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Stokes's Theorem problem

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    http://puu.sh/d5FpW.png [Broken]

    2. Relevant equations
    ∫∫(∇XF)⋅dS = ∫F(r(t))⋅r'(t)dt

    3. The attempt at a solution
    So I figured I'd have two line integrals (using Stokes's theorem.

    Paramatizing S1:

    I can think of the cylinder as the closed curve circle x²+y² = 9 on the x-y plane right?

    x = 3cost, y = 3sint, z = 0; 0 ≤ t ≤ 2pi

    or r(t) = (3cost)i + (3sint)j

    Then F = (27(cost)^3)i +(2187(sint)^7)j

    The dot product of F and r'(t) gives me

    -81sint(cost)3 + 6561(sint)^7cost

    Then I integrate this from t = 0 to t = 2pi and get 0.

    For S2:

    Similar to S1, I can think of this hemisphere as the circle

    x²+y² = 9 on the x-y plane with z = 8 right?

    Then paramatizing:

    r(t) = (3cost)i + (3sint)j + 8k ; 0≤t≤2pi

    I go through the same process as S1, but I don't get 0 I get -576 pi, which is obviously wrong.

    Would someone mind walking me through how to do this problem correctly?Am I applying Stokes's theorem incorrectly.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 25, 2014 #2

    LCKurtz

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    The only boundary of the silo/hemisphere is the circle in the xy plane, and you got zero for that.
     
  4. Nov 25, 2014 #3
    Hi LCKurtz, thanks for your reply.

    I was thinking I may only need to do the integral for S1 since, as you said, both surfaces share a circle in the xy plane as the boundary, but thought that made the problem too trivial.

    So instead of splitting the surface into two parts S1 and S2, I can think of it as one surface like the middle image in this picture: image001.png
    so what I calculated for S1 is for the entire surface as well?
     
  5. Nov 25, 2014 #4

    LCKurtz

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    Yes. One way to think of it is if that surface was flexible rubber or a soap bubble, it could contract down to just the disk inside that circle. Any way you deform the surface still has that boundary.
     
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