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  1. anonymous
    • one year ago
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    what is the question

  2. anonymous
    • one year ago
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    From Stoke's Theorem we have \[\iint_\limits S (\vec \nabla \times \vec F). \hat n \space ds=\oint_\limits C \vec F.d \vec r\] Where S is any surface and C is a closed curve around the surface I can see how this works out in the case of a 2D surface |dw:1443885818098:dw| But if we consider the surface of a 3D object, we have |dw:1443885892309:dw| Here S is the lateral surface of that figure What do we take for our C here??

  3. ganeshie8
    • one year ago
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    C is the closed curve which is the boundary of that surface |dw:1443886311804:dw|

  4. anonymous
    • one year ago
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    It isn't really enclosing the surface though??

  5. anonymous
    • one year ago
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    Oh, it must be bounded by C that's the only condition

  6. anonymous
    • one year ago
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    Suppose we want to find the surface integral of curl F both through the lateral surface and the circular surface, Would that be done something like.. \[\oint_\limits C \vec F.d \vec r+\iint_\limits R (\vec \nabla \times \vec F).\hat k \space dxdy\]|dw:1443886644120:dw|

  7. IrishBoy123
    • one year ago
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    the line integral and the surface integral are the same, that is what Stokes' theorem says

  8. IrishBoy123
    • one year ago
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    you know the [deformable] butterfly net analogy?

  9. anonymous
    • one year ago
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    Nope, but is my step correct for finding the surface integral of curl F through the whole surface(including circular top) |dw:1443887053149:dw|

  10. IrishBoy123
    • one year ago
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    yes but |dw:1443887366725:dw|

  11. anonymous
    • one year ago
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    Hm, that's easy to understand, because the curves bounding these surfaces are the same, so the line integral through these curves should be the same for a particular vector field!

  12. IrishBoy123
    • one year ago
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    and the flat net enclosed by the ring = the puffed out nets because all are enclosed by same ring.....and covered by the same line integral

  13. anonymous
    • one year ago
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    Wait does that mean for total surface we have, \[2 \times \oint_\limits{C} \vec F . d \vec r\] ??

  14. IrishBoy123
    • one year ago
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    at the risk of sounding agricultural, it comes in the hoop and goes out the net.

  15. IrishBoy123
    • one year ago
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    i can post an example that can be done several ways

  16. anonymous
    • one year ago
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    So it will be 2 times the line integral??? for total surface??

  17. anonymous
    • one year ago
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    If it's equal for the surfaces, since they have the same C

  18. anonymous
    • one year ago
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    then it just becomes 2 times for total surface?

  19. IrishBoy123
    • one year ago
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    no, they should net out to zero so take a sphere |dw:1443888358128:dw|

  20. IrishBoy123
    • one year ago
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    you should be able to do both surfaces using the loop so they are the same but they cancel out |dw:1443888386190:dw||dw:1443888398064:dw|

  21. anonymous
    • one year ago
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    |dw:1443888517480:dw|

  22. anonymous
    • one year ago
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    Applying stokes theorem on R and S, they convert to line integrals of the same curve, so just 2 times the line integral for total??

  23. anonymous
    • one year ago
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    Why do they cancel out instead of adding up??

  24. IrishBoy123
    • one year ago
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    because in the physical sense you are typically looking at the flux (like flow) of the curl of some field. so say from Maxwell you have \(\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}\) relating the curl of an eletric field with a magnetic field and its flux.

  25. anonymous
    • one year ago
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    Hmmm, I remember something similar from school \[e.m.f=-\frac{d \Phi_{B}}{dt}\] Faraday's law of electromagnetic induction

  26. anonymous
    • one year ago
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    Your equation looks different

  27. IrishBoy123
    • one year ago
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    yours is faraday's in simplest form - induced voltage from a changing magnetic field i quoted maxwell #3 in its differential form which is the basis of what you quoted my point though is that when we look at surfaces we are usually concerned with the rate at which something is flowing through it, be it a fluid or a magnetic field.

