# 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 :-)

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The boundary of a surface does not have to consist of a single curve. When you slice the torus vertically (through the center), there are two curves that make up the boundary. Picture a pair of pants. There are three curves that make up the boundary. Do you know how they should be oriented? – Thom Tyrrell Nov 15 '12 at 21:19
is it like with green's theorem where we can talk about a disk with a circular hole in its interior by going clockwise around the circle in the middle and counter clockwise on the outside? And so for the torus as long as we go in opposite directions around each curve we will have the right orientation – Ben Nov 15 '12 at 21:23
I'm not sure that the orientations on the half-torus need to be opposite. How would this work for a pair of pants? – Thom Tyrrell Nov 15 '12 at 21:29
awww ok sorry i changed the question a few times thought we were talking about a full torus. so it would just be in the same direction. so that whilst you travel around each circle the rest of the surface is to the left? – Ben Nov 15 '12 at 21:32
sorry to be a pain, what do you mean to the right? – Ben Nov 15 '12 at 21:45

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)

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Thank you very much :-) – Ben Nov 16 '12 at 11:11
@Ben You are welcome, if you are satisfied with my answer, would you kindly vote it up and accept it? – Shuhao Cao Nov 16 '12 at 16:33