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Surface Integral of a Cylinder!

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S denote the closed cylinder with bottom given by z=0, top given by z=4, and lateral surface given by the equation x^2 + y^2 = 9. Orient S with outward normals. Determine the indicated scalar and vector surface integral to ∫∫ x^2 i dS (I have tried to solve this problem, but i dont think i have done it correct, since i get the answer 0. Please if anybody can help, this task is very importaint to get right)

    2. Relevant equations
    [itex]\int\int[/itex]x f dS=[itex]\int\int[/itex] D f(X(s,t))[itex]||[/itex]Ts[itex]\times[/itex]Tt[itex]||[/itex]

    3. The attempt at a solution
    Since this is a cylinder, (i think) we can slice it in to 3 parts,
    S1 (lateral cylindrical surface), S2 (bottom disk) and S3 (top disk).
    I parametrized the three smooth pieces as follows:
    S1 (lateral cylindrical surface): x=3Cos(t), y=3Sin(t), z=t With bouderies 0[itex]\leq[/itex]s[itex]\leq[/itex]2π and 0[itex]\leq[/itex]t[itex]\leq[/itex]4

    S2 (bottom disk): x=sCos(t), y=sSin(t), z=0. With bounderies 0[itex]\leq[/itex]s[itex]\leq[/itex]3 and 0[itex]\leq[/itex]t[itex]\leq[/itex]2π

    S3 (top disk): x=sCos(t), y=sSin(t), z=4. With bounderies 0[itex]\leq[/itex]s[itex]\leq[/itex]3 and 0[itex]\leq[/itex]t[itex]\leq[/itex]2π

    Then I found the ||Ts[itex]\times[/itex]Tt[itex]||[/itex] for each S, that is
    S1||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(3Cos(s), -3Sins,0)

    S2||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(0,0,sCos^2 (t)+sSin^2 (t))

    S3||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(0,0,sCos^2 (t)+sSin^2 (t))

    Then I thought i could set in all that is in the parametrizised in the x^2 positision, dotted with the ||Ts[itex]\times[/itex]Tt[itex]||[/itex].
    S2 and S3 will then be 0 ( since ||Ts[itex]\times[/itex]Tt[itex]||[/itex] have both 0 in the x position)

    The last one S1
    [itex]\int\int[/itex] X^2 dS=[itex]\int\int[/itex] (3^2 Cos^2 (s), 0, 0) ×(3Cos(s),3Sin(s),0) ds dt (boundaries 0 and 4(dt), 0 and 2π(ds))

    =∫∫12 Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))

    =12∫∫ Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))

    =12∫ -3Sin(s)Cos^2 (s) dt (input 0 and 2π( for s)) (boundaries 0 and 4(dt)

    This gives =12∫ dt wich gives 0 <-- and i dont think this is the right answer. If anybody can help
     
  2. jcsd
  3. Nov 25, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Was this a typo? You should have x= 3Cos(s), y= 3sin(s), not "t".

    It is a bit confusing to use the same letters as before for the parameters, but change their meaning.

    Okay, so the "t" above was a typo. This is correct for the lateral surface.

    S2||Ts[itex]\times[/itex]Tt[itex]||[/itex] =(0,0,sCos^2 (t)+sSin^2 (t))[/quote]
    You do know that [itex]s cos^2(t)+ s sin^2(t)= s[/itex], right? :tongue:

    Actually, 0 is correct! This integral measures the flow of the vector quantity [itex]x^2[/itex]i into the cyinder. Over the top and bottom, which are parallel to the xy-plane, there is no flow through the surface. For the lateral surrface, the flow into the cylinder on the x< 0 side is cancelled by flow out of the cylinder on the x> 0 side.
     
  4. Nov 25, 2012 #3
    I Thank Thee!
    I did find though that I had some mistakes in my calculations:

    =∫∫12 Cos^3 (s) dsdt (boundaries 0 and 4(dt), 0 and 2π(ds))
    is ∫12 (1/12(9sin(s)+sin(3s)) dt =∫(9sin(s)+sin(3s))dt instead of =12∫ -3Sin(s)Cos^2 (s) dt which I wrote at the beginning.
    But =∫(9sin(s)+sin(3s))dt ( with boundaries 0 to 2π) also gives the answer 0.
     
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