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Outward flux

  1. Jan 15, 2014 #1
    Question: Find the outward flux of the vector field F = i-2j-2k across the surface S defined by z = 4-x2-y2 0≤z≤4

    At first, I used the Divergence Theorem to solve this problem. I took the divF and got the answer of 0. By definition, integrating 0 three times will still equal 0. Thus, the answer I wrote down currently is 0.

    Now there's also another way to find the outward flux, and that is by taking the double integral of Fn

    n = [r_r X r_θ]/[|r_r X r_θ|]
    dσ = |r_r Xr_θ|

    I found r_r X r_θ here from another problem: http://i.imgur.com/UMj72Ub.png

    Checking with WolframAlpha, I got this:
    1st integration: http://www.wolframalpha.com/input/?i=integrate+(-2r^2cos(theta)+4r^2sin(theta)-2r)+dr+from+0+to+2

    2nd integration: http://www.wolframalpha.com/input/?i=integrate+(-4/3)(-8sin(theta)+4cos(theta)+3)+from+0+to+2pi

    I would greatly appreciate it if someone can explain why the answer using the Divergence theorem is not equal to the Outward Flux when taking the double integral.

    Thanks,

    Differentiate1
     
    Last edited: Jan 15, 2014
  2. jcsd
  3. Jan 15, 2014 #2
    Never mind, figured out my mistake. I should've not used the parametrized form to take the double integral. I fixed this by making n = [-2x-2y]/|-2x-2y| and dσ = |-2x-2y| (I acquired these values from finding the gradient of z).
     
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