Why Do Orientations Affect Stokes' Theorem Results?

[mayaitagaki]

Hi everyone,

I have to prove this problem but I have no idea how to approach this problem. I tried something but it seems not working...

Suppose F is a vector field in R3 whose components have continuous partial derivatives. (So F satisfies the hypotheses of Stoke's Theorem.)

(a) Explain why any two surfaces S and S' oriented upward and with boundary the
unit circle x^2 + y^2 = 1 satisfy

Int Int_s (curl F dot d sigma) = Int Int_s' (curl F dot d sigma)

(b) Explain why any two surfaces S and S'', one oriented upward, the other oriented
downward, and both with boundary the unit circle x^2 + y^2 = 1 satisfy

Int Int_s (curl F dot d sigma) = - Int Int_s'' (curl F dot d sigma)

Since I don't know how to insert Greek characters, I attach 2 files here; one is the problem and another one is my attempt.

Thank you for help!

 

[rhoparkour]

Let me give you a hint. Stoke's theorem relates the surface integration of the curl of F with the boundary integration of JUST F, taking into account orientation (you should be able to answer how you do this, the trick is in relating the normal to the surface to a direction in boundary integration).

Both questions are essentially the same (since the value of these integrals depend, because of Stoke's theorem) only on the value of F on the boundary of the surface.

Please reply if this was helpful. I'd be happy to write a more detailed answer if you find yourself stuck.

 

[mayaitagaki]

Thank you for your fast reply,

I think I need a more detailed answer. When you check my attempt would you tell me where I should modify so that I can go further?

Thanks,
Maya

 

[rhoparkour]

Ok, here goes, I hope my LATEX is good enough.

Let me describe the situation we have here. We have a vector field F of class C1 defined on the whole of R3, two surfaces S and S' sharing the same boundary B (namely, the circle). For the sake of clarity, Stokes' theorem says:

$\iint_{S} \left( \nabla \times \mathbf{F} \right) \cdot \mathbf{n} \, dS = \int_\textrm{B} \mathbf{F} \cdot \mathbf{t} \, dl$

Where n is the unit normal of the surface and t is the unit tangent of the boundary of the surface.
n is chosen arbitrarily, but t must respect the orientation imposed by choosing n; t must obey the right hand rule: if n is in the direction of your thumb, t must point in the direction of your remaining fingers.

Alternately, we could arbitrarily choose t and get the direction of n respecting the right hand rule.

We have assumed implicitly that the tangent vector of the circle has been chosen.

a) Since the surfaces share boundary and have the same orientation, they are equal. Their normals are parallel, so are their tangents to their boundaries.

For the first surface,

$\iint_{S} \left( \nabla \times \mathbf{F} \right) \cdot \mathbf{n} \, dS = \int_\textrm{B} \mathbf{F} \cdot \mathbf{t} \, dl$

for the other one,

$\iint_{S^{'}} \left( \nabla \times \mathbf{F} \right) \cdot \mathbf{n} \, dS = \int_\textrm{B} \mathbf{F} \cdot \mathbf{t} \, dl$

b) Since the surfaces share boundary but have opposing orientation they must be equal to each other's negatives. Their orientation is opposite because they have opposing normals and the same tangents to their boundary.

As for your attempt, there is no need to write these integrals in any sort of coordinates, the question posed here is a matter of a subtlety that arises in Stoke's theorem: orientation. Basically, the n and t used in Stokes' theorem must express the same orientation of the surface.

I hope this was helpful, if you have any questions just ask and be happy.

*I'm not getting this latex feature, I may not be implementing it right, I'll try to fix it.
**Got it fixed.

 

[mayaitagaki]

thank you very much for your detailed answer

Now my question has got solved!

I really appreciate it!

Maya