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I Evaluation of a line integral

  1. Sep 14, 2017 #1
    calculate the line integral for a vector field F= -xy⋅j over a circle which is c(t)=costi+sintj,
    so I used x=cost y=sint and ∫(0 to 2pi) -(sintcost)(cost)dt=(cos^3(2pi)-cos^3(o))/3=0 now here is the problem, if this enclosed line integral is zero then why is the vector field not conservative????
     
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  3. Sep 14, 2017 #2
    dr(t)/dt=(-sint,cost) then dot with F gets -(sintcos^2t)
     
  4. Sep 14, 2017 #3
    cant find anything wrong with my calculation process
    and the textbook tells me that if enclosed line integral is zero then vector field must be conservative which means there is something with my work but I checked my work so many times and I just can't find it
     
  5. Sep 14, 2017 #4

    vanhees71

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    Why should it be conservative, if just the integral along one closed line vanishes? It's definitely not conservative, not even locally, because ##\vec{\nabla} \times \vec{F} \neq 0##.
     
  6. Sep 14, 2017 #5
    Thanks in advance for anyone who takes time to read it and would like to help me with the problem
     
  7. Sep 14, 2017 #6
    right, but the enclosed line integral is zero, which means either I am wrong with the integral calculation, or the vector field is conservative
     
  8. Sep 14, 2017 #7
    but I just can't find where I calculated wrong
     
  9. Sep 14, 2017 #8

    vanhees71

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    The line integral vanishes, but obviously not all line integrals. Try
    $$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j} $$
    instead.
     
  10. Sep 14, 2017 #9
    what do you mean it vanishes? is my line integral zero?
     
  11. Sep 14, 2017 #10
    but this doesn't describe the circle, does it?
     
  12. Sep 14, 2017 #11

    vanhees71

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  13. Sep 14, 2017 #12
  14. Sep 14, 2017 #13

    vanhees71

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    It describes a unit circle with its center in ##(0,1,0)##. You'll see that it's a closed line for which the line integral does not vanish, and that explains, why the field is not conservative. For a field to be conservative (in some region) all line integrals along any closed line must vanish (in this region).
     
  15. Sep 14, 2017 #14
    right, but I was calculating the line integral of a circle centered at (0,0,0)
     
  16. Sep 14, 2017 #15

    vanhees71

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    Yes, your line integral vanishes, as you have correctly evaluated yourself. If you try the other line integral instead, you'll see this one doesn't vanish, and thus the field cannot be conservative!
     
  17. Sep 14, 2017 #16

    vanhees71

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    Please think about what I try to explain with my answers!
     
  18. Sep 14, 2017 #17
    and if The region I calculated is a closed curve and it is zero, then it should be conservative, right?
     
  19. Sep 14, 2017 #18

    vanhees71

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    NO!!! The field is conservative in a region, if the line integrals along ALL closed lines in that region vanish. It's not sufficient that one special line integral along a special closed curve vanishes!
     
  20. Sep 14, 2017 #19
    yes of course, the integral I calculated from your region is definitely NOT zero. But the region I calculated IS zero, and the region is still closed curve. But the textbook tells me that if the Line integral of an enclosed curve is zero, then the vector field must be conservative.
     
  21. Sep 14, 2017 #20

    vanhees71

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    Which textbook? If it really claims what you say, find a better one!

    Just to give you another example for something really puzzling! Consider the scalar field
    $$\phi(\vec{x})=\arctan(y/x).$$

    (a) calculate the gradient field ##\vec{V}=-\vec{\nabla} \phi##.
    (b) Integrate this vector field along the unit circle ##\vec{x}(t)=\cos t \vec{i} + \sin t \vec{j}##.
    (c) Explain the (hopefully) surprising result!
     
    Last edited: Sep 14, 2017
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