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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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$$ \begin{align} &\int_0^{2\pi}\int_0^1\int_0^1xy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{1}\\ &=2\int_0^{2\pi}\int_0^1\int_0^yxy\sqrt{x^2+y^2-2xy\cos(\theta)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta\tag{2}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^2\sqrt{r^2y^2+y^2-2ry^2\cos(\theta)}\,\,y\,\mathrm{d}r\,\mathrm{d}y\,\mathrm{d}\theta\tag{3}\\ &=2\int_0^{2\pi}\int_0^1\int_0^1ry^4\sqrt{r^2+1-2r\cos(\theta)}\,\mathrm{d}y\,\mathrm{d}r\,\mathrm{d}\theta\tag{4}\\ &=\frac25\int_0^{2\pi}\int_0^1\sqrt{r^2+1-2r\cos(\theta)}\,r\,\mathrm{d}r\,\mathrm{d}\theta\tag{5}\\ &=\frac25\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}r^2\,\mathrm{d}r\,\mathrm{d}\theta\tag{6}\\ &=\frac2{15}\int_{-\pi/2}^{\pi/2}8\cos^3(\theta)\,\mathrm{d}\theta\tag{7}\\ &=\frac{16}{15}\int_{-1}^1(1-u^2)\,\mathrm{d}u\tag{8}\\[4pt] &=\frac{64}{45}\tag{9} \end{align} $$ 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 $u$

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  • $\begingroup$ +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
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    $\begingroup$ @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
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    $\begingroup$ 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

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