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A line integral

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let S be the spiral surface parametrized by {(x,y,z) | x=ucosv, y=usinv, z=v} where 0<u<1 and 0<v<Pi/2. Let F(x,y,z) = (z,x,y) and compute [tex]\oint F \cdot t dS[/tex].

    2. Relevant equations



    3. The attempt at a solution

    Let me explain where I'm not certain. I can compute the tangent vector t quite easily given the parametrization of S (using the chain rule) and parametrize F with x=ucosv, y=usinv, and z=v. Then I can put it all into the dot product, get a nice scalar equation, and do a surface integral over u and v - not bad. But is that the best way to approach it? Here's what I'm wondering: if I look at the spiral surface, it's not simply connected (right?) so I can't use Stoke's theorem. At least, that's what I think - but I seem to be having a hard time getting the simply connected concept into my head. Am I right (that the region isn't simply connected)? If I drew a curve around the spiral and tried to shrink it down, I'd have trouble when I got to the center of the spiral - so I think it's not simply connected...

    Basically, I'm looking for the most efficient way of solving the problem. But I'm not sure if that's my approach or not. Any comments?
     
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  3. Sep 29, 2009 #2

    LCKurtz

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    I am confused about your notation. The circuit integral suggests a line integral and yet you seem to be discussing a surface integral. What is the tangent vector t tangent to? And, given a circuit integral, what is the path? Given a surface, normally dS would refer to an area element. Is it to represent arc length, in which case you might mean t ds and the whole thing a line integral usually written as:

    [tex] \oint \vec F \cdot d\vec{R}[/tex]

    And to answer your other question, that portion of the surface certainly is simply connected, in fact the whole surface is.
     
  4. Sep 29, 2009 #3
    Oh, you're right - I don't know what I was thinking. I guess it's not even clear to me how to evaluate this as a line integral unless I can set either u or v constant. Now I am more confused.

    But I guess if it's simply connected (I think I'm beginning to see how it COULD be, thank goodness) I could use Stokes' theorem and instead do the integral [tex]\int\int \nablaxFdS[/tex] instead.
     
  5. Sep 29, 2009 #4
    oops - the Tex part didn't show up. I meant that instead I could do the surface integral of curl(F) instead of the line integral F dot t.
     
  6. Sep 29, 2009 #5

    LCKurtz

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    Well, if the line integral is around the boundary of that surface, Stokes seems like a good idea. You do have a nice paramaterization to work with.
     
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