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A question struggled me for a long time: If a $3D$ Klein Bottle represented by Robert Israel function. How to determine the exact boundary. I mean if you put it in a cuboid, what will the exact values of width, length, and height of the cuboid be?

$$\begin{align} x(u,v) =& -\frac{2}{15}\cos u(3\cos v - 30\sin u + 90 \cos^4 u\sin u\\ & - 60\cos^6 u \sin u + 5\cos u \cos v\sin u)\\ y(u,v) =& -\frac{1}{15}\sin u(3\cos v - 3\cos^2 u\cos v - 48\cos^4 u\cos v + 48\cos^6 u \cos v\\ &-60\sin u + 5\cos u\cos v\sin u - 5\cos^3 u\cos v\sin u \\ &-80\cos^5 u\cos v\sin u + 80\cos^7u \cos v \sin u )\\ z(u,v) =& \frac{2}{15}(3 + 5\cos u\sin u)\sin v \end{align} $$ for $0 \le u < \pi$ and $0 \le v < 2\pi$.

original image: https://i.stack.imgur.com/BM00U.png

achille hui
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DragonZ
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  • Your attempt to post a link didn't work. Perhaps you can type out Robert's description of that "Klein bottle." – Gerry Myerson Oct 25 '18 at 09:30
  • @DragonZ What do you mean? Explain please better about that we say. – Michael Rozenberg Oct 25 '18 at 09:38
  • The $z$-variable is pretty easy to handle. $5\cos u\sin u=(5/2)\sin2u$ lies between $-5/2$ and $5/2$, so $z$ lies between $-11/15$ (when $u=\pi/4$ and $v=3\pi/2$) and $11/15$ (when $u=\pi/4$ and $v=\pi/2$). The others look like they might get messy, once you've taken the partial derivatives and set them equal to zero and tried to solve for $u$ and $v$, but one can anyway make trivial estimates just based on $|\sin t|\le1$ and $|\cos t|\le1$: $|x|\le388/15$, $|y|\le332/15$. – Gerry Myerson Oct 26 '18 at 03:41
  • @GerryMyerson Thanks, I may try your boundary in my program to see whether it has any error. I really haven't touch Calculus for a long time and need to review for how to determine the boundary – DragonZ Oct 26 '18 at 10:23
  • @GerryMyerson Are you certain about the range is so large? The z-variable ranges 22/15=1.467 (from -11/15 to 11/15) which is easy to be determined and the range of x variable and y variable of your estimation seems too large. I tested in my program, the range of x should be about 3.712 and the range of y should be about 4.4, do you have any suggestions? – DragonZ Nov 03 '18 at 01:40
  • I never said the range was that large. I said those were trivial estimates on how large the range is. – Gerry Myerson Nov 03 '18 at 01:43
  • @GerryMyerson Then that is really a large estimation. Any idea of calculate it in accurate way? – DragonZ Nov 03 '18 at 01:49
  • As I wrote earlier, you can take partial derivatives, set them equal to zero, and try to solve. If you can't solve by hand, there are computer algebra programs like Maple and Mathematica and Matlab that might be able to solve for you. Maybe Wolfram Alpha can handle it. – Gerry Myerson Nov 03 '18 at 03:40
  • Making any progress with those partial derivatives? – Gerry Myerson Nov 04 '18 at 22:45
  • @GerryMyerson I need to review Calculus first, that's why I determine myself as a Programmer, not a mathematic researcher. – DragonZ Nov 05 '18 at 23:10
  • OK. Take your time. Keep us informed. – Gerry Myerson Nov 05 '18 at 23:12

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