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Stokes Theorem Boundry

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data
    This is a question about stokes theorem in general, not about a specific problem.

    Directly from lecture:
    "If S has no boundry (eg. if S is the boundry of a solid region) then [tex]\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = 0[/tex] "

    because apparently "no boundry C exists"

    However in the next example he gives S is a hemisphere or radius 1 above the XY plane; and he uses the projection of the hemesphere on the XY plane as the bondry curve C and it works out fine.
    Isnt the hemisphere a boundry of a solid region? So shoudn't the integral be zero? Why can you use the boundry of the projection of S on the XY plane as the curve C?

    This is hard to explain in words, if someone is so kind as to lok into this, you can see exactly what I am seeing here:
    http://webcast.berkeley.edu/course_details_new.php?seriesid=2008-D-54472&semesterid=2008-D
    Lecture 42, 4:00-5:00 is where he says the integral is zero, then at 9:40 he does the example with the hemesphere and it is not zero.


    2. Relevant equations

    [tex]\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = \int_{C}\stackrel{\rightarrow}{F}\bullet d\vec{r}[/tex]

    Thanks anyone willing to look/help :)
     
  2. jcsd
  3. Dec 19, 2008 #2

    tiny-tim

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    the boundary of a boundary is zero

    Hi swraman! :smile:

    (I haven't watched the webcast lecture you linked to but:)

    The boundary of a boundary is zero (empty).

    The boundary of a solid hemisphere is a hemipherical shell plus the "base".

    In your example, the surface S is only the hemipherical shell.

    So S is not the boundary of a solid object, and S's boundary is therefore not zero. :wink:
     
  4. Dec 19, 2008 #3
    Re: the boundary of a boundary is zero

    Thanks for hte reply.

    Im still a little confused; so is the shell of a full sphere considered to be a "boundry"? But a hemisphere's shell isnt?
     
  5. Dec 19, 2008 #4

    tiny-tim

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    The surface surrounding any solid object is a boundary …

    a spherical shell does surround a sphere, so it is the boundary of the sphere …

    but a hemi-spherical shell does not surround a hemi-sphere: the hemi-sphere is surrounded by the hemi-spherical shell and the base :smile:
     
  6. Dec 19, 2008 #5
    so...if the surface can be expressed as a single function it is a boundry?
     
  7. Dec 19, 2008 #6

    tiny-tim

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    Nooo … if the surface divides space into two, ie if it surrounds a solid volume, then it is a boundary.

    This is even if it's in two parts (i assume that's what you would call "two functions"), like a hemispherical shell and a disc. :smile:
     
  8. Dec 19, 2008 #7
    doesnt the hemisphere shell surround a solid region?
     
  9. Dec 20, 2008 #8

    tiny-tim

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    i'm not sure we're talking about the same thing …

    a solid hemisphere is surrounded by a curved piece and a flat piece …

    i'm calling the curved piece the "hemisphere shell", and the flat piece the "base" or the "disc" …

    so the hemisphere shell on its own is not the boundary of a solid region, and so its boundary C is not the boundary of a boundary. :smile:
     
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