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im currently looking at an application in which I have to calculate or at least approximate the surface integral of the absolute Gaussian curvature over a patch of a regular surface $S$ given by a parametrisation $\varphi:U\rightarrow \mathbb{R}^3$.

Because this seems very hard to do in an analytical fashion (as any formula for the Gaussian curvature is complicated). As im looking for the absolute curvature, the Gauss-Bonet theorem doesn't seem useful either. I've looked around a bit and finally found the following theorem:

The image of the Gauss Map (i.e. the unit normal map $N$) of $\varphi(R)$ for a region $R \subseteq U$ has area $\iint_R |K| d A$ (this is the exact surface integral im looking for). This is awesome as it allows me to at least approximate the integral easily.

I think understand the proof given for this theorem; however there's one thing I cant quite wrap my head around: The theorem also implies $\iint_R |K| d A$ is always bound by the surface area of the unit sphere, i.e. $4\pi r^2$, no matter how curvy and messed up the considered patch R is. How can that be? Am I missing something here? Since there is no bound on the Gaussian curvature itself, I don't understand how there can be one for the surface integral.

I'd be very grateful for any tips helping and/or correcting my intuition here :).

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  • $\begingroup$ As stated your theorem is wrong. You can get an embedded surface that looks like a ball folded over itself multiple times. Depending on the number of times folded over the integral can be arbitrarily large. Your theorem probably assume something like the Gauss map being non-degenerate or that the number is an upper bound for the area. $\endgroup$ – user225318 Apr 11 '15 at 19:52
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Hum, thinking about it a bit more, it is likely that there was a typo in your theorem. Without the absolute value signs, the statement that the area of the image of Gauss map is exactly $\int_R K \mathrm{d}A$ is precisely Gauss's statement of the Gauss-Bonnet theorem.

(See, e.g. Wikipedia.)

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  • $\begingroup$ Thanks, i checked the article. While this goes with my intuition, I still dont quite understand how $\iint_R | N_u \times N_v |$ du dv can only be positive and $\iint_R K d A$ can be negative, yet they'e supposed to be equal... am I missing something? $\endgroup$ – hodgepodge666 Apr 15 '15 at 7:13

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