# Exercise in differential geometry using Gauss-Bonnet

For a positive real number $$r$$ let $$M_{r}$$ be the regular surface

$$M_{r}=\{(x,y,z) \mid x^2+y^2=z0,y>0\}$$.

Let $$K$$ denote the Gaussian curvature of $$M_{r}$$. Determine

$$\int_{M_{r}}KdA$$ and $$\lim_{r \rightarrow \infty}\int_{M_{r}}KdA$$.

Solution; Using Gauss-Bonnet we know $$\int_{M_{r}}KdA=\frac{\pi}{2}$$, since the surface is the positive quadrant of a paraboloid.

But how does one solve the limit?

• If the sequence is constant, isn't the limit trivial? Aug 6, 2019 at 11:17
• @PinkPanther geometrically it should be something like an infinite paraboloid I suppose..but I dont know how to think about its edges. Aug 6, 2019 at 11:22
• That's not the question. You want to find $\lim_{r\rightarrow\infty}x_r$, where $x_r$ is constant, so the limit is the same constant Aug 6, 2019 at 11:25
• Your use of Gauss-Bonnet is wrong here. Your justification was quite glib, so I can't tell you exactly how you've gone wrong, but remember that it applies to compact surfaces without boundary.
– Rhys
Aug 6, 2019 at 11:38
• There are sort of two related comments here... one is that $M$, as defined, is not compact because it does not include the boundary at $z=r$. The other is to say that including the boundary at $z=r$ (i.e. changing the inequality to equality) does not change the value of the integral, but gives a compact surface with boundary. You can look up the technical definition of a boundary, but it is what you intuitively think of.
– Rhys
Aug 6, 2019 at 11:55

As a complement to the other answer here is a solution that uses Gauss-Bonnet:

Let $$S_{r}=\{(x,y,z) \mid x^2+y^2=z\leq r^2\}$$. By symmetry $$\int_{M_{r}}K\;dA=\frac 14\int_{S_{r}}K\;dA$$. Since $$S_{r}$$ is a compact two-dimensional Riemannian manifold by Gauss-Bonnet

$$\int_{S_{r}} K\;dA+\int_{\partial S_{r}}k_g\;ds=2\pi\chi(S_{r})$$

As $$S_{r}$$ is homeomorphic to a disc, $$\chi(S_{r})=1$$. The boundary $$\partial S_{r}$$ can be parametrized by the curve $$\gamma(t)= (r\cos t,r\sin t,r^2)$$. The unit tangent vector is $$T=(-\sin t,\cos t,0)$$ and the inward-pointing unit normal to the boundary $$\partial S_{r}$$ on the surface $$S_r$$ at $$\gamma(t)$$ is $$N=\frac {-1}{\sqrt{1+4r^2}}(\cos t,\sin t,2r)$$.

Hence $$\int_{\partial S_{r}}k_g\;ds=\int_{0}^{2\pi}\langle T',N\rangle \,dt=\int_{0}^{2\pi}\frac 1{\sqrt{1+4r^2}}\,dt=\frac {2\pi}{\sqrt{1+4r^2}}$$

which implies

$$\int_{S_{r}} K\;dA=2\pi\left(1-\frac{1}{\sqrt{1+4r^2}}\right)\;,\;\text{so}\;\int_{M_{r}}K\;dA=\frac{\pi}{2}\left(1-\frac{1}{\sqrt{1+4r^2}}\right)$$

• How come I get the wrong answer applying the formula$\int_{M}KdA=2\pi \chi(M)$? Aug 11, 2019 at 11:12
• i.e why does the integral of the geodesic curvature enter the equation? I have a theorem stating that for compact $M$ the formula in the other comment holds. Aug 11, 2019 at 18:02
• Then your formula is probably stated only for compact 2-dimensional surfaces without boundary. $M_r$ has no boundary but you cannot apply Gauss-Bonnet since it is not compact. $S_r$ however is compact with boundary and the geodesic curvature at each point of the boundary is $1/(r\cdot\sqrt{1+4r^2})$. Aug 11, 2019 at 19:01
• but how do we compare $M_{r}$ and $S_{r}$ in that case? Then they are not the same set? Aug 11, 2019 at 20:01
• Yes, (up to sign) it's that if you start with an parametrization of the boundary by arclength and $N$ is the Gauss map (for some chosen orientation). Actually i call $N$ what you would call $N\times T$ if the orientation is chosen correctly. If you don't use an arclength parametrization then you have to divide by the the speed of the parametrizing curve (which here is r ) to get the geodesic curvature. If you then integrate the geodesic curvature using that parametrization you have to multiply by the speed again so it cancells out. Aug 14, 2019 at 18:50

Just calculate: from Wikipedia (and many other places) we can find the Gaussian curvature of a surface expressed as the graph of a function $$z(x,y)$$: $$K = \frac{z_{xx} z_{yy} - z_{xy}^2}{(1+z_x^2+z_y^2)^2}$$ where subscripts denote partial derivatives. So we just need to integrate this over the given surface.

Cylindrical coordinates are natural here, so let $$x = R\cos(\theta), y=R\sin(\theta)$$, and use coordinates $$R,\theta$$ on $$M_r$$. The Gaussian curvature is $$K = \frac{4}{(1+4R^2)^2}$$ and the metric on $$M_r$$ becomes: $$ds^2 = dR^2 + R^2d\theta^2 + dz^2 = (1+4R^2)dR^2 + R^2d\theta^2$$ where we have used the equation of the surface in the last step. Hence the volume form is $$R(1+4R^2)^{\frac 12}d\theta dR$$.

So the final expression to evaluate is (bearing in mind that we only integrate over $$0\le\theta\le\pi/2$$ because this corresponds to $$x, y > 0$$): $$\int_0^{\pi/2}\!\!\!\!d\theta\int_0^r\!\!\!\!dR ~4R(1+4R^2)^{-\frac 32}$$ Integrate this however you like to get the answer: $$\frac{\pi}{2}\left(1-\frac{1}{\sqrt{1+4r^2}}\right)$$

The limit as $$r\rightarrow\infty$$ is then just $$\pi/2$$.

• My mistake, I missed that. Thank you
– Rhys
Aug 8, 2019 at 11:28