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Stokes theorem

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    verify Stokes theorem for the given Surface and VECTOR FIELD
    x2 + y2+z2=4, z≤0 oriented by a downward normal.
    F=(2y-z)i+(x+y2-z)j+(4y-3x)k

    2. Relevant equations
    ∫∫S Δ χ F dS=∫ ∂SF.ds

    the triangle is supposed to be upside down.
    3. The attempt at a solution

    myΔχF = 3i +2j-1k
    my N i got 2Xi + 2Yj and since Z≤0
    I'm not so sure what to next
    ∫∫(6X+4Y)dxdy
    in the example in the book it said something on a similar exercise that by the symter og D and that fact that 6x and 4y are odd functions are odd functions we have the double integral = 0

    is it correct and permanant to go with that fact because I also got 0 on the right hand side. ?
     
  2. jcsd
  3. Nov 7, 2011 #2

    LCKurtz

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    What triangle?? Your curl(F) is correct. But you have the bottom half of a sphere, oriented downward. Your normal must have a negative z component, so it is incorrect. The natural way to do the surface integral would be to parameterize the surface in spherical coordinates in terms of [itex]\phi,\,\theta[/itex]:
    [tex]\vec R = \vec R(\phi,\theta)[/tex]
    and use the formula:
    [tex]\iint_S \nabla \times \vec F \cdot d\vec S = \pm \iint_{(\phi,\theta)}
    (\nabla \times \vec F) \cdot (\vec R_\phi \times \vec R_\theta )\, d\phi d\theta[/tex]
    where the sign is chosen to agree with the orientation.
     
  4. Nov 7, 2011 #3

    rude man

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    Strangely worded, but my interpretation is that the surface in question is a disc located anywhere from z = 0 to z = -2 since the volume is a hemisphere of radius 2 located below z = 0 and centered on (0,0,0).

    So (curl F)•dS is integrated across that circle, with the direction of dS = -j. Of course, there are an infinite no. of such circles ranging from radius R = 0 to R = 2, but the equation for the circles will all be the same in terms of x, y and z, so no problem there.

    Meanwhile, F•ds is integrated along that same circle.

    It would probably be beneficial to convert to spherical coordinates if you're comfortable with that.

    Keep the right-hand rule in mind for both integrations so you get the sign of the two sides to be the same.
     
  5. Nov 7, 2011 #4
    I've used sperical coordinates on the right side and I get 0 unless I did something wrong,


    but I'm more concerned about the left side
    I've changed it a bit but the question still remains, can I just say its = 0 becoz the surface is symmetrical and odd functions have double integral = 0? it makes sense because I got 0 on the right side, but I'm asking can I just do it like that?
     
  6. Nov 7, 2011 #5

    HallsofIvy

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    The wording of the problem is not strange, but you declaration that, having been specifically told that the surface is a hemisphere, you "interpret" it to be a disc is very strange!
     
  7. Nov 7, 2011 #6

    LCKurtz

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    Among other things, [itex]\vec F\cdot ds[/itex] doesn't make any since it is a vector dot a scalar.
     
  8. Nov 7, 2011 #7

    LCKurtz

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    I'm guessing the answer is no. You haven't shown your work. z is a function of x and y on the surface. What do you get for [itex]\nabla \times \vec F \cdot d\vec S[/itex]? You can work the problem by using x and y as the parameters if you wish:

    [tex]\vec R(x,y) =\langle x, y, -\sqrt{4-x^2-y^2}\rangle[/tex]

    and use Rx and Ry in a similar formula as I gave earlier. But if you do that you will most likely want to change the integral at least to cylindrical before you are done. But whether it is "obviously zero" depends on how the integral sets up, which you haven't shown us.
     
  9. Nov 7, 2011 #8

    rude man

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    Fellas, fellas. Get real.

    Don't you think you might take a closer look at what Stokes' theorem says. I quote from Thomas:

    "The circulation of vector F around the boundary C of an oriented surface S in the direction counterclockwise with respect to the surface's unit normal vector n equals the integral of (curl F)•n over S. ∫F•dr = ∫∫(curlF)•n dσ." (Thomas uses r instead of s and for an element of area (disk area!).


    So, question to Hallslofivy is, define your boundary per the above if your surface is the whole hemisphere. And why do you think the problem included "oriented by a downward normal"? The hemisphere is x2 + y2 + z2 = 4, z <= 0, so its orientation is totally defined. Answer: it defines the disks I described, their normal being along the z axis and their periphery where the hemisphere's surface intersects the x-y plane a t each value of z.

    And LC Kurtz - ds IS a vector, its direction being tangential to C. As Thomas states.
     
    Last edited: Nov 8, 2011
  10. Nov 7, 2011 #9

    LCKurtz

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    Sorry "rude man", but I have taught enough Calc III that I don't need to "take a closer look" at Stokes' theorem and I'm sure HallsofIvy doesn't either. Your interpretation of the problem is bizarre. But I will let Halls address your comments to him. As far as your notation is concerned, you are apparently confused about the difference between the two types of line integrals

    [tex]\int_C \vec F \cdot d\vec R\hbox{ and } \int_C f(x,y,z)\, ds[/tex]

    I don't have a copy of Thomas handy at the moment, but your own quote above that I highlighted in red shows that he uses [itex]\int \vec F \cdot d\vec r[/itex]. He doesn't use ds there where he means dr and I don't believe you that he does elsewhere.
     
