I was reading this note about Surface Integrals and came across this paragraph:
Let S be a surface parameterized by $\mathbf X : D → \mathbf R^3$. A point $(s_0, t_0) \in D$, is mapped to $\mathbf X(s_0, t_0) \in\mathbb R^3$. An infinitesimal $dx × dt$ parallelogram at $(s_0, t_0) \in D$ has area $dx dt$. It’s mapped to an infinitesimal $\mathbf T_s(s_0, t_0)ds × \mathbf T_t(s_0, t_0)$ rectangle with area $||\mathbf T_s × \mathbf T_t|| ds dt$, which equals $\|\mathbf N\| ds dt$. We’ll call this infinitesimal parallelogram the surface area differential, denoted $dS$.
I'm quite new to the topic, and I kind of got confused here. When the note says "An infinitesimal $dx \times dt$ parallelogram... has area $dx dt$", are the $dx$'s supposed to be $ds$'s? Otherwise, if $dx$ is right, could someone help me understand why we use $dx$?
I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds \times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds \times T_t(s_0, t_0)dt$.