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Multivariable surface integral

  1. Jun 10, 2007 #1
    a) Find the area of the part of the surface S = {x^2+ y^2+ (z-1)^2 = 4, 0 ≤ z ≤ 1}.

    Note that this is part of the sphere of radius 2 with center (0,0,1).
     
  2. jcsd
  3. Jun 10, 2007 #2
    Oh noes, a multivariable surface integral!
    Well.. the coolest way (in my opinion) to do a question like this is to use spherical co-ordinate parametrization..
    So try the substitution:
    z-1 = 2 cos phi
    x = 2 cos theta sin phi
    y = 2 sin theta sin phi
    (Since r = 2 in these cases)
    Think you can work from there?
    This should be in the homework help forum, by the way.
     
  4. Jun 10, 2007 #3
    so what should the limits of phi be?? I did it by using polar coordinates but I still got the wrong answer :mad: :mad:
    Thanks for your help though
     
  5. Jun 11, 2007 #4
    You're dealing with a sphere-- so you should be using spherical co-ordinates! (Polar would be too time-consuming)
    Obviously we're dealing with the top half, 0<z<1, thus we have the limits:
    0<theta<2pi
    0<phi<pi/2
    Since you want the surface area of the thing, you want:
    [tex]\int_{S} dS = \int_{0}^{\pi/2} \int_{0}^{2\pi} ||r_{\theta} \times r_{\phi}|| d\theta d\phi[/tex]
    Where r is the position vector corresponding to the parametrization I provided above and subscripts denote the respective partial derivatives.
     
  6. Jun 11, 2007 #5

    HallsofIvy

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    The surface area of a sphere of radius r is [itex]4\pi r^2[/itex]. What is the area of half of a sphere of radius 2?
     
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