Filling in a missing detail in a Gauss-Bonnet problem The problem I'm working on is as follows

Let $M \subseteq \mathbb{R}^3$ be a non-compact orientable surface without boundary which coincides with the $(x, y)$-plane outside of the ball of radius $10$ centered at the origin. Prove that if the Gaussian curvature $K$ of $M$ is everywhere non-negative, then $K$ is everywhere $0$.

Here's what I've got so far.
Let $\alpha : [0, 20 \pi ] \to M$ be the closed unit-speed curve $$\alpha(s) = \left( 10 \cos \frac{s}{10}, 10 \sin \frac{s}{10}, 0 \right) ,$$ and let $R$ be the surface (with boundary) enclosed by $\alpha$. Gauss-Bonnet tells us that
$$\iint_{R} K \mathrm{d} M + \int_\alpha \kappa_g \mathrm{d} s = 2 \pi \chi(R) .$$
We know that along the curve $\alpha$ we have $\textrm{T}_{\alpha(s)} M = \{ (x, y, 0) : x , y \in \mathbb{R} \}$, and our Darboux frame is
\begin{align*}
T(s) & = (- \sin (s / 10) , \cos (s / 10) , 0) , \\
T’(s) & = \frac{1}{10}(- \cos (s / 10) , - \sin (s / 10) , 0) \\
 & = \frac{1}{10} N(s) .
\end{align*}
So $\kappa_g \equiv \frac{1}{10}$, meaning
\begin{align*}
\iint_{R} K \mathrm{d} M + \int_{\alpha} \kappa_g \mathrm{d} s & = 2 \pi \chi(R) \\
= \iint_{R} K \mathrm{d} M + 2 \pi \\
\Rightarrow \iint_{R} K \mathrm{d} M & = 2 \pi (\chi(R) - 1) .
\end{align*}
Now what I want to do is place an upper bound on $\chi(R)$ such that $\chi(R) \leq 1$, meaning that $\iint_{R} K \mathrm{d} M \leq 0$. If $K$ is everywhere nonnegative, this would preclude $K$ taking any positive values on $R$, since for $K$ to take a positive value would force the integral to be positive. I can then argue that $K$ vanishes everywhere on $M$, since outside of $R$, the surface $M$ is just the $(x, y)$-plane.
My trouble is that I don't understand the Euler characteristics of surfaces with boundaries -or really even non-compact surfaces- well enough to see why this bound should be, assuming I'm on the right track in the first place. Can someone explain this step to me?
Thanks!
EDIT: I see now that this method is misguided. Specifically, we can’t guarantee that $R$ is compact, meaning we can’t necessarily invoke GB. For example, we could consider the case where $M$ is the $(x, y)$-plane minus the origin. Then $R$ would be a punctured disc, thus not compact. Some other technique is needed.
 A: Consider first the local form of the Gauss-Bonnet theorem. The local Gauss-Bonnet theorem relies exclusively on intrinsic quantities: the metric tensor, the geodesic curvature, and angles on the surface. More exactly the external angles are summed up the $k$ vertices of the boundary $\partial R$. In order to extend the local formulation to the global one, we need to introduce the triangulation of a surface. The triangulation of a surface consists of a network of a finite number of regular curve segments on the surface such that any point on the surface either lies on one of the curves or lies in a region that is bounded by precisely three curve segments.
Now divide R into triangles $T_i$ such that each side of a triangle is the side of precisely one other triangle. In this way each edge in this subdivision is
the side of precisely two triangles. Note that all of this can be done without referring to the metric structure of the manifold. Parametrizing the boundary curves of the triangles to run counter clockwise shows that when two polygons have a side in common these sides are parametrized in opposite directions. The global Gauss-Bonnet theorem introduces the Euler characteristic given by:
\begin{equation}
\chi = F − E + V
\end{equation}
where F denotes the number of polygons in the triangulation, E the number of
edges (each edge is the common side of two polygons but only counted once), and
V the number of vertices (each vertex is met by any number of polygons, but only
counted once). Now the Gauss-Bonnet integral can be decomposed according to the triangular subdivision. By doing that is is possible to prove the global Gauss-Bonnet Theorem. You can for instance have a look to the Grant Rotskoff's note. Remember the the vertices V of planar curves, are the points  where the first derivative of curvature is zero.
