Prove that the Möbius strip is not oriented but its boundary is closed and positive oriented.
Solution:
We kick things off by giving the visualization of Möbius strip.
Now by picking an arbitrary vector on the surface and rotating it wee see that the vector after it completes its rotation will return back to its initial position; only this time it will look at the opposite direction. Hence the Möbius strip is not oriented.
A parametrization of this curve is given by:
$$x = \left [R + s \cos \frac{t}{2} \right] \cos t, \;\; y = \left [R + s \cos \frac{t}{2} \right] \sin t, \;\; z = s \sin \frac{t}{2}$$
Under that parametrization the boundary which is (clearly) closed is positively oriented.
Solution:
We kick things off by giving the visualization of Möbius strip.
Now by picking an arbitrary vector on the surface and rotating it wee see that the vector after it completes its rotation will return back to its initial position; only this time it will look at the opposite direction. Hence the Möbius strip is not oriented.
A parametrization of this curve is given by:
$$x = \left [R + s \cos \frac{t}{2} \right] \cos t, \;\; y = \left [R + s \cos \frac{t}{2} \right] \sin t, \;\; z = s \sin \frac{t}{2}$$
Under that parametrization the boundary which is (clearly) closed is positively oriented.
P.S: For $s=0$ Möbius strip is transformed into a circle (fair and square)
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