I want to visualize Ricci flow solution on the following sphere 

   Let $r> 0$   

   $L = \{ (x cos \theta, x sin \theta, x) | r < x < R \}$

   $S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$ 

   $T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$ 


   If $R$ is sufficiently large, then we have a two dimensional sphere $U$ enclosed by
   $L$, $S$, and $T$.


   The Ricci flow solution on $U$ shrinks fastly around region covered by $S$, but 
   the region covered by $L$ remains unchanged, since the Gaussian curvature is 0. 

   This confuses me. Where is wrong ? 


   MOTIVATION : I want to know the Ricci flow on a orbifold $O$ which is smooth except 
   one point. In generally, is there a solution on $O$ ? 
   Around a singuler point, the curvature is very large so that Ricci flow solution on $O$ 
   shrinks to the singular point fastly. 

   If we consider normalized Ricci flow on $O$, the solution goes to a "canonical"
   orbifold ? 

   What I say is that if $O$ is a two sphere with exactly one point singularity, then 
   the solution goes to $lim_{r \rightarrow 0} U$  


   Anything related with my opinion is welcome.          

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   I have a second question. First notice the following. We can define 
   $L_c = \{ (x cos \theta, x sin \theta, cx) | r < x < R \}$ 
   so that we have $U_c$. In addition $lim_{r \rightarrow 0} U_c$ is an 
   orbifold for some $c$ 


   Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ? 

 
   Question 2: In the following paper, a space with curvature $ \geq k$ is defined.


   M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded
   below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51.

   
  Is a space with curvature $ \geq k$ is a limit of manifolds ?    


  Thank you for your attention.