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

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

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?

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  • $\begingroup$ If the sequence is constant, isn't the limit trivial? $\endgroup$ Aug 6 '19 at 11:17
  • $\begingroup$ @PinkPanther geometrically it should be something like an infinite paraboloid I suppose..but I dont know how to think about its edges. $\endgroup$
    – user123124
    Aug 6 '19 at 11:22
  • $\begingroup$ 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 $\endgroup$ Aug 6 '19 at 11:25
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    $\begingroup$ 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. $\endgroup$
    – Rhys
    Aug 6 '19 at 11:38
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    $\begingroup$ 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. $\endgroup$
    – Rhys
    Aug 6 '19 at 11:55
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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

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

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  • $\begingroup$ How come I get the wrong answer applying the formula$\int_{M}KdA=2\pi \chi(M)$? $\endgroup$
    – user123124
    Aug 11 '19 at 11:12
  • $\begingroup$ 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. $\endgroup$
    – user123124
    Aug 11 '19 at 18:02
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    $\begingroup$ 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 $1/(r\cdot\sqrt{1+4r^2})$. $\endgroup$ Aug 11 '19 at 19:01
  • $\begingroup$ but how do we compare $M_{r}$ and $S_{r}$ in that case? Then they are not the same set? $\endgroup$
    – user123124
    Aug 11 '19 at 20:01
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    $\begingroup$ 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. $\endgroup$ Aug 14 '19 at 18:50
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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 $\pi/2$.

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  • $\begingroup$ My mistake, I missed that. Thank you $\endgroup$
    – Rhys
    Aug 8 '19 at 11:28

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