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Stokes' Theorem parameterization

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    HUT71do.png

    2. Relevant equations

    eq0002MP.gif

    3. The attempt at a solution
    I only know that they gave the parameterization of the circle: r(t) = <cost, sint, 2>.
    My problem is, did they already give the curl of F in the line integral? I don't understand why dx, dy, and dz are separated like that.
     
  2. jcsd
  3. Apr 23, 2016 #2

    vela

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    If ##\vec{F} = F_x\,\hat{i}+F_y\,\hat{j}+F_z\,\hat{k}## and ##d\vec{r} = dx\,\hat{i} + dy \,\hat{j} + dz\,\hat{k}##, what's ##\vec{F}\cdot d\vec{r}## equal to?
     
  4. Apr 23, 2016 #3
    It would equal Fxdx + Fydy + Fzdz, but wouldn't it be simplified to just F?
     
  5. Apr 23, 2016 #4

    vela

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    What do you mean "simplified to just F"?
     
  6. Apr 23, 2016 #5
    For example, (F/dx)(dx), so the dx would cancel out. That happens to each. Actually, wouldn't it be 3F?
     
  7. Apr 23, 2016 #6

    vela

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    ##\vec{F}## is a vector-valued function. ##F_x## is the x-component of ##\vec{F}##, not the derivative of ##F## with respect to ##x##. Besides, the expression ##F/dx## is meaningless.
     
  8. Apr 23, 2016 #7
    Oh right, so do I just take the derivative of the parameterization which will give me values equal to dx, dy, dz and plug that into the original integral? Do I also plug in the parameterization into F?
     
  9. Apr 23, 2016 #8

    vela

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    Sounds like a plan.
     
  10. Apr 23, 2016 #9
    How would you go about simplifying the integral though? Some parts are fine but how about e^(-cos^2(t)/2), for example?
     
  11. Apr 23, 2016 #10

    vela

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    You don't. The problem is asking you to evaluate the integral using Stoke's theorem.
     
  12. Apr 24, 2016 #11
    So is the answer just:
    t=0 to t=2pi (-e-cos2t/2sint+2sin2t+e-sin2t/2cost+4cos2t)dt
     
  13. Apr 24, 2016 #12

    LCKurtz

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    No. That isn't using Stoke's theorem. The other side of Stoke's theorem is$$
    \iint_S \nabla \times \vec F \cdot d\vec S$$Do that.
     
  14. Apr 24, 2016 #13
    So do I find the curl of F first, plug in the parameterization into the curl, then multiply it with the normal vector using dot product?
    I'm having a hard time relating these things I think.
     
  15. Apr 24, 2016 #14

    LCKurtz

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    Yes

    Not sure what you mean by "the" parameterization. You are given the parametrization of a curve enclosing an area. You are going to need to parameterize that area and use that. You will need two parameters for a surface.

    You will have to show us what you do before we can tell if you are doing it correctly.
     
  16. Apr 24, 2016 #15
    So far, I found the curl to be <-x, -y, 2z+2>. Do I find the normal vector by using the circle's parameterization?
     
  17. Apr 24, 2016 #16

    LCKurtz

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    Let's call that vector you got ##\vec V = \langle -x,-y,2z+2\rangle##, which looks correct for the curl. So now you need to integrate ##\vec V\cdot d\vec S## over the described area up in the ##z=2## plane. Your next issue is to figure out ##d\vec S##. How do you find the area vector? If you look at the geometry for this problem, it should be easy to figure out a unit normal vector and an area element. You can worry about the integral after you figure out the integrand.
     
  18. Apr 24, 2016 #17
    I think what I am confused about is do you use F or the circle to find the unit normal vector?
     
  19. Apr 24, 2016 #18

    LCKurtz

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    You are given a curve which encloses a surface. Surfaces have normal vectors. It doesn't have anything to do with the vector ##\vec F##. I suggest you draw a picture of your surface.
     
  20. Apr 24, 2016 #19
    Is the unit normal vector just <cost, sint, 2>?
     
  21. Apr 25, 2016 #20

    LCKurtz

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    No. That isn't even a unit vector. It is a parametric equation of your circular boundary. Surely your text tells you how to calculate a normal vector to a surface. Or you could just draw a picture of the surface for this problem and look at it.
     
    Last edited: Apr 25, 2016
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