Bound on the surface integral of the absolute Gaussian curvature given by surface of unit sphere? 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 :).

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.)
