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Surface integrals and flux

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data
    calculate the upward flux of f(x,y,z) = <yz,2x+y,y^2+z>

    Let S be the portion of the cylinder z=4-y^2 lying in the first octant to the right of the plane y=4.
    a parametrization into the u v plane is:r(u,v)=(u,v,4-v^2)
    region is a rectangle in the uv plane with bounds, (0,0) , (0,2) and (4,0)

    2. Relevant equations
    [tex]\int[/tex][tex]\int[/tex] F [tex]\bullet[/tex](ru [tex]\times[/tex]rv) dA



    3. The attempt at a solution

    ru x rv = 0i + 2vj + 1k

    so then i have [tex]\int[/tex] <4v-v^3, 2u+v, v^2 + 4 - v^2 > x <0 + 2v + 1 >

    [tex]\int[/tex] (2v^2 + 4uv + 4)dudv
    4[tex]\int[/tex] (v^2/2 + uv + 1) dudv


    i know i can't go to polar so i integrate

    is this the correct integral?

    0 to 2, outside integral, 4* 0 to 4 inside integral ( v^2/2 + uv + 1 ) dudv ?
     
  2. jcsd
  3. Dec 4, 2008 #2
  4. Dec 5, 2008 #3

    Defennder

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    You're not writing it out properly. It's supposed to be (0 2v 1)^T There isn't any "+".

    How did you get v^2 ? And I still don't understand your notation. Why is there 2 separate integrals instead of a single double integral?
     
  5. Dec 5, 2008 #4
    I'm sorry there's not. That's just the next step to solving the integral.

    then i take the dot product of <4v-v^3, 2u+v, v^2 + 4 - v^2 > dot <0, 2v,1 >

    I get < 0, 4uv + 2v^2, v^2 -v^2 + 4 > which simplifies into 2v^2 + 4uv + 4

    so i integrate between...

    [tex]\int[/tex] 0 to 2 and [tex]\int[/tex] 0 to 4

    integrand of : 2v^2 + 4uv + 4 (dudv)
     
  6. Dec 5, 2008 #5

    Defennder

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    Well ok, I can get that. So what's the problem now?
     
  7. Dec 5, 2008 #6

    HallsofIvy

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    So it is
    [tex]\int_{v= 0}^2\int_{u= 0}^4 (2v^2+ 4uv+ 4)dudv[/tex]?

    Looks pretty straightforward to me. What is the problem? That can, of course, be separated into
    [tex]2\int_0^4 du\int_0^2 v^2dv+ 4\int_0^4 udu\int_0^2 vdv+ 4\int_0^4du\int_0^2dv[/tex]
     
  8. Dec 5, 2008 #7

    Defennder

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    For some reason the "u=0" and "v=0" isn't showing up properly in Latex.
     
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