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During evaluation of different special cases of Gaussian curvature for the surface of revolution in Ch. 7 (page 152) of Elementary Differential Geometry by Pressley, it comes to the case when $\sigma(u,v)=(\text{cos}\ u\ \text{cos}\ v,\text{cos}\ u\ \text{sin}\ v,u)$. The book claims that it is not a surface at all. Why it is not a surface? What is it then?

EDIT - I add the text from the book:

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  • $\begingroup$ Why the edit removing important information from the question? Please do not repeat this unfortunate move. $\endgroup$ – Did Oct 8 '15 at 20:05
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Was it "not a surface" or "not a smooth surface" or "not a closed surface"? This is the surface of rotation of the graph of $y = \cos z$ about the $z$-axis. It is singular at z= $\frac \pi 2 + k\pi$. But other than that it qualifies as a surface.

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  • $\begingroup$ "does not give a surface". I add the text. Thanks $\endgroup$ – user231343 Oct 3 '15 at 5:05
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    $\begingroup$ It is the case $a = 0$ that does not give a surface, not the formula you provided. When $a = 0, f(u) = 0$ and $\sigma(u,v) = (0,0, g(u))$ which is evidently a vertical line, not a surface. $\endgroup$ – Paul Sinclair Oct 3 '15 at 5:21
  • $\begingroup$ Yes! thanks a lot :) $\endgroup$ – user231343 Oct 3 '15 at 5:42

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