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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3$\begingroup$ 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.)? $\endgroup$– MartyCommented Jun 2, 2010 at 18:16
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$\begingroup$ I mean the latter (a multivariable calculus course for engineers). $\endgroup$– PandoraCommented Jun 2, 2010 at 19:06
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$\begingroup$ This morning I'm trying to get $$\int_0^1\int_0^1\int_0^1\frac{dxdydz}{1-\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}}$$ but I cann't find a closed-form (I use a CAS, but the integrals are tedious). I don't know if this integral, or its generalization, is in the literature. My motivation was to do variants of an integral representation for the Apéry's constant that involves the quantity $xyz$ in the denominator, thus the geometric mean $(x^3y^3z^3)^{1/3}$. Feel free to try yourself variants for different combinations of generalized means (the example that I evoke is using the harmonic mean) $\endgroup$– user142929Commented Nov 24, 2019 at 10:10
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12 Answers
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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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$\begingroup$ Only a single integral is needed if you use spherical symmetry. $\endgroup$ Commented Oct 19, 2012 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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1$\begingroup$ I remember seeing an elementary solution of this one in a certain installment of the famous Scientific American column by the late Martin Gardner. $\endgroup$ Commented Jun 2, 2010 at 22:18
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1$\begingroup$ If somebody knew in which number this appeared- I am interested. $\endgroup$ Commented Jun 3, 2010 at 7:50
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3$\begingroup$ @Xandi or @JHS: Since I don't feel like committing suicide, can one of you give me the best method for solving the problem? $\endgroup$ Commented Aug 10, 2010 at 20:02
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$\begingroup$ @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. $\endgroup$ Commented Aug 11, 2010 at 3:36
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$\begingroup$ 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. $\endgroup$ Commented Aug 11, 2010 at 8:04
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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 $x^4 + y^4 \leq 1.$
answered Jun 2, 2010 at 20:42
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$\begingroup$ In Whittaker and Watson, end of the Gamma function chapter, by the way. $\endgroup$ Commented Jun 2, 2010 at 21:31
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$\begingroup$ 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. $\endgroup$ Commented Jun 2, 2010 at 22:23
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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.
answered Jun 2, 2010 at 19:12
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$\begingroup$ The question "U(3) Sato-Tate measure" is a quite different example. $\endgroup$ Commented Jun 2, 2010 at 21:00
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$\begingroup$ 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. $\endgroup$ Commented Aug 11, 2010 at 3:45
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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].
answered Jun 2, 2010 at 23:08
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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.
answered Jun 2, 2010 at 18:56
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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 $.
Good thing about this one is that there may exist a CAS out there that does not give you the answer at once.
answered Jun 2, 2010 at 21:05
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$\begingroup$ Indeed, Mathematica 7 will handle the integral in $u$, and then in $v$, but can't handle the resulting integral in $w$. $\endgroup$ Commented Jul 6, 2010 at 20:41
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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.
answered Jun 3, 2010 at 3:25
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2$\begingroup$ Is the formula a special case of a general phenomenon? $\endgroup$ Commented Jun 3, 2010 at 3:28
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1$\begingroup$ 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. $\endgroup$ Commented Jun 3, 2010 at 4:40
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4$\begingroup$ 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/PolyasRandomWalkConstants.html $\endgroup$ Commented Jun 3, 2010 at 7:23
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$\begingroup$ @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. $\endgroup$ Commented Jun 3, 2010 at 16:28
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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 $n$.
answered Jul 6, 2010 at 19:58
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$\begingroup$ I don't think you really need induction... after all, you have $\int_{[0,1]^n} \left( 1 - x_1 \cdots x_n \right)^{-1}dx_k = \int \sum_{i=0}^{\infty} (x_1 \cdots x_n)^i dx_k = \ldots = \sum_{i=1}^{\infty} 1/i^n$ $\endgroup$ Commented Jul 6, 2010 at 22:22
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$\begingroup$ 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... $\endgroup$ Commented Oct 19, 2012 at 14:29
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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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$\begingroup$ See mathdl.maa.org/images/upload_library/22/Ford/Wagon601-617.pdf for more proofs. $\endgroup$ Commented Aug 11, 2010 at 6:27
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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 $x_1^{n-1}x_2^{n-2}\dots x_{n-1}$.)
answered Jun 4, 2010 at 0:23
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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: $Vol(Fl_R^3)= Vol(RP^1) Vol(RP^2) = \frac{Vol(S^1)}{2}\frac{Vol(S^2)}{2}$
