# Integral of Gaussian curvature over S

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.

## 1 Answer

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?

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