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

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1. anonymous

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

find the angle first

3. Michele_Laino

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

since we have this situation: |dw:1435851886306:dw| so theta runs from pi/3 to 2*pi/3

5. Michele_Laino

I called \theta with u

6. Michele_Laino

the radius of our sphere is: $\Large r = \sqrt 3$

7. dan815

yes thats right spherical looks nice

8. dan815

|dw:1435852161889:dw|

9. dan815

|dw:1435852254510:dw|

10. dan815

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

yes! I used (r, u, v)

12. dan815

should probably show him how u got the equations for (x,y,z) just so he is clear on that

13. dan815

|dw:1435852565980:dw|

14. dan815

|dw:1435852666953:dw|

15. Michele_Laino

by definition the spherical or polar cordinates are like below: |dw:1435852590928:dw|

16. Michele_Laino

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

|dw:1435852829873:dw|

18. dan815

are you there alien?

19. Michele_Laino

here is the surface we have to parametrize: |dw:1435852996763:dw|