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Stokes' Theorem

  1. Feb 20, 2005 #1
    Hi, I have this vector calculus question to do, and I can't seem to get it right! Could someone take a look for me?

    Q. The vector A(r) = (y,-x,z). Verify Stokes' Theorem for the hemispherical surface |r|=1, z>=0.

    A. I considered, the line integral about the circle in the xy plane (All interior boundaries cancel). Use polar co-ordinates, take the integral of 2sintcost.dt from 0->2pi thus getting an answer of zero.

    Then, find curl of A = -2k and dot this with the unit normal = -2. I know I now need to take the surface integral but I'm not sure how I proceed? Neither am I convinced I even understand what I'm doing!

    Please help :uhh:
  2. jcsd
  3. Feb 20, 2005 #2


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    Who's "k"...?And why doesn't the unit normal have modulus =1...?

  4. Feb 20, 2005 #3


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    Firstly, this line integral is incorrect. The circle may be parametrised as (x, y, z) = (cosФ, sinФ, 0)
    d/dФ (x, y, z) = (-sinФ, cosФ, 0)

    So the line integral is:
    Int{0 -> 2π} (y, -x, z) . (-sinФ, cosФ, 0) dФ
    = Int{0 -> 2π} (sinФ, -cosФ, z) . (-sinФ, cosФ, 0) dФ
    = Int{0 -> 2π} -1 dФ
    = -2π

    The problem here is you're dotting curl(A) with the wrong unit vector (the one pointing in the k direction), when you should really be dotting it with the unit vector which is normal to the hemispherical surface.

    The unit hemispherical surface may be parametrised as:

    (x,y,z) = (sinθcosФ, sinθsinФ, cosθ)

    (note that θ is the polar angle, Ф is the azimuthal angle)

    let A = d/dθ (x,y,z) = (cosθcosФ, cosθsinФ, -sinθ)
    let B = d/dФ (x,y,z) = (-sinθsinФ, sinθcosФ, 0)

    A x B (cross product)
    = (sin^2(θ)cosФ, sin^2(θ)sinФ, sinθcosθ)

    and finally,
    let C = curl(A) . (A x B) = (0,0,-2) . (sin^2(θ)cosФ, sin^2(θ)sinФ, sinθcosθ)
    = -2sinθcosθ
    = -sin2θ

    Compute the double integral of C with limits θ: 0 -> π/2 and Ф: 0 -> 2π, and you get -2π as required, matching up with the result of the line integral previously calculated.
    Last edited: Feb 20, 2005
  5. Feb 20, 2005 #4
    Thanks ever so much, that cleared up most of it for me, except there's one line in which I'm not entirely sure what you're doing and why:

    I see how you have calculate A & B, but don't understand their relevance to the problem? Are you using them to calculate the unit normal? If so, could you explain to me why they give the unit normal when crossed together? Thanks :tongue:

    Additionally, can anyone recommend a good Vector Calculus book for someone with limited understand of Vectors that goes through everything in detail with plenty of Physical Examples (ie. designed for a physicist rather than a mathmo?) Thanks
    Last edited: Feb 20, 2005
  6. Feb 20, 2005 #5


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    A and B are vectors which are both tangential (but not necessarily perpendicular to eachother) to the surface of the hemisphere. A x B is a vector which is perpendicular to both A and B (and hence NORMAL to the hemispherical surface). The magnitude of A x B is the area of the parallelogram formed by putting the vectors A and B together (head to tail). So you can see that F . (A x B) dθdФ (where F is any vector field) is the component of F which is normal to the surface, multiplied by an infinitesimal area (formed by minature parallelograms superimposed on the surface). Integrate this over the appropriate limits for the variables θ and Ф to get the final answer. That's probably the best explanation I can give. :smile:

    As far as textbooks for learning vector calculus go, I'm not aware of any physics oriented texts that go deeply into vector calculus most I've seen only superficially cover it (e.g. Griffiths). The text we used was "Calculus of vector functions" by Williamson, Trotter & Crowell (a pure maths oriented textbook - but still worth a look).
    Last edited: Feb 20, 2005
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