  28. IrishBoy123
    • one year ago
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    scroll down here to equation 7 http://www.maxwells-equations.com/faraday/faradays-law.php

  29. anonymous
    • one year ago
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    Shouldn't it be a double integral on the right side of equation 7

  30. IrishBoy123
    • one year ago
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    there's a lot of variation in notation. i think the \(\int_S \dots dS \) is enough for some people to signify a double integral

  31. anonymous
    • one year ago
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    Oh, cuz I thought about it like this \[\vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t}\] \[(\vec \nabla \times \vec E).d \vec A=-\frac{\partial \vec B}{\partial t}.d \vec A\]\[\iint (\vec \nabla \times \vec E). \space d \vec A=-\iint \frac{\partial \vec B}{\partial t}. \space d \vec A\]\[\oint \vec E.d \vec l=-\iint \frac{\partial \vec B}{\partial t}.d \vec A\]

  32. anonymous
    • one year ago
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    Electromagnetism is fascinating, I can't wait to study it in later semesters, the moment when you realize ahh you can apply the math you've learned so beautifully

  33. IrishBoy123
    • one year ago
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    that's pretty good!

  34. anonymous
    • one year ago
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    I think I'll leave discussion about finding total flux over that surface, it still looks a bit confusing and I doubt I'd use it right now, most of the problems require me to find the flux over the lateral surfaces (they mainly want to see how u can apply the theorems etc.) so my exams will be more theorems(statements) and problems based. My main doubt was how to find C for a 3D figure I quote from my syllabus "The emphasis of course is on applications in solving problems of interest to physicists. The students are to be examined entirely on the basis of problems, seen and unseen."

  35. IrishBoy123
    • one year ago
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    try this. i could really do with the practise too but i've got a lot of rugby to watch this evening:-) so i'll have a go later.... it is set up to be do-able in a few ways, so might be interesting.

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  36. IrishBoy123
    • one year ago
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    oh, i mean example 1

  37. anonymous
    • one year ago
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    \[\vec V=4y \hat i+x \hat j+2z \hat k\] S is \[x^2+y^2+z^2=a^2\] \[z \ge 0\] |dw:1443891158189:dw| Hmm one way we can do is convert into line integral using stokes' theorem! \[\iint_\limits{S}(\vec \nabla \times \vec V).\hat n \space d \sigma=\oint_\limits C \vec V.d \vec r\] C is the circular loop in the xy plane therefore z=0, dz=0 \[\iint_\limits S=\oint_\limits{C}4ydx+xdy=\int\limits_{0}^{2\pi} (-4a^2\sin^2(\theta)+a^2\cos^2(\theta)) d \theta\]\[\iint_\limits{S}=\frac{a^2}{2}\int\limits_{0}^{2\pi}(5\cos(2\theta)-3)d \theta=\frac{a^2}{2}[\frac{5\sin(2 \theta)}{2}-3 \theta]_{0}^{2\pi}=-3 \pi a^2\] 2nd we can find curl then take the projection on xy plane \[\iint_{S}=\iint_{R}(\vec \nabla \times \vec V).\hat n \frac{dxdy}{\hat n . \hat k}\] \[\vec \nabla \times \vec V=-3 \hat k\] \[\hat n=\frac{x \hat i+y \hat j+z \hat k}{\sqrt{x^2+y^2+z^2}}\]\[(\vec \nabla \times \vec V).\hat n=-\frac{3z}{\sqrt{x^2+y^2+z^2}}\]\[\frac{1}{\hat n . \hat k}=\frac{\sqrt{x^2+y^2+z^2}}{z}\] \[\iint_\limits{S}=\iint_\limits{R}-\frac{3z}{\sqrt{x^2+y^2+z^2}}.\frac{\sqrt{x^2+y^2+z^2}}{z}dxdy=-3\iint_\limits{R}dxdy=-3\pi a^2\]

  38. anonymous
    • one year ago
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    @IrishBoy123

  39. IrishBoy123
    • one year ago
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    yes! i think this is a really good book, you will know the maths already: http://www.amazon.co.uk/A-Students-Guide-Maxwells-Equations/dp/0521701473 there is a pdf somewhere on the net so you can try it first and see if you like it. by the time you finish it you will see how Maxwell worked out the speed of light whilst sitting at his desk doing some maths! amazing stuff really.

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