Can I apply the divergence theorem to compute the flux of the curl of this vector field?

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Homework Statement:

Let Σ = [(x, y, z) ∈ R3 : z=-1/4+√(x^2+y^2) 0 <= z<=1/2]
be a surface oriented
so that the normal versor of Σ forms an obtuse angle with the fundamental versor of the z–axis.
Compute the flux of the curl of the vector field
F(x, y, z) = 2yz *exponontiel(x^2+z^2),3z*exponontiel(x^2+y^2),(0,5x+z)exponontiel(x^2+y^2)

Relevant Equations:

Divergence theorem
Good day all
my question is the following

Is it correct to (after calculation the new field which is the curl of the old one)to use the divergence theroem on the volume shown on that picture?
1598617194382.png

The divergence theorem should be applied on a closed surface , can I consider this as closed?
Thanks a lot!
 

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  • #3
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Homework Statement:: Let Σ = [(x, y, z) ∈ R3 : z=-1/4+√(x^2+y^2) 0 <= z<=1/2]
be a surface oriented
so that the normal versor of Σ forms an obtuse angle with the fundamental versor of the z–axis.
Compute the flux of the curl of the vector field
F(x, y, z) = 2yz *exponontiel(x^2+z^2),3z*exponontiel(x^2+y^2),(0,5x+z)exponontiel(x^2+y^2)
Relevant Equations:: Divergence theorem

Good day all
my question is the following

Is it correct to (after calculation the new field which is the curl of the old one)to use the divergence theroem on the volume shown on that picture?

The divergence theorem should be applied on a closed surface , can I consider this as closed?
Thanks a lot!
I might be reading the problem incorrectly, but when I read flux of the curl, that suggests an integral of the form:
[tex] \iint_S (\nabla \times \vec F ) \cdot \vec{dS} [/tex]

I am not fully convinced that the surface illustrated is closed as the tops are open? (i.e. am I correct in thinking that the surface does not span across that?) The illustrations (namely the paths) seem to suggest the use of Stokes' theorem to me, but perhaps I am mis-interpreting the photo.

If you were to use Gauss' theorem, then we would have [itex] \nabla \cdot (\nabla \times \vec F ) [/itex], which is 0 for any vector field [itex] \vec F [/itex]. This would suggest the answer would be 0, but I'm not sure that is what the problem is looking for.

I will keep reading to see whether I can make sense of the illustrations.
 
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  • #4
benorin
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##\Gamma _1, \text{ and } \Gamma _2## are the top and bottom and they're closed evidently
 
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  • #5
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Thanks a lot for you answer this is exactely what i'm asking for
the exercice describe a cone, and there is then they asked us to calculate the curl of the field over a specific region, my question is : Can I suppose that region closed between (z=0 and z=sqrt(x^2+y^2)-1/4 ) to see it a locked volume even though materially the initial solid is not ( we have a cone)?
 
  • #6
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let me rephrase my question
we can use the divergence theorem in a sphere because it's geometrically closed
can we use the divergence on a topless and botomless cylinder? ( just suppose the exsitance of a virtual walls?
 
  • #7
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let me rephrase my question
we can use the divergence theorem in a sphere because it's geometrically closed
Yes

can we use the divergence on a topless and botomless cylinder? ( just suppose the exsitance of a virtual walls?
No, I don't think so. Including the extra surfaces would add extra contributions into your surface integral.
 
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  • #8
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thanks a lot !!!
 
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  • #9
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thanks a lot !!!
No problem.

However, there may be tricks involved by using extra surfaces and then doing other integrals to get rid of the extra contributions. Not sure whether that will be easier or not here. For example, if were to include those extra surfaces (top and bottom) and then evaluate the total surface integral (perhaps with gauss theorem) to get some result [itex] R_{total} [/itex]. Then we need to remove the contributions of the bottom and top due to 'outward' pointing vectors. Then we would have two more surface integrals (top and bottom) which we need to evaluate (perhaps with Stokes' Theorem) to yield [itex] R_{bottom} [/itex] and [itex] R_{top} [/itex]. Then we could do: [tex] R_{final} = R_{total} - R_{top} - R_{bottom} [/tex]
That should yield the correct result, but am not sure whether that would shorten or elongate the working time...

Hope that makes some sense. If not, am happy to explain further, but it was only another (potentially longer) method of reaching the solution.
 
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  • #10
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thanks a million it does really help
 
  • #11
etotheipi
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Yes a closed surface is a surface with no boundaries. The surface in your problem has two boundaries, so it is not closed. But you can still use @Master1022's approach of considering a closed surface consisting of the given surface joined to two other surfaces with boundaries ##\Gamma_1## and ##\Gamma_2## respectively, and then subtracting off the flux through these two surfaces :smile:
 
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  • #12
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Thanks a lot :)
 
  • #13
pasmith
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There seems to be some confusion here.

You are asked to compute [tex]
\int_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S}.[/tex] That can be done using Stokes's Theoem: [tex]
\int_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S} = \oint_{\partial \Sigma} \mathbf{F} \cdot d\mathbf{x}.[/tex] Here [itex]\partial \Sigma[/itex] consists of two disjoint curves, oriented as shown in the diagram.
 
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  • #14
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yes true , thanks a lot, but I was also trying to see if I can use the divergence theorem with the new field (the curl of the old one) and the volume would be the closed volume shown on the picture
 
  • #15
benorin
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Go For It!
 
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