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Surface integral of a vector field w/o div. theorem

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data
    First a thanks for the existence of this site, i find it quite useful but had no need to actually post till now.

    I am stuck on the following problem in "introduction to physics"
    We should calculate the [itex]\oint[/itex] [itex]\vec{v}[/itex].d[itex]\vec{A}[/itex]

    of a object with the following parameters Vol = x^2+y+2<4, 0<z<4-x^2-y^2
    with [itex]\vec{v}[/itex] being x+y,y+z,x+z)
    with and without the div. theorem,

    2. Relevant equations
    Divergence theorem,div = 3 - simple cylinder integral = possible solution =32pi
    Parametrization of volume as
    2cos(u)sin(v)
    2sin(u)sin(v)
    4cos(v)
    u = (0,2pi) v= (0,pi/2)
    Surface integration using [itex]\vec{v}[/itex](u,v).ru x rv

    3. The attempt at a solution
    I have 32Pi as a solution using the div theorem,
    and i am in over 2 pages of writing now without using it,
    Am i missing something or is this the kind of problem that only exist "to show you how useful something is" and i really should come to some horrible solution that is taking a double integral of something over 50 symbols long ?

    Thanks in advance.
     
  2. jcsd
  3. Jan 5, 2012 #2
    Yep, it's one of those - "To see how nice the theorem is, do it the hard way too", after one more page calculus and wolfram alpha both results match (dunno if i can delete the topic or close it first post here (technically second :) )
     
  4. Jan 5, 2012 #3

    LCKurtz

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    Is that first inequality supposed to be ##x^2+y^2<4##?
    That is not a correct parameterization. You don't have a sphere, you have an upside down paraboloid.
    I don't think your volume is expressed correctly even with my assumption above. You have apparently worked the volume integral for the region under the paraboloid and above the xy plane. But what does the cylinder have to do with it? Something is wrong with your statement of the problem.
     
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