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Homework Help: Flux Through a Cylinder

  1. Dec 11, 2017 #1
    1. The problem statement, all variables and given/known data
    In the attached image.

    2. Relevant equations
    Gradient(x, y, z) * <f, g, h> = <fx, gy, hz>

    3. The attempt at a solution
    Because the cylinder's not capped, I know that all the flux will be in the radial direction. So, I can find a normal vector by finding the gradient of the cylinder: n = <2x, 0, 2z>/(2sqrt(x^2+z^2)) = <x, 0, z>/sqrt(x^2+z^2)

    Now, I want to put this in terms of t (the angle) and h (y):
    r(t, y) = <acos(t), h, asin(t)>
    Where: y: (-2, 2) and t: [0, 2pi)

    Now we can rewrite the integrand:
    <acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> * <acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> dS
    =(a^2cos^2(t) + a^2sin^2(t))/a dS
    =a dS

    Now, the only thing I'm confused by (assuming everything else is right), is what to do with dS. I know it needs to be put in terms of dt and dh (where I already have the limits of integration), but I am unsure of how to perform this conversion.
     

    Attached Files:

  2. jcsd
  3. Dec 11, 2017 #2

    Charles Link

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    Isn't the vector ## \vec{E }## of the field used in this flux equation simply a unit vector in the radially outward direction? If it is, (and I don't know that I should be giving you this much of the answer, but the problem is almost trivial), isn't the flux equal to the surface area?
     
  4. Dec 11, 2017 #3
    I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.
     
  5. Dec 11, 2017 #4

    Charles Link

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    You don't need to do any conversion. Surface area ## S=\pi a^2 h ##.
     
  6. Dec 11, 2017 #5
    Okay. Just curious though, how would I perform that conversion in a less trivial case?
     
  7. Dec 11, 2017 #6

    Charles Link

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    Surface integrals in the general case can take much effort to evaluate. The problem, I think, was designed to keep it simple.
     
  8. Dec 11, 2017 #7

    Orodruin

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    In the general case, where you have a surface parametrised by the two variables ##t## and ##s##, the directed surface element is given by
    $$
    \newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
    d\vec S = \vec n \, dS = \dd{\vec x}{t} \times \dd{\vec x}{s} dt\, ds,
    $$
    where ##\vec x## is the position vector (which on the surface is parametrised by ##t## and ##s##.
     
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