1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parametrizing Surfaces and Curves

  1. May 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the surface:
    x^2 + y^2 + z^2 = 1 but x + y + z > 1 (actually greater than/equal to)

    I'd like to parametrize both this portion of the sphere and I'd like to find a parameterization of the boundary of the surface (that is, the intersection of the above sphere and plane).

    3. The attempt at a solution
    Parametrizing the surface is killing me. If the portion of the sphere were rotated so that it was pointing towards the z-axis it would be parametrized simply by polar coordinates with theta going from 0 to 2Pi and Phi going from 0 to arcsin(3^(-1/2)), but for my problem I need to compute a certain flux (i would need to transform everything, which isn't even taught in this course). The surface isn't a function of any of the coordinate axes, and spherical parameters are useless.

    The curve is also lost to me, I'm not sure how to satisfy both x^2 + y^2 + z^2 =1 and x + y + z = 1 with equations involving just one parameter.
     
  2. jcsd
  3. May 1, 2009 #2
    My two cents:

    I would parametrize it in polar coordinates. I'm not so sure if this is right but I'll give it a whirl.

    Let:
    x = rcos[tex]\Theta[/tex]
    y = rsin[tex]\Theta[/tex]
    z = [tex]\sqrt{1-r^2}[/tex]

    Thus we can write the parametrization as:

    r(r,[tex]\Theta[/tex]) = (rcos[tex]\Theta[/tex])i + (rsin[tex]\Theta[/tex])j + ([tex]\sqrt{1-r^2}[/tex])k

    I hope this helps!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Parametrizing Surfaces and Curves
  1. Parametrize this curve (Replies: 14)

  2. Parametrizing a curve (Replies: 3)

  3. Parametrizing a Curve (Replies: 0)

Loading...