks!

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Just since terminology differs in different countries: when you say "second semester analysis students", do you mean math concentrators at a university, taking a rigorous (proof-based) analysis sequence? Or do you mean students taking a multivariable calculus course (learning computational skills with Greens/Stokes theorem, etc.)? – Marty Jun 2 2010 at 18:16
I mean the latter (a multivariable calculus course for engineers). – Irene Adler Jun 2 2010 at 19:06

## 12 Answers

Ask the students what is the average distance from a point in the ball of radius R to the center of the ball. You will get the answer R/2. You then point out that the volume within R/2 of the center is much smaller than the remaining volume. Next some bright student proposes the radius that equates the volume inside and outside the shell. You explain why that does not give the correct answer either, which is more subtle. You commend the student who equated volumes for a good proposal, and ask if anyone has a better one. Then it is time to set up the very simple triple integral that gives the answer, and solve it by the method of spherical shells, or by spherical coordinates. And you compare the correct answer with the answer obtained by equating volumes, and see that they are quite close.

The emphasis in this example is more on setting up the right triple integral to answer the question, and not so much on evaluating a complicated integral.

You should let the students have a little time to ponder. I let them have about three minutes for each stage, because I don't want to lose the slower ones.

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Only a single integral is needed if you use spherical symmetry. – Richard Eager Oct 19 at 15:56

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I have seen students close to suicide over the following one: For $i=1,2,3$ denote by $C_i$ be the solid cylinder in $\mathbb R^3$ with axis given by the standard basis vector $x_i$ and radius 1. Compute the volume of $C_1\cap C_2 \cap C_3$.

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I remember seeing an elementary solution of this one in a certain installment of the famous Scientific American column by the late Martin Gardner. – J. H. S. Jun 2 2010 at 22:18

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If somebody knew in which number this appeared- I am interested. – Xandi Tuni Jun 3 2010 at 7:50

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@Xandi or @JHS: Since I don't feel like committing suicide, can one of you give me the best method for solving the problem? – André Henriques Aug 10 2010 at 20:02
@André: It actually works to integrate, as a function of radius, the surface area, whose smooth pieces come from cylinders cut by planes of symmetry and therefore unroll to sine curves. – Tracy Hall Aug 11 2010 at 3:36
I remember in high school trying (without success) to convince one of the engineers where I worked that specifying a quarter-round at the corner in each of the $X$, $Y$, and $Z$ directions of a blueprint was not technically the same (rather, only by convention) as putting a section of sphere there--that if you just cut along the projections, there would still be a distinct (if blunt) corner. – Tracy Hall Aug 11 2010 at 8:04

I do not know the exact location in his Collected Works but Dirichlet found the $n$-volume of

$$x_1, x_2, \ldots, x_n \geq 0$$ and $$x_1^{a_1} + x_2^{a_2} + \ldots + x_n^{a_n} \leq 1 .$$

For example with $n=3$ the volume is $$\frac{ \Gamma \left( 1 + \frac{1}{a_1} \right) \Gamma \left( 1 + \frac{1}{a_2} \right) \Gamma \left( 1 + \frac{1}{a_3} \right) }{ \Gamma \left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) }$$

Note that this has some attractive features. The limit as $a_n \rightarrow \infty$ is just the expression in dimension $n-1,$ exactly what we want. Also, we quickly get the volume of the positive "orthant" of the unit $n$-ball by setting all $a_j = 2,$ and this immediately gives the volume of the entire unit $n$-ball, abbreviated as $$\frac{\pi^{n/2}}{(n/2)!}$$

I think he also exactly evaluated the integral of any monomial $$x_1^{b_1} x_2^{b_2} \cdots x_n^{b_n}$$ on the same set.

So the question would be: given, say, positive integers $a,b,c,$ find the volume of $x,y,z \geq 0$ and $x^a + y^b + z^c \leq 1.$ If you like, fix the exponents, the triple $a=2, b=3, c=6$ comes up in a book by R.C.Vaughan called "The Hardy-Littlewood Method," page 146 in the second edition, where he assumes the reader knows this calculation.

