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?12

Use Stokes' Theorem to find the circulation of F =4yi +3zj +7xk around the triangle obtained by tracing out the path (6,0,0) to (6,0,5), to (6,5,5) back to (6,0,0). Circulation = ∫CF ⋅dr

  • one year ago
  • one year ago

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  1. ?12
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    so the curl is -3i-7j-4k

    • one year ago
  2. wio
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    If you want to go from line integral to surface integral, shouldn't you need the anticurl?

    • one year ago
  3. ?12
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    What is the anticurl?

    • one year ago
  4. wio
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    Nevermind, they gave the anticurl

    • one year ago
  5. ?12
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    I see

    • one year ago
  6. wio
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    I mean A such that \[ B = \nabla \times A \]

    • one year ago
  7. wio
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    But apparently they gave it to you already.

    • one year ago
  8. ?12
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    do you know how I would set up the line integral. I know that I have to use the points to make an equation of the triangle but after that I'm not sure

    • one year ago
  9. wio
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    So to use stokes theorem, you need to parametrize the surface.

    • one year ago
  10. wio
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    They don't want you to set up the line integral, but I can help you do that if you want.

    • one year ago
  11. ?12
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    The surface is the triangle?

    • one year ago
  12. ?12
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    I want to be able to do it using stoke's theorem/how they want it

    • one year ago
  13. wio
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    Here is stokes theorem: \[ \iint_S\mathbf{F}\cdot d\mathbf{S}=\int_{\partial S}\nabla \times \mathbf{F}\cdot d\mathbf{r} \]

    • one year ago
  14. wio
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    Now, the only way to use stokes theorem here is to do the left (surface integral) side. If you did the right (line integral side) would you really be using stokes?

    • one year ago
  15. ?12
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    No. I guess I was making a mistake. So how would I parametrize the surface? After that would the surface integral require a normal?

    • one year ago
  16. ?12
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    I'm getting line integrals and surface and green's theorem andstoke's theorem confused since we are barely covering them.

    • one year ago
  17. wio
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    You can easily find two parallel vectors for the triangle right?

    • one year ago
  18. ?12
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    the yz-plane?

    • one year ago
  19. wio
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    No, you do it by finding a vector between two points.

    • one year ago
  20. ?12
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    Between two points. So would I have to use points not contained in the triangle?

    • one year ago
  21. ?12
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    ?

    • one year ago
  22. wio
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    oh wow, you're right.

    • one year ago
  23. wio
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    yeah so we want the bounds on the yz plane.

    • one year ago
  24. ?12
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    okay so since I know the yz plane is parallel to the trianle would I be able to choose any 2 points there?

    • one year ago
  25. ?12
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    oh so it is a projection onto the yz plane

    • one year ago
  26. ?12
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    of the triangle, right?

    • one year ago
  27. wio
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    |dw:1362708042969:dw|

    • one year ago
  28. ?12
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    okay so does that tell me what the limits of integration for z are going to be or am I supposed to use it to parametrize the triangle?

    • one year ago
  29. wio
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    You could parametrize it as \[ \Phi(u,v) = (6,u,v),\quad 0\le u \le 5,\quad u\le v \le 5 \]

    • one year ago
  30. ?12
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    Oh I see. Because the x value is constant, right? Then from there will I have to calculte the normal to dot it with the curl of F?

    • one year ago
  31. wio
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    In this case \(\mathbf{F}\) doesn't need to be messed with, you only take the curl to do the line integral part.

    • one year ago
  32. ?12
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    So I just dot the parametrization of the triangle with the curl and then I just use the limits of integration for u and v?

    • one year ago
  33. wio
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    How many times do I have to say you don't take the curl?

    • one year ago
  34. ?12
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    Sorry I keep looking up at the equation. Okay. What would be the next step in my process?

    • one year ago
  35. wio
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    \[ \iint_S \mathbf{F}\cdot d\mathbf{S}= \iint_D\mathbf{F}\circ\Phi(u,v)\cdot \frac{\partial \Phi}{\partial u}\times \frac{\partial \Phi}{\partial v}dudv \]

    • one year ago
  36. wio
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    \[ \mathbf{F}\circ \Phi(u,v) = \mathbf{F}(x(u,v),y(u,v), z(u,v)) \]

    • one year ago
  37. wio
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    \(D\) corresponds to the bounds of \(u,v\).

    • one year ago
  38. ?12
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    okay so it is the vector field f in terms of u and v dotted with the normal of the parametrization?

    • one year ago
  39. wio
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    Yes.

    • one year ago
  40. ?12
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    Okay I'm going to work it out really quick.

    • one year ago
  41. ?12
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    Okay so when I did the normal I got <1,0,0> which, then dotted with <4u,3v,42> So I got 4u then I took the double integral \[\int\limits_{0}^{5}\int\limits_{u}^{5}4udv du\] that became \[\int\limits_{0}^{5}20u-4u^2du\] from there I got \[10u^2-4/3u^3\] and my answer was 250/3. Did I mess up somewhere because my answer seems incorrect? Was my set up correct?

    • one year ago
  42. wio
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    You have to consider the orientation.

    • one year ago
  43. ?12
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    the orientation is counter is wingspanwise so that means the normal is supposed to be supposed to be negative, right, or the parametrization of the triangle?

    • one year ago
  44. wio
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    |dw:1362709722289:dw|

    • one year ago
  45. wio
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    Yes it should be negative.

    • one year ago
  46. ?12
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    the normal?

    • one year ago
  47. wio
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    yeah.

    • one year ago
  48. ?12
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    Okay I change the normal to negative and then my answer just ended up being -250/3 but it is still wrong. Does it look like I did a mistake in setting it up?

    • one year ago
  49. wio
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    Do you know what the answer is?

    • one year ago
  50. ?12
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    Okay I got it I decided to use the parametrization that you gave me. Then I took the gradient of it and dotted it with the curl I got. so it was <-3,-7,-4> dotted with the gradient of <6,-u,v> which gave me 7-4 which is 3. Then I used all the limits of integration you gave me and got the correct answer which is 75/2 . I'm not sure if this is another part of the formula but it worked. Thank you very much! Would you gladly help me with one more wuestion if I post it up?

    • one year ago
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