I'm studying differential geometry using the book "Differential Geometry of Curves & Surfaces - Manfredo P. do Carmo", and the following doubt came to me.
Let $S\subset \mathbb{R}^3$ a surface, considere $x:U \rightarrow S$ be an isothermal parametrization at a point $p ∈ S$, and let $R ⊂x(U)$ be a simple region without vertices, containing $p$ in its interior.
Let $α: [0,l] →x(U)$ be a curve parametrized by arc length $s$ such that the trace of $α$ is the boundary of $R$. Let $w_0$ be a unit vector tangent to $S$ at $α(0)$ and let $w(s)$, $s ∈ [0,l]$, be the parallel transport of $w_0$ along $α$. Note that
\begin{align} 0 &= \int_{0}^{l} \left[\frac{Dw}{ds} \right]ds\\ &= \int_{0}^{l} \frac{1}{2\sqrt{EG}} \left\{G_u \frac{dv}{dt} - E_v \frac{du}{dt} \right\}ds + \int_{0}^{l} \frac{d\varphi}{ds}ds\\ &= - \int \int_R K d\sigma + \varphi(l) - \varphi(0) \end{align}
where $\varphi = \varphi(s)$ is a differentiable determination of the angle from $x_u$ to $w(s)$, $E = \langle x_u,x_u\rangle$, $G = \langle x_v,x_v\rangle$, $K$ is the Gaussian curvature and $\left[\frac{Dw}{dt}\right]$ is the algebraic value of the covariant derivative of $w$ at $t$.
It follows that $\varphi(l)−\varphi(0)$ = $\Delta \varphi$ is given by
$$\Delta \varphi = \int \int_R K dS. $$
I would like to know if $R\subset x(U)$ be a simple region WITH vertices the result still holds. If so how to prove it?