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