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Flux Integral

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data

    flux.png

    2. Relevant equations

    I know that flux is ∫∫F[r(t,s)]dotted[∂r/∂s x ∂r/∂t]dtds

    3. The attempt at a solution

    So I parameterized the equation y=x^2+z^2

    as:

    x = s
    y = s^2 + t^2
    z = t



    r(t,s) = [s, s^2+t^2, t]
    I take the partial derivative of r with respect to s and t and then take the cross product.

    I then that F[r(s,t)] and dot that with the cross product above.

    I then need the boundaries which I find to be 1/sqrt(2) ≤ s ≤ 1/sqrt(8) and 1/sqrt(2) ≤ s ≤ 1/sqrt(8)

    After integrating I get the answer 35/384 which says it's incorrect. I did the problem all over again and got the same exact answer. Any help?
     
  2. jcsd
  3. Dec 5, 2012 #2

    LCKurtz

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    Have you checked whether ##r_s\times r_t## is in the direction consistent with the given orientation of the surface? Also, just curious, why you rename the variables instead of just using ##x## and ##z## as the parameters.
     
  4. Dec 6, 2012 #3

    HallsofIvy

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    Doing this problem in the same way you describe, I get [itex](21/16)\pi[/itex] rather than your answer. You will need to describe more exactly what you did. What did you get for [itex]d\vec{S}[/itex]? What limits of integration did you use?
     
  5. Dec 6, 2012 #4
    LCKurtz, my rs×rt does point in the direction I believe it is supposed to. Also, the reason I renamed the variables was just my preference I guess. When setting up the vector, I like to think of each component that acts like the x, y, and z. It's probably unnecessary but that what I did.

    Let me give more detail into my work.

    - So I parameterized the equation y=x^2+z^2

    x = s
    y = s^2 + t^2
    z = t

    - I then get r(t,s) = [s, s^2 + t^2, t]

    - I then take the partial derivative with respect to both variables
    ∂r/∂s = [1, 2s, 0] and ∂r/∂t = [0, 2t, 1]

    - I then take the cross product of these to get,

    [2s, -1, 2t]

    - Now this cross product is oriented away from the y axis at all times, unless t and s are 0 I suppose...

    - I then use this formula ∫∫F[r(t,s)]dotted[∂r/∂s x ∂r/∂t]dtds

    - ∫∫ [s+t, 1, t]dotted[2s,-1,2t]dtds

    -Which equals

    ∫∫(2t^2+2ts+2s^2-1)dtds

    -I use the limits of 1/sqrt(2) ≤ s ≤ 1/sqrt(8) and 1/sqrt(2) ≤ s ≤ 1/sqrt(8) because these translate to match up with my parametrized equations and the limitation that 1/4 ≤ y≤ 1
     
  6. Dec 6, 2012 #5

    LCKurtz

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    Ignore that post I just did if you saw it. Be back later.
     
  7. Dec 6, 2012 #6

    LCKurtz

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    You are OK to there. But those limits describe a square. You need to change the integral to polar coordinates. Your t and s parameters should describe the region between two circles in the st (same as xz) plane.
     
  8. Dec 6, 2012 #7
    In doing that,

    would the ∫∫(2t^2+2ts+2s^2-1)dtds turn into ∫∫((2r^2+2*r*sin(θ)*r*cos(θ)-1)*r)drdθ with boundaries of 1/4≤ r ≤ 1 and 0 ≤ θ ≤ 2pi ?
     
  9. Dec 6, 2012 #8

    LCKurtz

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    Those are the limits for ##r^2##, not ##r##.
     
  10. Dec 6, 2012 #9
    Ahh, yes I almost messed that on up big time. So it should be 1/2 ≤ r ≤ 1
     
  11. Dec 6, 2012 #10
    Thanks so much LCKurtz, that seemed to be what I was doing wrong and I got the correct answer.
     
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