This came back to mind because of a recent closed question on the area of $x^4 + y^4 \leq 1.$

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In Whittaker and Watson, end of the Gamma function chapter, by the way. – Charles Matthews Jun 2 2010 at 21:31
Thank you, Charles. I actually copied this as a few pages out of Dirichlet's collected works in translation. That was years ago and I can't find the pages. – Will Jagy Jun 2 2010 at 22:23

Coffee mug problem: the volume of the part of a cylinder cut off by a plane, representing the volume of the coffee you have left when you first see the bottom of the mug. (Plane is horizontal, cylinder tilted, plane touches the two circles bounding the cylinder). Much integration.

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The question "U(3) Sato-Tate measure" is a quite different example. – Charles Matthews Jun 2 2010 at 21:00
I always thought it was more interesting to ask the same problem for when you can see precisely half the bottom, since then it can't be solved in three seconds using symmetry.  The again, it's a useful drill to complete a series of excruciating integrals, only to be shown the immediate answer for each one coming from symmetry.  – Tracy Hall Aug 11 2010 at 3:45

The students can do $$\iiint_{[0,1]^3}\frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{(1-(1-xy)z)^{n+1}}\ dx\ dy\ dz,$$ at least for $n=0$ and $n=1$. These are the integrals producing Apery's approximations to $\zeta(3)$; the original paper is [F. Beukers, Bull. London Math. Soc. 11 (1979) 268--272].

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This is one of my all time favourites (quoted from Problems and Theorems in Analysis by Polya and Szego).

The 3D domain $\mathcal D$ is defined by the inequalities $$-1\leq x,y,z\leq 1,\quad -\sigma\leq x+y+z\leq \sigma.$$ Show that the volume of $\mathcal D$ is $$\iiint\limits_{\mathcal D}dx\, dy\, dz=\frac{8}{\pi}\int\limits_{-\infty}^{\infty} \left(\frac{\sin t}{t}\right)^3\frac{\sin \sigma t}{t} dt.$$

Hint. Show first that the number of integer lattice points that satisfy the conditions $$-n\leq x,y,z\leq n,\quad -s\leq x+y+z\leq s$$ for some $n$, $s\in\mathbb N$, is equal to $$\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\left(\frac{\sin \frac{2n+1}{2}t}{sin\frac {t}{2}}\right)^3\frac{\sin \frac{2s+1}{2}t}{\sin \frac{t}{2}} dt.$$

Edit. Check out also the lecture notes by Oliver Knill for some classical examples of volume computation and fabulous illustrations.

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I guess you can find plenty of them in the Problem Department of the American Mathematical Monthly. For instance, here is a nice one by M. Hajja and P. Walker:

Evaluate $\displaystyle \int_{0}^{1}\int_{0}^{1} \int_{0}^{1} (1+u^{2}+v^{2}+w^{2})^{-2} du dv dw$.

Good thing about this one is that there may exist a CAS out there that does not give you the answer at once.

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Indeed, Mathematica 7 will handle the integral in $u$, and then in $v$, but can't handle the resulting integral in $w$. – Kevin O'Bryant Jul 6 2010 at 20:41

Not suitable for any student, but I can't resist giving it:

$\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dxdydz}{3-\cos{x}-\cos{y}-\cos{z}} = \frac{\sqrt{6}}{96}\Gamma(\tfrac{1}{24})\Gamma(\tfrac{5}{24})\Gamma(\tfrac{7}{24})\Gamma(\tfrac{11}{24})$,

proven by G. N. Watson in 1939.

