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If I start with an infinite flat sheet of graph paper, and in polar coordinates cut out a piece according to: $r>0, \ \ -f(r) < \theta < f(r)$

Now I want to stitch the remaining graph paper together, by associating each point $(r,f(r))$ to $(r,-f(r))$ on the seam.

How do I calculate what the curvature is of the stitched up paper along the seam?

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    Take a look at [this answer by Deane Yang](http://mathoverflow.net/questions/16850/curvature-and-parallel-transport/17099#17099) at MathOverflow to get an idea of how curvature can be computer via parallel transport of a vector around a closed loop. On a two-dimensional surface this computation simplifies drastically (see, e.g. exercise 12 in [this problem set](http://www.dpmms.cam.ac.uk/study/II/DifferentialGeometry/2010-2011/dg2e4.pdf); note that the computation is still morally valid if your manifold is only $C^1$, as it would be in your case).2011-04-30

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Perhaps your description of what you wanted to do doesn't match with what you intended. If we go backwards, i.e., start with a right circular cone, and make some squiggly cut in it, from base to apex, and then flatten that out we will get a disk with a wedge with squiggly sides cut out. The two sguiggly sides match up.

Assume that if we did the cut with a straight line, then we'd get a wedge described by the lines $\pm \tilde{\theta}$. Now, we can imagine replacing those straight lines by the squiggly one described by your function. To keep from making a mess, we'd need constraints on the function, corresponding to your squiggly line on the cone not wrapping around and intersecting itself. Say f'(0) = 0, and $|f(r)| < \pi r$, but we might have to think about that more. Now we have a wedge cut out by curves described by $\pm \tilde{\theta} + f(r)$. In other words, we're removing the piece according to $ -\tilde{\theta} + f(r) \leq \theta \leq \tilde{\theta} + f(r), $ which is not the same as you have in your question. If we get it right, and we have a seam with no curvature, then it will be represented by a curve on the cone we get, so it makes sense to go backwards like this.

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    won't be $r_0$ when the cusp gets in the way. I am not sure why @Willie suggested that we have a $C^1$ manifold here. Maybe he will elaborate on that.2011-04-30