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Surface integral in Spherical Polar

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Attached.

    2. Relevant equations

    cd7a7dabc3ed67196d363a39be943c45.png
    01218db9301c3665178919b26218da4b.png
    d9beecbcecba1ca83ecd273a33330938.png
    1f5ce7228bc649add6fcd17c1ddbf174.png

    3. The attempt at a solution
    Hi,

    Ok, so for the first part of this question it asks to evaluate the integral of the dot product between A and dS. The magnitude of dS is as shown above, and it is in the radial direction in spherical polar coordinates. Here, A(r) is the unit vector of θ in spherical polar. And since we know that the polar direction θ and the radial direction are orthogonal to each, their dot product must = 0. Therefore, the integral of A.dS must be equal to zero. Is this line of thought correct?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Mar 2, 2015 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, the direction of [itex]d\underline{S}[/itex] is not "in the radial direction". The "vector differential of area" is [itex](a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2 sin(\phi)cos(\phi)\vec{k})d\theta d\phi[/itex]. That is not a multiple of a radial vector which would be of the form [itex]r cos(\theta)sin(\phi)\vec{i}+ r sin(\theta)sin(\phi)\vec{j}+ r cos(\phi)\vec{k}[/itex].
     
  4. Mar 3, 2015 #3
    Thanks for the reply. But if I'm thinking about this right, couldn't you factor out of that vector the scalar a^2*sin(α) which would leave the unit vector in the radial direction? Please see the attached.
     

    Attached Files:

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