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I'm studying Stokes'theorem. My textbook as an example with the surface $S = \{(x,y,z) \in \mathbb{R^3}: x=1-y^2-z^2; z>0;x>0\}$ with orientability given by the norm n with first component positive.

with boundary described by the lines: $\Gamma_1= \{(x,y,z) \in \mathbb{R^3}: x=1-y^2; z=0;x>0\}$ and $\Gamma_2= \{(x,y,z) \in \mathbb{R^3}: y^2+z^2=1; z>0;x=0\}$

My textbook simply says immediately that the orientation of the lines in the boundary is the one in the following image because it needs to match S (with orientation given by the norm I mentioned).

enter image description here

How can you know that? Does anyone has any tip?

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    You've got two good answers, but as a practical matter, a connected curve has precisely two orientations (directions of traversal). If you're stuck in a particular example, just pick one arbitrarily and see if it's compatible with the choice of surface normal. If it's not, the other orientation must be correct (though it's a good idea to check positively, as well). – 2017-01-05

3 Answers 3

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Do you know the "right hand rule"?

Make a fist with your right hand.

Then stick your thumb up.

Now, with your right hand in that configuration, align it with the picture so that your thumb points in the same direction as $v$, and so that the curl of your fisted fingers aligns tangentially with the boundary of the surface.

The direction of your fingers will then align with the required orientation of the boundary.

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Look at the surface $S$ from the tip of the chosen normal. The boundary $\partial S$ then has to go counterclockwise around $S$, as indicated in the figure.

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You already got two good answers, but here's one more: if you're walking on $\partial S$ and the normal field points outward, then $S$ must be on your left.