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Let f: $\Bbb R^2 \rightarrow \Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 \geqq 1.$ Consider the surface $S$ in $\Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 \leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.

I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.

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By the local Gauss-Bonnet theorem, $$ \int_{S} K \, \mathrm dA = 2\pi -\int_\gamma k_g \, \mathrm ds, $$ where $\gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2\geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $\gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $\gamma$, and parametrizing $\gamma$ accordingly.

Can you proceed?

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  • $\begingroup$ I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature? $\endgroup$ – jman63 Dec 7 '18 at 0:52
  • $\begingroup$ Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that. $\endgroup$ – MSobak Dec 7 '18 at 7:20

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