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Show that the cylinder $(x,y,z) \in R^3; x^2+y^2=1 $is a regular surface and find parameterizations whose coordinate neighborhoods cover it.

I'm going to be honest I saw this answer but I don't quite understand it. I am familiar with the propositions that I am given but not sure how its applied.

Proposition 2: If $f: U \subset R^3 \rightarrow R $ is a differentiable function and $a \in f(U)$ is a regular value of $f$, then $f^{-1}(a)$ is a regular surface in $R^3$.

Proof:Define this function $f(x,y,z)= x^2+y^2+z^2-1$. Then the cylinder is the set $f^{-1}(0)$. Computing all partial derivatives, this result is obtained.

$\frac{\partial f}{\partial x}=2x, \frac{\partial f}{\partial y}=2y, \frac{\partial f}{\partial z}=2z.$

It is clear that all partial derivatives are zero if and only if x=y=z=0 or $(0,0,0)$. Further checking shows that $f(0,0,0) \neq 0$, which means that $(0,0,0)$ does not belong to $f^{-1}(0)$. Hence for all $u \in f^{-1}(0)$, not all of partial derivatives at $u$ are zero. By proposition 2, the cylinder hence is regular surface.

Futhermore the cylinder can be parameterized as $g(u,v)=(cos u, sinu,v)$ where $u,v \in \mathbb{R}$

Okay so I'm not sure why we define the function as $f(x,y,z)= x^2+y^2+z^2-1$; is it because we are told it is in $R^3$ so we have to include $z$? I really want to understand all the steps to this. If someone can help I would really appreciate it.

Ruth Gutierrez
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1 Answers1

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  1. The function should be $$f(x,y,z)=x^2+y^2-1$$ In this case $(f_x,f_y,f_z)$ does not vanish on the cylinder
  2. In the above case we consider the cylinder as a level surface of the function $f$ while the parametrization is a different presentation of the surface as you did.
Semsem
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    So I don't understand; are you saying that how the function was defined is wrong? – Ruth Gutierrez Mar 04 '14 at 07:25
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    If your function is zero then $x^2+y^2+z^2=1$ which is a sphere not a cylinder@RuthGutierrez – Semsem Mar 04 '14 at 10:15
  • ok so does everything stay the same other then defining the function as $x^2+y^2+z^2-1$. We still get the partial $(2x,2y)$ and have $(0,0)$ making it a regular surface right, and have parametrization of $(cosu,sinu)$? – Ruth Gutierrez Mar 04 '14 at 15:13
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    defining the function as $x^2+y^2-1\to \nabla f=(2x,2y,0)$ which is not zero for any point on $x^2+y^2-1=0$ i.e. the surface is regular. – Semsem Mar 04 '14 at 15:17
  • Ok right. So really its almost the same information with the exception of the partials right? Am I right about the parametrization that it is $(cosu,sinu)$? – Ruth Gutierrez Mar 04 '14 at 16:43
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    The parametrization should be $r(u,v)=(\cos u,\sin u, v)$ as you did in your question. Note that $r$ is defined on ${(u,v)\in\mathbb{R}: -\pi< u < \pi }$ – Semsem Mar 04 '14 at 16:56
  • Yes I figured the parametrization didn't change. I'm getting this for the most part but I'm still confused on one thing, how do you show that the partial is not zero for any point on the function. I'm sure you can do like a matrix right and find the determinate?[ \left( \begin{pmatrix} 2 & 0 \ 0 & 2 \ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \end{pmatrix} \right)] – Ruth Gutierrez Mar 04 '14 at 17:01
  • Then eventually you get like $4+2\frac{\partial f}{\partial x}^2+2\frac{\partial f}{\partial y}^2 + \frac{\partial f}{\partial x}^2\frac{\partial f}{\partial y}^2 - \frac{\partial f}{\partial x}^2\frac{\partial f}{\partial y}^2 \neq 0$ – Ruth Gutierrez Mar 04 '14 at 17:07
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    you have to check regularity by one of the two separate different ways. One of them is to use the proposition, as $\nabla f=0$ only when $x=y=0$. But $f^{-1}(0)$ are those points $(x,y,z)$ where $x^2+y^2-1=0$ and (0,0) is not one of them i.e. $\nabla f$ does not vanish – Semsem Mar 04 '14 at 17:15
  • Second way is to use the parametrization and show that $r_u×r_v\ne 0$ – Semsem Mar 04 '14 at 17:18
  • Ok this makes more sense. Thank you so much – Ruth Gutierrez Mar 04 '14 at 19:16