Let $T: [0,2] \times [0,2\pi] \to \mathbb{R}^3$ be defined by $T(r,\theta) = (\cos\theta, \sin\theta, r)$, which parametrizes a cylinder with radius $1$ and height $2$.
To find the boundary curve of this surface (or any other surface), a common approach is to find where each segment of the domain ended up by parametrizing each segment as $\sigma$ and then composing $T(\sigma)$. However, this method does not take into account the effect of "gluing" two edges together.
For example, if we take a rectangular ribbon and glue two opposite edges together to form a cylinder the new cylinder has two boundary curves instead of four like the original paper ribbon, the method i described earlier, would still count those two glued segments as a part of the boundary curve even though its not.
How can I fix the method in order to consider "glued" edges?
concrete example with the cylinder:
- $s_1(t)=(t,0), t\in[0,2]$
- $s_2(t)=(2,t), t\in[0,2\pi]$
- $s_3(t)=(2-t,2\pi),t\in[0,2]$
- $s_4(t)=(0,2\pi-t),t\in[0,2\pi]$
each $s_n$ segment corresponds to a side of the domain rectangle, or the "edges" of the rectangle, now if I compose $T(s_n)$ i should obtain the boundary curve of the surface T :
- $T(s_1)=(0,0,t), t\in[0,2]$,
- $T(s_2)=(cos(t),sin(t),0), t\in[0,2\pi]$,
- $T(s_3)=(0,0,2-t),t\in[0,2]$ and
- $T(s_4)=(cos(t),-sin(t),0),t\in[0,2\pi]$.
Both $T(s_2)$ and $T(s_4)$ belong in the boundary curve of T, though $T(s_1)$ and $T(s_3)$ do not.