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Is the formula a special case of a general phenomenon? – Mariano Suárez-Alvarez Jun 3 2010 at 3:28
I don't believe so. – David Hansen Jun 3 2010 at 3:30

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It is! This integral appears in recent study of random walks, and there a more general integral is given. Don't ask me details; they are in recent papers of Jon Borwein on random walks, which are available from his site. – Wadim Zudilin Jun 3 2010 at 4:40

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Indeed, this integral gives (up to some constants) the probability that a 3-dimensional random walks ever returns to the starting point. The famous result of Polya is that this probability is 1 for dimensions 1,2 and strictly less than 1 for dimensions 3 and over. This fact is related to the fact that the values analogous integral for two variables ($\frac{1}{2-\cos(x)-\cos(y)}$) or one variable $\frac{1}{1-\cos(x)}$ are infinite. mathworld.wolfram.com/… – Guy Katriel Jun 3 2010 at 7:23
@Wadim: What I meant was, I don't know of any more general family of integrals this one fits into, which can be evaluated in terms of gamma functions. – David Hansen Jun 3 2010 at 16:28

A good example that my Honors Calc prof gave me is the following: $$\zeta(n) = \int_{[0,1]^n}\frac{d\boldsymbol x}{1-x_1\dotsm x_n}$$ The proof is an easy induction on $n$.

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I don't think you really need induction... after all, you have $\int_{[0,1]^n} \left( 1 - x_1 \cdots x_n \right)^{-1}dx_k = \int \sum_{i=0}^{\infty} (x_1 \cdots x_n)^i dx_k = \ldots = \sum_{i=1}^{\infty} 1/i^n$ – Eben Freeman Jul 6 2010 at 22:22
Well anywhere you see products or sums of arbitrarily many variables (let alone multiple integrals of arbitrary multiplicity) there's an implicit use of induction to define and manipulate them... – Noam D. Elkies Oct 19 at 14:29

This riddle is better phrased in term of double integral but can work for cubes instead of rectangles.

Here is the riddle: Call a rectangle integral if the length of (at least) one of its edges is an integer.

Prove that a rectangle that can be tessellate by integral rectangles if also integral rectangle.

It is a huge hint to say that it has something to do with double integral so you may want to avoid it in the beginning.

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See mathdl.maa.org/images/upload_library/22/Ford/… for more proofs. – Yuval Filmus Aug 11 2010 at 6:27
mathoverflow.net/questions/14212/… – J. H. S. Nov 8 2010 at 1:26

Feynman's formulae for products. $$\iiint\limits_{[0,1]^3}\frac{x_1^2 x_2}{[(a_4-a_3)x_3 x_2x_1+(a_3-a_2)x_2 x_1+ (a_2-a_1)x_1+a_1]^4}dx_1dx_2dx_3=$$ $$=\int\limits_{0}^{1}dx_1\int\limits_{0}^{1-x_1}dx_2\int\limits_{0}^{1-x_1-x_2}dx_3 \frac{1}{[(a_3-a_4)x_3+(a_2-a_4)x_2+(a_1-a_n)x_1+a_4]^4}=\frac{1}{3!a_1a_2a_3a_4}.$$

There are several ways to compute the integrals, the simplest one is probably by induction in dimension (an obvious generalization of the formula holds true for any $n=\dim [0,1]^n$ with the numerator in the first integral being $x_1^{n-1}x_2^{n-2}\dots x_{n-1}$.)

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The volume of the real flag manifold $Fl_R^3 = O(3)/Z_2^3$ can be obtained on one hand by the explicit integration on the $O(N)$ invariant volume element on its big cell:

$\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3(1+x_1^2+(x_3-\frac{x_1x_2}{2})^2)^{-1} (1+x_2^2+(x_3+\frac{x_1x_2}{2})^2)^{-1}$

This integral is not hard to evaluate using elementary integration techniques. The result $2\pi^2$ can also be obtained from: $Vol(Fl_R^3)= Vol(RP^1) Vol(RP^2) = \frac{Vol(S^1)}{2}\frac{Vol(S^2)}{2}$

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