find parametrization of the surface...

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find parametrization of the surface...

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

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since we have this situation: |dw:1435851886306:dw| so theta runs from pi/3 to 2*pi/3
I called \theta with u
the radius of our sphere is: \[\Large r = \sqrt 3 \]
yes thats right spherical looks nice
|dw:1435852161889:dw|
|dw:1435852254510:dw|
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)
yes! I used (r, u, v)
should probably show him how u got the equations for (x,y,z) just so he is clear on that
|dw:1435852565980:dw|
|dw:1435852666953:dw|
by definition the spherical or polar cordinates are like below: |dw:1435852590928:dw|
from picture above, we can write: \[\Large \left\{ \begin{gathered} x = r\sin u\cos v \hfill \\ y = r\sin u\sin v \hfill \\ z = r\cos u \hfill \\ \end{gathered} \right.,\quad \begin{array}{*{20}{c}} {0 \leqslant u \leqslant \pi } \\ {0 \leqslant v \leqslant 2\pi } \end{array}\]
|dw:1435852829873:dw|
are you there alien?
here is the surface we have to parametrize: |dw:1435852996763:dw|

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