  11. Nov 8, 2011 #10

    rude man

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    Stokes' theorem is a bedrock of electro-magnetism, with which I am intimately familiar, having dealt with the subject for 40 years. (It's a present from a fellow named Mike Faraday). So I know that whether I call a vector dx i + dy j + dz k = ds or dr don't make a whole lot of difference. I am sorry you don't seem to understand that I meant ds as an elemental vector.

    Just out of curiosity - may I inquire as to where you are/have been teaching?

    EDIT:
    Sorry I accused you of avoiding the question of what is C. I am waiting with bated breath for hallsofivy's response. But, come to think of it - would you care to answer it anyway?
     
    Last edited: Nov 8, 2011
  12. Nov 8, 2011 #11

    LCKurtz

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    C is the boundary of the hemisphere. It is the curve x2+y2=4 in the xy plane oriented clockwise as viewed from the positive z axis.

    When you use non-standard notation by using symbols that are generally used differently it causes confusion. ds is standardly used for arc length. Either dS or dσ are commonly used for the element of surface area. [itex]d\vec r[/itex] is sometimes written as idx + jdy + kdz. You will also see the relation between ds and [itex]d\vec r[/itex] written as [itex]d\vec r =\hat T ds[/itex] where [itex]\hat T = \frac{d\vec r}{ds}[/itex] is the unit tangent vector to the curve. It is best not to mix up the notation.

    You can click on my name if you wish to see my profile.
     
  13. Nov 8, 2011 #12

    LCKurtz

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    Commenting additionally on this post: Yes, you could replace the surface of integration with the disk in the xy plane with dS oriented by -j. That is a consequence of Stokes' theorem. But it is not what the OP was asked to do. He was asked to verify Stokes' theorem by working both sides of the equation, which requires integrating over the hemisphere itself for the surface integral. I have no idea what your comment about the infinite number of circular cross sections is supposed to be addressing. Care to explain?
     
    Last edited: Nov 8, 2011
  14. Nov 8, 2011 #13
    thanx I solved it doing it your by changing to spherical and got 0 on both sides, but the question still remains, is there a theory that says something about if you have an odd function and a symetric surface then the integral is automatically 0. Like no matter what function as long as it's odd and there is symmetry the integral is Zero, thanx for the help =) but I need to understand incase in an exam i get something similar.
     
  15. Nov 8, 2011 #14

    LCKurtz

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    Yes, there are cases where you can tell the integral is 0 because of symmetry as long as you are careful and know what you are doing. What you are talking about is discussed in

    http://en.wikipedia.org/wiki/Multiple_integral

    Scroll down to "Use of Symmetry" section.
     
  16. Nov 8, 2011 #15

    rude man

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    LC - I have learned a thing or two about Stokes since my last post. I admit to never having had to deal with a non-planar surface.

    I completely agree with your last post - and I have struggled a bit to rationalize whether or not it's acceptable to prove the theorem by doing the contour integral and then comparing that using a planar (degenerate, i.e. disc) surface with the -j normal, or whether one really has to perform the entire surface integration as you state. I will not argue with your conclusion, but I still think it's an interesting speculation.

    You state that C is x^2 + y^2 = 4, z = 0 . What has finally dawned on me is that ANY contour hugging the hemispherical surface is an acceptable contour, since the surface integral is an entity unto itself and so must by the theorem be independent of C as long as C is closed and always on the surface.

    So to "prove" the theorem rigorously, one would have to verify the contour integral for all possible contours, which sounds kinda tedious. So I am wondering: if it's acceptable to pick the most convenient contour, as you have done, is it not equally justifiable to pick the most convenient surface defined by that convenient contour? Like, what's sauce for the goose is sauce for the gander? Your comments appreciated.

    Oh yeah, my infinite number of circular cross-sections: you picked one cross-section, to wit, x^2 + y^2 = 4, z = 0. I extended that to all cross-sections x^2 + y^2 = R(z), --2 < z <= 0, that's all. All those contours are equally qualified to be used in "proving" the theorem. Ergo the infinite number of planar surfaces, aka discs.

    As to symbol convention - we physicists/engineers prefer to reseve r for radius, and s for arc length, which we have no qualms interpreting as a vecor or a length, depending.

    PS Your qualifications are hereby vetted, and I apologize for questioning them.
     
  17. Nov 8, 2011 #16

    LCKurtz

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    That simply isn't true. Stokes' theorem applies to an open surface and its boundary. There is only one boundary and for this hemisphere it is the curve C described above.
    There is only one relevant "contour", it being the boundary or "edge" of the given surface. The fact that for calculation purposes you can replace the given surface with another surface sharing the same boundary and orientation is a consequence of the validity of Stokes' theorem and wouldn't be used in the proof of the theorem.

    If you want to see a proof, there is an online pdf file giving a proof at:
    http://www.math.ncku.edu.tw/~rchen/Advanced Calculus/divergence theorem.pdf

    Interestingly enough in that argument you will see both s and [itex]\vec s[/itex] used (and carefully distinguished) with
    [tex]d\vec s = \frac{\partial \vec r}{\partial s}ds[/tex]
    which, in his setting is the equivalent of [itex]\hat Tds[/itex] I mentioned earlier.
     
  18. Nov 8, 2011 #17

    rude man

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    OK, I think I see it all now. A closed surface always has both sides of the Stokes equation = 0. The contour defines the open section of the surface.

    Still - the problem never specified that the plane part of the hemisphere was the open part. Couldn't it have been a pinhole near z = -2?

    Never mind.

    Sorry about my pigheadedness. Sometimes one learns by making a fool of oneself.
     
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