Exercise in differential geometry using Gauss-Bonnet For a positive real number $r$ let $M_{r}$ be the regular surface
$M_{r}=\{(x,y,z) \mid x^2+y^2=z<r^2, x>0,y>0\}$. 
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? 

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 $\pi/2$. 
