How to find the boundary curve of a surface, like the Möbius strip? I feel like I am missing a key piece of intuition in trying to understand this. I have just recently started using Stoke's theorem and I struggle to see what the boundary curve of surfaces are. In some cases it is easy... like a hemisphere for example. But what about the boundary curve of the surface given below
Could someone explain what this boundary curve of the surface below and just some basics of how to find it. I have read and re-read my book but all  of the examples in there are things like a hemisphere.
Wait so i'm not so sure it is a torus anymore, it's been a long day, the surface is given by:
$$x=[1+u\cos (t)]\cos (2t), \quad y=[1+u\cos (t)]\sin (2t), \quad z=u\sin (t)$$
$$- \frac{1}{2} \le u \le \frac{1}{2} \quad ,  0 \le t \le \pi$$
Thanks :-)

This is a Möbius strip, following is what I drew in MATLAB based on your parametrization:

The boundary is obtained just by, first taking $u$ be $1/2$, for the angle part, imagine we start from one point on the boundary, we will get to the other side when $t$ changes from $0$ to $\pi$, then back to where we start after $t$ becomes $2\pi$. Therefore, the boundary curve should be:
$$x(t)=[1+\cos (t)/2]\cos (2t), \quad y(t)=[1+\cos (t)/2]\sin (2t), \quad z(t)=\sin (t)/2$$
$$0 \le t \le 2\pi$$
$\newcommand{\v}[1]{\boldsymbol{#1}}$
Also another way to extract the boundary curve would be to let $u$ be $1/2$ and $-1/2$, and let $t$ changes from $0$ to $\pi$, two ways are equivalent.
Now suppose we would like to apply Stokes theorem for a vector field $\v{F}$ on this surface:
$$
\int_{M} \nabla \times\v{F} \cdot \v{n} \,dS = \oint_{\gamma} \v{F} \cdot d\v{r}
$$ 
where $\v{r}(t) = \langle x(t),y(t),z(t)\rangle$, and the right hand side would be
$$
\oint_{\gamma} \v{F} \cdot d\v{r} = \int^{2\pi}_0 \v{F}(x(t),y(t),z(t))\cdot \v{r}'(t)\,dt
$$
where $\gamma$ is the directioned boundary curve parametrized above.

Attached the MATLAB code snippet you could try by yourself:
[u t] = meshgrid(-0.5:0.01:0.5,0:pi/100:pi);
x = (1+u.*cos(t)).*cos(2*t);
y = (1+u.*cos(t)).*sin(2*t);
z = u.*sin(t);
c = sqrt(x.^2+y.^2+z.^2);
surf(x,y,z,c,'facealpha',0.5,'edgealpha',0.3)

