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Outward Flux Through Cylinder

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the outward flux of [itex]F = 3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}[/itex] through the base of a cylinder centered at the origin with radius [itex]\mathrm{R}[/itex] and height [itex]\mathrm{2H}[/itex].
    2. Relevant equations

    3. The attempt at a solution

    I am unsure of how to tell if I have calculated the outward flux, and not entirely confident that it makes sense to have 'H' in my final answer. I'd really appreciate it if you could take a look.

    [itex]\Phi = \int_{s} \mathbf{F}. \mathbf{n} dA[/itex]

    [itex]\mathbf{F}. \mathbf{n} = (3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}).(\mathbf{e_{z}}) = 2z[/itex]

    [itex]\Phi = \int_{s} 2z dA[/itex]

    [itex]\Phi = \int_{s} 2z \rho d\rho d\phi[/itex]

    [itex]\Phi = \int_{0}^{2\pi} \int_{0}^{R} 2z \rho d\rho d\phi[/itex]

    [itex]\Phi = \int_{0}^{2\pi} z \rho^{2} d\phi = z \rho^{2} [\phi]^{2\pi}_{0}[/itex]

    [itex]\Phi = 2 \pi z \rho^{2}[/itex]

    [itex]\Phi = -2 \pi H R^{2}[/itex]

    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2


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    Seems ok. Except I would say the OUTWARD normal on the bottom of the cylinder is ##-\mathbf{e_z}##. You could actually have done this calculation in your head, since you are integrating over a disk and the integrand is constant.
  4. Feb 28, 2015 #3
    Ah, of course! I should have picked up on that.

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