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anonymous

  • one year ago

find parametrization of the surface...

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  1. anonymous
    • one year ago
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    spherical band. the portion of the sphere x^2 + y^2 + z^2 = 3 between the planes z = sqrt(3) / 2 and z = - sqrt (3) / 2

  2. dan815
    • one year ago
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    find the angle first

  3. Michele_Laino
    • one year ago
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    I think that the requested parametrization is given by the subsequent formulas: \[\Large \left\{ \begin{gathered} x = \sqrt 3 \sin u\cos v \hfill \\ y = \sqrt 3 \sin u\sin v \hfill \\ z = \sqrt 3 \cos u \hfill \\ \end{gathered} \right.,\quad \begin{array}{*{20}{c}} {\frac{\pi }{3} \leqslant u \leqslant \frac{{2\pi }}{3}} \\ {0 \leqslant v \leqslant 2\pi } \end{array}\]

  4. Michele_Laino
    • one year ago
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    since we have this situation: |dw:1435851886306:dw| so theta runs from pi/3 to 2*pi/3

  5. Michele_Laino
    • one year ago
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    I called \theta with u

  6. Michele_Laino
    • one year ago
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    the radius of our sphere is: \[\Large r = \sqrt 3 \]

  7. dan815
    • one year ago
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    yes thats right spherical looks nice

  8. dan815
    • one year ago
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    |dw:1435852161889:dw|

  9. dan815
    • one year ago
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    |dw:1435852254510:dw|

  10. dan815
    • one year ago
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    when you think about parametrization you are basically thinking, how can i define all the points on this surface (x,y,z) or (r,phi,theta) or (u,v,w)

  11. Michele_Laino
    • one year ago
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    yes! I used (r, u, v)

  12. dan815
    • one year ago
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    should probably show him how u got the equations for (x,y,z) just so he is clear on that

  13. dan815
    • one year ago
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    |dw:1435852565980:dw|

  14. dan815
    • one year ago
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    |dw:1435852666953:dw|

  15. Michele_Laino
    • one year ago
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    by definition the spherical or polar cordinates are like below: |dw:1435852590928:dw|

  16. Michele_Laino
    • one year ago
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    from picture above, we can write: \[\Large \left\{ \begin{gathered} x = r\sin u\cos v \hfill \\ y = r\sin u\sin v \hfill \\ z = r\cos u \hfill \\ \end{gathered} \right.,\quad \begin{array}{*{20}{c}} {0 \leqslant u \leqslant \pi } \\ {0 \leqslant v \leqslant 2\pi } \end{array}\]

  17. Michele_Laino
    • one year ago
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    |dw:1435852829873:dw|

  18. dan815
    • one year ago
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    are you there alien?

  19. Michele_Laino
    • one year ago
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    here is the surface we have to parametrize: |dw:1435852996763:dw|

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