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Parameterizing and finding its boundaries (vector calculus)

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data

    I was studying for my finals, and then the only thing that I got stuck on was parameterizing a surface and finding the area of the surface.

    and my problem states that x2 + y2 + z2 = 4
    and z [tex]\geq[/tex][tex]\sqrt{2}[/tex]

    2. Relevant equations

    so when parameterizing a sphere, it comes out to be
    x = r sin[tex]\phi[/tex]cos[tex]\theta[/tex]
    y = r sin[tex]\phi[/tex]sin[tex]\theta[/tex]
    z = r cos[tex]\phi[/tex]

    3. The attempt at a solution

    so i'm sure that [tex]\theta[/tex] goes from 0 to 2[tex]\pi[/tex]. But i was getting confused what boundaries should be for [tex]\phi[/tex].

    What i tried is that
    since z [tex]\geq[/tex][tex]\sqrt{2}[/tex] and should be less than 2 since 2 is the radius, i put [tex]\sqrt{2}[/tex] [tex]\leq[/tex] z [tex]\leq[/tex] 2.
    which z = 2 cos[tex]\phi[/tex]
    so when i solved it, it came out to be [tex]\pi[/tex]/4 [tex]\leq[/tex] [tex]\phi[/tex] [tex] leq[/tex] 0.

    but it just doesn't make sense to me how the boundary can go backwards. can someone explain to me~??
  2. jcsd
  3. Sep 6, 2008 #2
    oops, last one, i meant to say

    phi is greater than or equal to pi/4
    phi is less than or equal to 0
  4. Sep 6, 2008 #3


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    Homework Helper

    When you have [tex]\cos \phi > \frac{1}{\sqrt{2}} [/tex], you should note that the graph of cos x decreases continuously from 1 to 0 in the interval 0 to pi/2. Which is why you have to reverse the inequality when taking the arc cos the interval for phi.
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