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Where have I gone wrong with this surface integral problem?

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral ∫F.dS where F = xi - yj + zk and where the surface S is of the cylinder defined by x^2+y^2≤4, and 0≤z≤1. Verify your answer using the Divergence Theorem.

    2. Relevant equations



    3. The attempt at a solution

    I parametrized the surface in terms of θ and z: r=(2cosθ, 2sinθ, z). I found dr/dθ X dr/dz=(2cosθ, 2sinθ, z). (How do I know which way round to do the cross product?). I rewrote F as (2cosθ, -2sinθ, z). I then did the following integral:
    ∫∫(2cosθ, -2sinθ, z).(2cosθ, 2sinθ, z)dθdz and got ∫∫4(cosθ)^2-4(sinθ)^2 dθ dz=0

    But using the divergence theorem: divF=1, so I found the answer to be the volume of the cylinder, 4pi.

    Where have I gone wrong?

    Thanks in advance! :-)
     
  2. jcsd
  3. Apr 12, 2012 #2

    Office_Shredder

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    Did you include the top and bottom surfaces of the cylinder?
     
  4. Apr 12, 2012 #3
    Ah, I forgot those... Thanks for pointing that out. I'll try again now and see if I get the right answer. Could someone explain how you work out which way round to do the cross products?
     
  5. Apr 12, 2012 #4
    Thanks, I got the right answer. :-)

    I'm still confused about the cross product thing I mentioned before though...
     
  6. Apr 12, 2012 #5

    HallsofIvy

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    Your original problem, "Evaluate the surface integral ∫F.dS where F = xi - yj + zk and where the surface S is of the cylinder defined by x^2+y^2≤4, and 0≤z≤1" isn't complete. You need to specify the orientation of the surface as well. Here, as for general closed surfaces, you can specify either "oriented by outward pointing normals" or "oriented by inward pointing normals".

    Changing the order of a cross product only changes the direction of the normal vector. In your example, you get [itex]<2cos(\theta), 2sin(\theta), z>[/itex] If you check different values of [itex]\theta[/itex], for example, for [itex]\theta= 0[/itex] so that is [itex]<2, 0, z>[itex] or [itex]\theta= \pi/2[/itex] so that is [itex]<0, 2, z>[/itex] you can see that the vector is pointing away from the the z-axis, which is the axis of the cylinder, and so is "outward pointing". If you did the cross product in the other order, you would get [itex]<-2cos(\theta) -2sin(\theta), -z>[/itex] so that every vector is pointing toward the axis. That normal is "inward pointing" and integrating with that would give the negative of the integral using the "outward pointing" normal.

    That also, by the way, means that when you integrate over the two ends of the cylinder, you must make the same "outward" or "inward" choice. If you used outward pointing normals for the cylinder, then on the plane z= 1, you must use [itex]dS= <0, 0, 1> dxdy[/itex] so it is pointing upward (out of the cylinder) and on the plane z= 0 you must use [itex]dX= <0, 0, -1>dxdy[/itex] pointing downward (but still out of the cylinder).
     
  7. Apr 13, 2012 #6
    Thanks so much; this makes perfect sense now. :-)
     
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