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Stokes' Theorem Problem

  • Thread starter BigFlorida
  • Start date
  • #1
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Homework Statement



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) >


Homework Equations


F[/B] is the vector field.

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.
 

Answers and Replies

  • #2
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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)
 
  • #3
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@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!
 
  • #4
STEMucator
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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:
  • #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?
 
  • #6
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@BigFlorida: I do not agree with Zondrina that his argument "technically" uses Stokes' theorem.
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.
 
  • #7
LCKurtz
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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.
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.
 
  • #8
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@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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