0

RMM mathematical magazine question: $$I=\int_0^1\int_0^1\int_0^1\sqrt{x^2+y^2+z^2}dxdydz=?$$ By symmetry $$I=\frac18\int_{-1}^1\int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2+z^2}dxdydz$$ Let us consider the solid $S:$$0\leq x\leq1$, $0\leq y\leq x$, $y\leq z\leq 1.$ There are 24 pieces like this piece and by symmetry and $\frac{24}8=3,$ $$I=3\iiint_S\sqrt{x^2+y^2+z^2}dxdydz.$$ Converting to cylindrical coordinates looks nice: $$I=3\int_0^{\frac\pi 4}\int_0^{\sec\theta}\int_{r\sin\theta}^1 r\sqrt{r^2+z^2}dzdrd\theta.$$ But, I gave up after İ performed the first iteration.

What do you think about my way? Can this integral be evaluated?

Thanks in advance.

Bob Dobbs
  • 15,712
  • 1
    You should not use this cylindrical transformaition but the polar transfomation instead $x = r\cdot \cos(\theta), y = r\cdot \sin(\theta) \sin(\varphi), y = r\cdot \sin(\theta) \cos(\varphi)$ – NN2 Oct 25 '24 at 14:19
  • @NN2 You mean spherical? Would it be better? – Bob Dobbs Oct 25 '24 at 14:31
  • @BobDobbs At least $(x,y,z) \mapsto \sqrt{x^2+y^2+z^2}$ has a spherical symmetry. – mathcounterexamples.net Oct 25 '24 at 14:35
  • @Andreas: This question doesn't require a particular method to solve it, like the question you have mentioned. – FDP Oct 25 '24 at 15:23
  • 1
    @FDP True. My point is that the cited question has answers which SOLVE the problem addressed by OP, so we do not need do treat OP's question again. – Andreas Oct 25 '24 at 15:45
  • I see. It is solved by a famous theorem. Thanks for the comments. – Bob Dobbs Oct 25 '24 at 16:08
  • Mathematica says: $\frac{\sqrt{3}}{4}-\frac{\pi }{24}+\frac{1}{3} \ln \left(2+\sqrt{3}\right)+\frac{1}{72} \ln \left(3650401+2107560 \sqrt{3}\right)\approx 0.960592$ – Mariusz Iwaniuk Oct 27 '24 at 12:21
  • @MariuszIwaniuk https://www.wolframalpha.com/input?i=%282%2B%5Csqrt3%29%5E12 – Bob Dobbs Oct 27 '24 at 14:10

0 Answers0