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Correct Application of Divergence Theorem?

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    http://img593.imageshack.us/img593/5713/skjermbilde20111204kl11.png [Broken]

    3. The attempt at a solution
    I thought it seemed appropriate to use divergence theorem here: I have,
    [tex]div F = 0 + 1 + x = 1+x[/tex]
    I let that 0≤z≤c. If,

    [tex]x/a + y/b = 1[/tex]then y=b(1-x/a)
    [tex]x/a +z/c = 1 [/tex]then x=a(1-z/c)

    I have,

    [tex]\int_0^a \int_0^{c(1-x/a)} \int_0^{b(1-x/a)} (1+x) dydzdx[/tex]
    [tex]\int_0^a \int_0^{c(1-x/a)} (x+1) b(1-x/a) dzdx[/tex]
    [tex]\int_0^a (x+1) b(1-x/a) c(1-x/a) dx[/tex]

    But here I get stuck because I can't integrate this. Do I have the wrong bounds or am I missing something?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 4, 2011 #2

    Char. Limit

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    You can integrate this just fine, once you distribute everything. Your integral at the end is:

    [tex]bc \int_0^a \left(1+x\right) \left(1 - \frac{x}{a}\right)^2 dx[/tex]

    That's going to factor out to a cubic, which is easily integrable.
     
  4. Dec 4, 2011 #3

    LCKurtz

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    Your upper limit on the dy integral is wrong. y should go from y = 0 to the y on the plane. Solve the equation of the plane for y. It should have both x and z in it.
    It is wrong because of the above comment, but of course you should be able to integrate something like that. It is just a polynomial. All you have to do is multiply it out first.
     
    Last edited by a moderator: May 5, 2017
  5. Dec 4, 2011 #4
    So does y=b(1- z/c - x/a)?
     
  6. Dec 4, 2011 #5

    Char. Limit

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    That sounds more likely to be correct, yes.
     
  7. Dec 5, 2011 #6
    Okay. Let's see if I can do this. As above, I'll integrate in the order dydzdx.

    [tex]\int_0^{b(1-z/c-x/a)}(1+x) dy = (x+1) b(-x/a-z/c+1)[/tex]
    [tex]= (1+x)*b \int_0^{c(1-x/a)} (-x/a-z/c+1) dz [/tex]
    [tex]= (1+x)*b*[(1/2)*c*(x^2/a^2-1)+c*(1-x/a)][/tex]
    [tex]\int_0^{a} (1+x)*b*[(1/2)*c*(x^2/a^2-1)+c*(1-x/a)] dx= (1/24)*a*(a+4)*b*c[/tex]

    Hmm. I still feel like I missed something. Does it look about right?
     
    Last edited: Dec 5, 2011
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