Let $\mathbf{r}:(a,b)\times (0,1)\to\mathbb{R}^2\subseteq\mathbb{R}^3$ be a **injective** application, given by:


$$\mathbf{r}(u,v)=A(u)+v\cdot (B(u)-A(u)), \forall\ (u,v)\in (a,b)\times (0,1)$$


where $A,B:(a,b)\to\mathbb{R}^2$ are two functions of class $C^1((a,b))$.


Is it true that $\dfrac{\partial\mathbf{r}}{\partial u} (u\_0,v\_0)\times\dfrac{\partial\mathbf{r}}{\partial v} (u\_0,v\_0)\neq (0,0), \ \forall\ (u\_0,v\_0)\in (a,b)\times (0,1)$?


All I want is to show that the set $\mathbf{r}((a,b)\times (0,1))$ is a regular surface, so that I can use the well-known formula to compute its area. I could not find a couterexample of the statement I conjectured above.


I have asked some professors of differential geometry at my faculty, but I haven't got any answers. I'm interested in finding its area, because it gives a formula for the area between the tracks of a bycicle. You can also see this: [Old Peano theorem (demonstration is missing details)](https://mathoverflow.net/questions/203079/old-peano-theorem-demonstration-is-missing-details)


Do you know any way of finding the area of this set that avoid showing that it is a regular surface?



Area of a plane surface that gives a lot of theoretical problems
You don't need the map $r$ to be differentiable and the jacobian to be non-zero to compute the area of the image r((a,b)×(0,1)). It suffices that $r$ is injective Lipschitz. This guarantees that the Jacobian is defined almost everywhere and the usual formula holds. So even if r((a,b)×(0,1)) is not regular, you are ok. See e.g. the following [page](http://www.encyclopediaofmath.org/index.php/Area_formula) under the heading "volume of submanifolds".


For more information, search for geometric measure theory. 


