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Verify the divergence theorem for a cylinder

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data

    Verify the divergence theorem if [itex]\textbf{F} = <1-x^{2}, -y^{2}, z >[/itex] for a solid cylinder of radius 1 that lies between the planes z=0 and z=2.

    2. Relevant equations
    Divergence theorem


    3. The attempt at a solution

    I can do the triple integral part no problem. Where I run into issues is the surface integral part.

    Parametrizing a cylinder is done by <rcost,rsint,z>, correct?

    So looking at the top part I want it oriented in the positive k direction to get flow OUT of the cylinder - hence S: <cost,sint,2> because it lies in the plane z=2.
    Similarly for the bottom S: <cost,-sint,0>

    Now for the side, the side never has any k components so S: <cost,sint,0>

    Now lets look at the top again. If I take the derivatives with repect to Z and try to cross them I end up with 0 net flow in every direction, it does not agree with the triple integral and isn't correct.
    I have a feeling that I'm not parametrizing my cylinder correct, I remember it being a special case.

    Another thing that I tried was parametrizing the top as <cost,sint,z>
    Then the dS vector becomes
    <cost,sint,0>
    Now it's showing no component in the z direction - clearly it should have a component in the z direction?!
     
  2. jcsd
  3. May 5, 2014 #2

    Dick

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    Clearly dS along the top should be in the +z direction. It should be <0,0,1>dA. How are you getting that it isn't? The coordinates along the top are <rcos(t), rsin(t), 2> take the r derivative and the t derivative and cross them.
     
  4. May 5, 2014 #3
    Ahh I see I thought it was
    <1cost,1sint,z> I didn't realize that you allowed R to vary as it is fixed in the problem.
     
  5. May 6, 2014 #4

    Dick

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    The outer radius of the cylinder is fixed at 1. The coordinate r isn't fixed along the top surface. It ranges from 0 to 1. Only z is fixed. Along the sides r is fixed and z isn't.
     
  6. May 6, 2014 #5

    HallsofIvy

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    No, S can have any value of z between 0 and 2: S: <cos(t), sin(t), z>. A two dimensional surface always requires two parameters! The NORMAL VECTOR has no k component:
    writing [itex]\vec{r}= cos(t)\vec{i}+ sin(t)\vec{j}+ z\vec{k}[/itex], we have [itex]\vec{r}_t= -sin(t)\vec{i}+ sin(t)\vec{j}[/itex] and [itex]\vec{r}_z= \vec{k}[/itex] and their cross product is [itex]cos(t)\vec{i}+ sin(t)\vec{j}[/itex]

    The top is NOT "<cos(t), sin(t), z>". That has r fixed at 1 and both t and z varying so is the cylindrical side. The top has both r and t varying and z fixed at 2: <r cos(t), r sin(t), 2> (the bottom is <r cos(t), r(sin t), 0>).
    The derivative with respect to r, for the top, is <cos(t), sin(t), 0> and the derivative with respect to t <-r sin(t), cos(t), 0>. Their cross product gives [itex]\vec{n}dS= r \vec{k} drdt[/itex]. Similarly, [itex]\vec{n}dS[/itex] is [itex]-r\vec{k}drdt[/itex] for the bottom.
     
    Last edited: May 6, 2014
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