Consider the following triple integral:

$$\int_0^{2\pi}\int_0^1 \int_0^1 xy\sqrt{x^2 + y^2 -2xy\cos(\theta)} \, dx \, dy \, d\theta$$

A solution was provided to this integral by Jack D'Aurizio here, but both his solutions required rather sophisticated methods, like elliptic integrals and special function expansions. What I've been struggling with-and would like a second opinion on-is whether or not this integral can be solved in closed form using very simple calculus techniques,like a standard change of variables to plane polar in the xy domain or spherical coordinates in $R^3$.

My labor over the last 2 days,multiple false starts and geometric arguments in the domain seem to indicate that the answer is no because there's no way to set up the integral without introducing a term of $\sqrt {\sin (ax)}$ or $\sqrt {\cos(ax)}$ at some point. Therefore, some special functional substitution or numerical method solution is needed.

Or am I wrong?

  • 1
    $\begingroup$ You probably noticed that yourself, but the argument of the square root very much looks like en.wikipedia.org/wiki/Law_of_cosines. Maybe this could be helpful for some geometric argument. $\endgroup$ – PhoemueX Mar 8 '15 at 21:20
  • 1
    $\begingroup$ I am really interested in elementary answers. By the way, a suggestion may be to give $$\int_{0}^{2\pi}\int_{0}^{1}t\sqrt{1+t^2-2t\cos\theta}\,dt\,d\theta$$ to integrate, that is just a double integral leading to exactly the same problem. $\endgroup$ – Jack D'Aurizio Mar 8 '15 at 21:20
  • $\begingroup$ @PhoemueX I saw that,but not how to use it to simplify the problem. Unless you just want to do a naked substitution of c for the whole kit and kaboodle and I don't see how that gets us anything simpler that makes sense. I'll have to go through it carefully,thanks for the hint! $\endgroup$ – Mathemagician1234 Mar 8 '15 at 21:54
  • $\begingroup$ @JackD'Aurizio Now THAT is a great idea,I never thought of it! The brilliance of this substitution is that it leaves the limits of integration unchanged! I'll try both suggestions and see which one gives a simpler solution-if any. $\endgroup$ – Mathemagician1234 Mar 8 '15 at 21:55

$$ \begin{align} &\int_0^{2\pi}\int_0^1\int_0^1xy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{1}\\ &=2\int_0^{2\pi}\int_0^1\int_0^yxy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{2}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^2\sqrt{r^2y^2+y^2-2ry^2\cos(\theta)}\,\,y\,\mathrm{d}r\,\mathrm{d}y\,\mathrm{d}\theta\tag{3}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^4\sqrt{r^2+1-2r\cos(\theta)}\,\mathrm{d}y\,\mathrm{d}r\,\mathrm{d}\theta\tag{4}\\ &=\frac25\int_0^{2\pi}\int_0^1\sqrt{r^2+1-2r\cos(\theta)}\,r\,\mathrm{d}r\,\mathrm{d}\theta\tag{5}\\ &=\frac25\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}r^2\,\mathrm{d}r\,\mathrm{d}\theta\tag{6}\\ &=\frac2{15}\int_{-\pi/2}^{\pi/2}8\cos^3(\theta)\,\mathrm{d}\theta\tag{7}\\ &=\frac{16}{15}\int_{-1}^1(1-u^2)\,\mathrm{d}u\tag{8}\\[4pt] &=\frac{64}{45}\tag{9} \end{align} $$ Explanation:
$(2)$: the integral is the same for $x\lt y$ as for $x\gt y$, so assume $x\lt y$ and multiply by $2$
$(3)$: substitute $r=\frac xy$
$(4)$: collect the $y$s and switch the order of integration
$(5)$: integrate in $y$
$(6)$: $(5)$ is the distance from $(1,0)$ integrated over the unit disk centered at $(0,0)$;
$\hphantom{(6):}$this is the same as $r$ integrated over the unit disk centered at $(1,0)$,
$\hphantom{(6):}$whose equation is $r\le2\cos(\theta)$ for $\theta\in[-\pi/2,\pi/2]$
$(7)$: integrate in $r$
$(8)$: substitute $u=\sin(\theta)$
$(9)$: integrate in $u$

  • $\begingroup$ +1 EXCELLENT computation-I was looking to use standard coordinate systems like plane polar for the xy-region or spherical for the entire region using $\theta$ for the vertical angle above the xy plane generated by cos($\theta$). It didn't occur to me to use polar coordinates in a more unorthodox manner-I should have.Any alternative solutions to this problem? $\endgroup$ – Mathemagician1234 Mar 9 '15 at 2:55
  • $\begingroup$ Looking over my own computations, I just realized I got as far as your equation (5) before getting stuck. : ( $\endgroup$ – Mathemagician1234 Mar 9 '15 at 3:05
  • 1
    $\begingroup$ @Mathemagician1234: I was going down the wrong path after $(5)$ until I took a walk with my dog, and realized while we were out that centering the integration on $(1,0)$ would simplify things greatly. $\endgroup$ – robjohn Mar 9 '15 at 3:09
  • 1
    $\begingroup$ See,that's the difference between someone who merely understands basic techniques and someone with a truly open mind and who's able to see the unorthodox facts that supply the key to a problem. Even the best mathematicians might not be able to see the relevance of that geometric fact,even though of course they understand what you did once you did it. I smacked my head and did a Homer Simpson once I understood what you did. Which once again demonstrates what too many of us forget once we begin rigorous math: WHENEVER POSSIBLE,DRAW A PICTURE. $\endgroup$ – Mathemagician1234 Mar 9 '15 at 3:16
  • $\begingroup$ (+1), but this answer deserves much more. $(5)\to(6)$ is the definitive trick, maybe it answers to a recent question of mine, too. Bravo, @robjohn. $\endgroup$ – Jack D'Aurizio Mar 9 '15 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.