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What is the value of the line integral

  1. Apr 23, 2006 #1
    Suppose F=F(x,y,z) is a gradient field with [itex]F=\nabla f[/itex], S is a level surface of f, and C is a curve on S. What is the value of the line integral [itex]\int_C F dr[/itex]?

    I know the answer is 0 but I cannot visualize why it would be zero?
  2. jcsd
  3. Apr 23, 2006 #2
    I can't remember what a level surface is right now, so please remind me what it is? I seem to recall it is some function set equal to a constant. But honestly don't remember.

    The fundemental theorem of line integrals is:
    [tex] \int_C \nabla f \cdot d\vec r = f(\vec r(b)) - f(\vec r(a)) [/tex]

    Since you said [itex] \vec F [/itex] is a vector field with [itex] \vec F = \nabla f [/itex] you know that the curl is 0. Right?
    So since the curl is 0, and a potential function exists (which is f(x,y,z)=f).

    I have no idea if this will help, because I really don't remember what a level surface is. I could speculate, or just look it up...
  4. Apr 23, 2006 #3
    Ok. I looked it up. So level surface would be if you set f(x,y,z) equal to some constant. Now, you have a curve that lies on this surface.

    Ummm... is that the full question?
    This kinda seems harder then I originally thought it was going to be. Might need to use Stoke's thm, or wait for someone else to give you some better advice :) Cause I don't know why it would be zero either. (but shoot, that doesn't mean anything)

    *sorry dude.

    Personally, I would make up some examples and test them out to get a better understanding.
  5. Apr 24, 2006 #4


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    Try Stoke's theorem.

    Hint: curl of the gradient ?
  6. Apr 24, 2006 #5


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    Stoke's works nicely. Actually, in this simple case, it's just the "fundamental theorem of calculus": since F= grad f, the integral of F on any path is independent of the path- it's just f evaluated at the endpoints. But this is a closed path. We can take the two endpoints to be the same.
  7. Apr 24, 2006 #6
    Just out of curiosity. How do you know it is a closed path? That greatly simplfies it. Let's just say it is not... then in general, it would not be zero right?
  8. Apr 24, 2006 #7


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    This automatically shows that the result is zero because the curve C is on a level surface of the potential function (as stated in theoriginal post). That means that all the points on that surface have the same value of "f". therefore the end and starting points of the curve have the same value of f therefore the integral is zero.
  9. Apr 24, 2006 #8
    Ahh yeah. That makes sense.

    When I originally read the problem I was thinking that the fundemental theorem would easily show that it is zero... But, then I wasn't sure if it was a closed path or not. Didn't see how to get to that step from the question. Got it now.
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