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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?
  2. jcsd
  3. Sep 29, 2009 #2


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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


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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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