I think I have proved the following version of Stokes' theorem:
Teorem 1: Let $\beta: [0,4] \to \mathbb{R}^2$ be the curve given by \begin{equation} \beta(t) = \begin{cases} (t,0) & \mbox{if } t\in[0,1], \\ (1,t-1) & \mbox{if } t\in[1,2], \\ (3-t,1) & \mbox{if } t\in[2,3], \\ (0,4-t) & \mbox{if } t\in[3,4]. \end{cases} \end{equation} This is the boundary of the square $D = [0,1] \times [0,1]$, oriented counterclockwise with speed 1. Suppose $\sigma: D \to \mathbb{R}^3$ is extensible to a function of class $C^2$ on an open set $A \subseteq \mathbb{R}^2$ containing $D$. If $\vec{F}: \mathbb{R}^3 \to \mathbb{R}^3$ is of class $C^1$ and $\alpha = \sigma \circ \beta$, then \begin{equation}\label{E:Stokes} \int_{\alpha} \vec{F} \cdot d\vec{r} = \iint_\sigma \mathrm{curl}\,\vec{F} \cdot d\vec{S}. \end{equation}
Apparently, I didn't need the usual hypothesis that $\sigma$ is a regular (or smooth) surface, in a sense that $$ \left\lVert \frac{\partial \sigma}{\partial u}(u,v) \times \frac{\partial \sigma}{\partial v}(u,v) \right\rVert \neq 0$$ for all $(u,v) \in D$. Is it in fact possible? Or must I have made some mistake?