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Stokes' Theorem Problem

  1. Aug 17, 2015 #1
    1. The problem statement, all variables and given/known data

    Evaluate the line integral of F dot dr by evaluating the surface integral in Stokes' Theorem with an appropriate choice of S. Assume that C has a counterclockwise orientation.

    F
    = (y2, - z2, x); C is the circle r(t) = < 3cos(t), 4cos(t), 5sin(t) >


    2. Relevant equations
    F
    is the vector field.

    3. The attempt at a solution
    I found the curl of F to be <2z, -1, -2y> and I computed the line integral and got 15pi, which is the correct solution, but I have to use Stokes' Theorem to arrive at that solution. My main problem is I have no clue how to solve for the normal vector to dot with the curl of F. The main thing throwing me off is that C is already parameterized and I cannot see how to un-parameterize it to get the equation for the circle out.

    Thank you in advance.
     
  2. jcsd
  3. Aug 17, 2015 #2
    Well, first off, <3cos(t),4cos(t)5sin(t)> isnt a circle-its more elliptical. You dont need to "un-parameterize" it. You need to first find a normal vector to the plane of the ellipse. (which you should use as your "capping" surface)
     
  4. Aug 17, 2015 #3
    @DeldotB Yeah, I could not get my head around why the book was calling it a circle. Thank you though, I shall give it a go!
     
  5. Aug 17, 2015 #4

    Zondrina

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    So you wish to compute the line integral of the vector field using stokes theorem. Notice you have ##\vec r(t)## given as a function of ##t## only. Computing a surface integral will not be possible with the given ##\vec r(t)## as is.

    If instead the surface was given in two variables, for example ##\vec r(\theta, \phi)##, then you would be able to compute a vector normal to the surface ##\vec n##. Then you would be able to use a surface integral to compute the result.

    As it stands, you should evaluate the line integral using:

    $$\oint_C \vec F(x, y, z) \cdot d \vec r = \int_a^b \vec F( \vec r(t) ) \cdot \vec r'(t) \space dt$$

    This is Stokes' theorem, simply backwards:

    $$\iint_S \text{curl}(\vec F(x, y, z)) \cdot d \vec S = \oint_C \vec F(x, y, z) \cdot d \vec r = \int_a^b \vec F( \vec r(t) ) \cdot \vec r'(t) \space dt$$

    So you are technically using Stokes' theorem.
     
    Last edited: Aug 17, 2015
  6. Aug 19, 2015 #5

    LCKurtz

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    @BigFlorida: I do not agree with Zondrina that his argument "technically" uses Stokes' theorem. Here are some suggestions. You have ##\vec R(t) = \langle 3\cos t, 4\cos t, 5\sin t\rangle##, and you have calculated ##\nabla \times \vec F = \langle 2z, -1, -2y\rangle##. You can see right away from ##\vec R(t)## that ##4x=3y## so that tells you the curve is in that plane. Also, since your curl came out in terms of ##y## and ##z##, that suggests using them as parameters. Again, looking at ##\vec R(t)## you can see that ##\frac{y^2}{16} + \frac{z^2}{25} = 1##, so the projection of the curve on the ##yz## plane is an ellipse. You could parameterize the interior of that ellipse in the ##yz## plane in a polar coordinate like transformation $$y = 4r\cos\theta,~z = 5r\sin\theta,~0\le r \le 1,~0\le\theta\le 2\pi$$If you want to parameterize the portion of the plane inside the given curve, you just need to add the parameterization for ##x## which which you know is ##x = \frac 3 4 y = 3r\cos\theta##.

    Hopefully that helps you with the parameterization of your surface. Can you take it from there?
     
  7. Aug 19, 2015 #6

    Zondrina

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    It does 'technically' use the theorem, in another form. You can of course go the other way and use a surface, which your post has outlined.

    I just wanted to say you can't use ##\vec r(t)## as is, or else you could not compute the surface integral.
     
  8. Aug 19, 2015 #7

    LCKurtz

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    Quoting from the original post: "Evaluate the line integral of F dot dr by evaluating the surface integral in Stokes' Theorem with an appropriate choice of S". Your post is irrelevant to that problem.
     
  9. Aug 20, 2015 #8
    @LCKurtz Wow, thank you so much. That makes complete sense, I did not even think of projecting it onto the yz-plane. I can definitely take it from there. Thank you again!
     
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