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Integrate divergence of a vector over an area

  1. Oct 22, 2014 #1
    Hello, I'm hoping somebody can give me some insight on how to solve this problem. This was a solid mechanics exam question and I wasn't able to finish it because I'm rather weak in math.

    1. The problem statement, all variables and given/known data

    WOxeM7f.jpg

    2. Relevant equations
    Recall divergence theorem for part ii. ∫div(V)dA = ∫V⋅ndS where n is normal to the surface.

    3. The attempt at a solution
    Part i.
    Not an issue, I can solve it easily.

    Part ii.
    I can apply divergence theorem no problem.
    For the n vector field I get:
    n1 = -e2
    n3 = -e1
    n2 = .707(e1 + e2)

    The problem is that I've been out of school for quite a while, and I can't remember how to successfully integrate over the surface (perimeter) now. I realize I need to break it up in to parts for each line of the surface, but it's been years since I've taken calculus.
     
    Last edited: Oct 22, 2014
  2. jcsd
  3. Oct 22, 2014 #2

    RUber

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    For c1, you will have ##\int_0^1 V\cdot n_1 \, dx_1##.
    For c3, you will have ##\int_0^1 V\cdot n_2 \, dx_2 ##.
    For c2, it gets a little tougher.
    ##\int_{c2} V\cdot n_2 dc2##
     
  4. Oct 22, 2014 #3
    For c2, should I do a double integral (x1 and x2, both from 0 to 1) of V⋅n2?
     
  5. Oct 22, 2014 #4

    RUber

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    It might be easier to directly compute the Area integral
    ##\int_0^1 \int_0^{1-x_1} div(V) dx_2dx_1 ##
     
  6. Oct 22, 2014 #5

    RUber

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    I don't think that is right, you would be taking the integral over an area. you want the integral over the line.
     
  7. Oct 22, 2014 #6
    Thanks RUber, I think you've provided the help I needed to finish it. It was required to perform divergence theorem for the exam though, I'll give it a shot with both methods and see how the solutions compare.
     
  8. Oct 22, 2014 #7

    RUber

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    Nevermind this...that doesn't make any sense.
    ---
    Try integrating over x1, and define x2 in terms of x1 along the line. That should do the trick.
     
  9. Oct 22, 2014 #8
    I solved it, I get 1/2 using both methods. Thanks again for the help RUber, it helped set me on the write path.
     
  10. Oct 22, 2014 #9

    RUber

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    When I worked it out, I still had the ##\frac 1 {\sqrt{2}}## in the answer.
     
  11. Oct 22, 2014 #10
    I'm not sure where the square root comes from, but I believe the accepted solution was 1/2 when we went over the exam (he just didn't take the time to go over something trivial like integration since most of my peers are coming straight out of undergrad >.<). I calculated it both by using divergence theorem to integrate over the line (more complicated) and by integrating the divergence over the area directly.
    FmqC9lS.jpg
     
  12. Oct 22, 2014 #11

    RUber

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    Aha, I forgot to multiply by the ##\sqrt{2}## in the dS term.
    Good work.
     
  13. Oct 23, 2014 #12

    BvU

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    Why (in post #7) "Nevermind this...that doesn't make any sense." ?
    I thought I had div V = 3x2 and that would be a simple integral, so I'm grossly overlooking sonething ?
     
  14. Oct 23, 2014 #13

    RUber

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    Thanks BvU, I was clearly overthinking things and convinced myself I was wrong.
     
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