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anonymous
 one year ago
find parametrization of the surface...
anonymous
 one year ago
find parametrization of the surface...

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0spherical 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5I think that the requested parametrization is given by the subsequent formulas: \[\Large \left\{ \begin{gathered} x = \sqrt 3 \sin u\cos v \hfill \\ y = \sqrt 3 \sin u\sin v \hfill \\ z = \sqrt 3 \cos u \hfill \\ \end{gathered} \right.,\quad \begin{array}{*{20}{c}} {\frac{\pi }{3} \leqslant u \leqslant \frac{{2\pi }}{3}} \\ {0 \leqslant v \leqslant 2\pi } \end{array}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5since we have this situation: dw:1435851886306:dw so theta runs from pi/3 to 2*pi/3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5I called \theta with u

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5the radius of our sphere is: \[\Large r = \sqrt 3 \]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0yes thats right spherical looks nice

dan815
 one year ago
Best ResponseYou've already chosen the best response.0when 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)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5yes! I used (r, u, v)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0should probably show him how u got the equations for (x,y,z) just so he is clear on that

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5by definition the spherical or polar cordinates are like below: dw:1435852590928:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5from picture above, we can write: \[\Large \left\{ \begin{gathered} x = r\sin u\cos v \hfill \\ y = r\sin u\sin v \hfill \\ z = r\cos u \hfill \\ \end{gathered} \right.,\quad \begin{array}{*{20}{c}} {0 \leqslant u \leqslant \pi } \\ {0 \leqslant v \leqslant 2\pi } \end{array}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5dw:1435852829873:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5here is the surface we have to parametrize: dw:1435852996763:dw
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