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Divergence theorem with inequality

  1. May 10, 2017 #1
    1. The problem statement, all variables and given/known data

    F(x,y,z)=4x i - 2y^2 j +z^2 k

    S is the cylinder x^2+y^2<=4, The plane 0<=z<=6-x-y

    Find the flux of F

    2. Relevant equations

    3. The attempt at a solution

    What is the difference after if I change the equation to inequality?

    For example :
    x^2+y^2<=4, z=0

    x^2+y^2<=4 , z=6-x-y

    x^2+y^2=4, z=6-x-y

    thanks!
     
    Last edited: May 10, 2017
  2. jcsd
  3. May 10, 2017 #2

    Math_QED

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    Make a sketch and everything will become clear.
     
  4. May 10, 2017 #3

    LCKurtz

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    Answer to what??

    One inequality gives the surface of the cylinder and the other gives the solid cylinder.
     
  5. May 10, 2017 #4
    Sorry, I made some mistake previously and I modified the question.
    If I want to find the flux of the bottom face, top face, side surface of cylinder and the tangent surface of the cylinder,
    How can I express it in inequality?
     
  6. May 10, 2017 #5

    LCKurtz

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    Bottom: ##\vec R(x,y) = \langle x,y,0\rangle,~0\le x^2+y^2\le 4##
    Top: ##\vec R(x,y) = \langle x,y,3\rangle,~0\le x^2+y^2\le 4##
    Side: ##\vec R(\theta,z) = \langle 2\cos\theta,2\sin\theta,z\rangle,~0\le \theta\le 2\pi,~0\le z\le3##
    I have no idea what you mean by the "tangent surface".
    You could, and likely should, use polar coordinates for the first two instead of ##x## and ##y## parameters.
     
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