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Tommynaut

  • one year ago

Let F = y^2 ~i + x ~j + z ~k. Let S be the curved surface of the cylinder x^2 + y^2 = 1 for 0 <= z <= 3. Calculate the outwards flux of F through S. I used the divergence theorem to get my solution, by finding divF = 1. I thought that would mean that the flux would be the volume of the cylinder, but the answer is actually 0. Could someone please explain the proper process? Oh, and is there a way to use latex when asking questions?

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  1. Empty
    • one year ago
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    Yeah, like this for example and the code: $$\nabla \cdot F = 0 $$ ``` $$\nabla \cdot F = 0 $$ ```

  2. Empty
    • one year ago
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    Actually wait, you mean specifically while asking questions. No, it's not really possible to access the equation editor but you can still type latex, it sucks I agree. But whatever let me try help you with your actual problem, one sec haha.

  3. Tommynaut
    • one year ago
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    Ok cheers haha

  4. Empty
    • one year ago
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    Ahhh ok the problem I believe you're having is that the divergence theorem applies to closed shapes, not open ones.

  5. Tommynaut
    • one year ago
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    But surely the cylinder can be imagined to be closed? I really didn't think it made a difference, especially as we're looking for the flux through the curved surface anyway.

  6. Tommynaut
    • one year ago
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    And how would the question then be done?

  7. anonymous
    • one year ago
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    You can only use the divergence theorem if the surface is closed.

  8. anonymous
    • one year ago
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    I don't think the top and bottom are included.

  9. anonymous
    • one year ago
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    It'd parametrize the cylinder as: \[ \{(\cos t, \sin t, z)|0\leq t\leq 2\pi,0\leq z\leq 3\} \]

  10. dan815
    • one year ago
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    you can apply div theorem, and them subtract the flux out of the top and bottom circles, if that simplfies it, it might be simplied to do the flux out of hte top and bottom as they are planar and perp to z so only the z component would matter

  11. IrishBoy123
    • one year ago
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    you should have gotten 3pi from div theorem surely. then the field on bottom plate = zero => zero flux normalised field at top plate = > z = 3, which is 3 pi also so net through curved surface = 0 which makes sense as the field is symmetrical in the x-y plane

  12. Tommynaut
    • one year ago
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    I did indeed get 3pi originally. I don't really understand the concepts you guys are explaining but thank you anyway. I found a method of solution that uses cylindrical coordinates and the unit normal outwards, followed by a double integral.

  13. IrishBoy123
    • one year ago
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    i hope this helps

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  14. IrishBoy123
    • one year ago
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    my point being that, in my experience at least, doing these types of things is v often about not doing them. ie spotting symmetries in the set up, or lucking out with divergence or stoke's theorem, the alternative being to do a complete pig of a double integral....:p you can often also have more confidence in your answer if you have simplified it down in one of these ways.

  15. Tommynaut
    • one year ago
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    Thank you for that! The notation you use is a bit different to what I've been taught but it makes sense nonetheless :)

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