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Surface element in cylindrical coordinates

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data

    [tex]\vec J_b = 3s \hat z [/tex]
    [tex]\int \vec J_b \, d\vec a [/tex]

    I need to solve this integral in cylindrical coordinates. It's the bound current of an infinite cylinder, with everything done in cylindrical coordinates and s is the radius of the cylinder. The answer should end up with a phi unit vector, but I just don't see how to get the actual unit vector.

    2. Relevant equations


    According to a pdf on MIT's site, the area element is:
    [tex] d \vec a = s d\phi dz \hat s [/tex]
    There's a caveat, though, as the page says there's a choice of direction. Using this way makes the most sense to me, but if I use this, the dot product between the s and z unit vectors will cause the whole thing to go to zero.
    http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf

    I've found another site that says to use the following, but I don't see the reasoning as to where the phi hat unit vector comes from.

    [tex] d \vec a = s d\phi \hat z [/tex]

    3. The attempt at a solution

    [tex]\int \vec J_b \, d\vec a = \int_0^s \int_0^{2\pi} s \hat z \cdot sdsd\phi \hat z = \frac{2 \pi}{3} s^3 \hat \phi [/tex]

    Using the first equation for the area element doesn't get me where I need to be, but the second one does.

    Any help in figuring this out would be appreciated.
     
  2. jcsd
  3. Nov 23, 2015 #2

    andrewkirk

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    Over what surface are you integrating? Your expression ##\int \vec J_b \, d\vec a ## does not provide that crucial information.

    In your first equation in 2, the integration variables are ##\phi## and ##z##, which suggests the integral is on the surface of the cylinder.

    But in the equation in 3, the integration variables are ##\phi## and ##s##, which suggests the integral is a cross sectional disc bounded by the cylinder.
     
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