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Homework Help: Stokes Theorem

  1. Aug 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Use Stokes Theorem to compute
    [tex]\int_{L}^{} y dx + z dy + x dx[/tex]
    where L is the circle x2 + y2 + z2 = a2, x + y + z = 0

    3. The attempt at a solution
    I feel like this problem shouldn't be that hard but I can't get the right answer: (pi)a2/3.

    I calculated the curl of F as: -(i + j + k)
    and the normal vector as:
    [tex]\frac{i + j + k}{\sqrt{3}}[/tex]

    [tex]\int_{L}^{} y dx + z dy + x dx = \int \int -(i + j + k) \cdot (\frac{i + j + k}{\sqrt{3}}) ds = -\frac{3}{\sqrt{3}} \int \int ds[/tex]

    Here's where I'm stuck. I think the integral should just be the area of the circle (pi*a2) but maybe I'm thinking about it wrong. Thanks.
  2. jcsd
  3. Aug 14, 2008 #2


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    Homework Helper

    Is this integral supposed to be

    \int_{L}^{} y dx + z dy + x dz
    [/tex] ?

    [last variable of integration being z, not x?]

    I think you'll find that, although the plane goes through the origin, it does not intersect the sphere at an equator, so the area of the circle is not [tex]\pi \cdot a^2[/tex].

    EDIT: Here's a way to see that. Write out the equation
    x + y + z = 0 using spherical coordinates. You can divide through by r, since we know that isn't zero on the circle of intersection. You now have

    [tex]sin \theta cos \phi + sin \theta sin \phi + cos \theta = 0 [/tex]

    Make a convenient choice of the azimuthal angle [tex]\phi[/tex], say, 0º. You now have an equation in [tex]\theta[/tex] only, and can see immediately that the polar angle (or co-latitude) can't be 90º, so the intersection circle is not an equator of the sphere. Where must it be and what is the circle's area there?
    Last edited: Aug 14, 2008
  4. Aug 15, 2008 #3
    Yes, thanks.

    I'm still confused, though. I thought that if a plane cut through the center of a sphere then it was a great circle, hence its area was a quarter of the surface area of the sphere.
  5. Aug 15, 2008 #4
    You're right linearfish. Because the sphere's center is at the origin and the plane intersects the origin, the intersection must, necessarily, be a great circle.

    EDIT: Of course, it is not the "equator" of the sphere (in the xy-plane).
  6. Aug 15, 2008 #5
    Okay, I get that. So is my calculation off somewhere else?
  7. Aug 15, 2008 #6
    Consider: what is the surface? Is it the interior of the circle? Or is it the half-sphere? Both surfaces have the same boundary.
  8. Aug 15, 2008 #7


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    Sorry, yes, I really should have checked my development of that thought with more points on the sphere. So ignore post #2 as the ravings of a sleep-deprived person...

    The problem is with n ds , which is normal to the surface of the hemisphere bounded by the inclined great circle, and so isn't [tex]
    \frac{i + j + k}{\sqrt{3}}[/tex].
  9. Aug 15, 2008 #8
    When one sleep deprived person is trying to converse with another, chaos is bound to ensue. I appreciate the help from both of you. I'll admit though that I'm still at a loss. I know what the normal should be so that the problem works out but I don't know how to get that.

    (I will also admit that I took vector calculus 6 years ago and I'm not doing such a great job of reteaching myself).
  10. Aug 15, 2008 #9
    Sorry, Stokes theorem will give the same answer irrespective of the surface. One problem is that the normal is probably negative. Besides that you've the done it correctly. To see why, rotate the space so that [tex]\nabla\times \vec{F} = -\sqrt{3}\vec{k}[/tex]. Now the circle lies in the xy plane. Parameterize the surface in the new coordinate system: [tex]\vec{\phi} = (rcos\theta,rsin\theta,0)[/tex]. If you carry out the surface integral explicitly, you get the same answer, [tex]\sqrt{3}\pi a^2[/tex].
  11. Aug 15, 2008 #10
    That makes sense. It's not the answer in my notes but maybe I wrote it down wrong.
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