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Computing the surface integral of a parabloid

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Ra02W.png

    2. Relevant equations



    3. The attempt at a solution

    4pTcm.png

    I am having difficulty understanding how the author determined the limits of integration of ##R##. The author used ##\theta=\pi/3\quad to\quad \theta=\pi/2## and ##r=1\quad to\quad r=1##. More accurately, I'm not even sure how the author graphed ##R## in the diagram. Where did he get the shape of ##R## from?
     
  2. jcsd
  3. Dec 8, 2013 #2

    haruspex

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    R is the projection of the paraboloid onto the XY plane. The surface boundary is given in the question by the lines x=0, y = x√3, z=1. Since 2z = 1 + x2+y2 for the surface, that last translates into 1 = x2+y2 for R. So R's boundary consists of two straight lines and an arc of a circle.
    If θ is measured from the +ve x-axis, what are its values at the two straight lines?
     
  4. Dec 8, 2013 #3

    LCKurtz

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    Well Haruspex! I'm shocked, SHOCKED!!
     
  5. Dec 8, 2013 #4

    haruspex

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    Only if you touch the -ve at the same time?
     
  6. Dec 9, 2013 #5
    Edit: never mind.
     
    Last edited: Dec 9, 2013
  7. Dec 9, 2013 #6
    Edit: never mind again.
     
  8. Dec 9, 2013 #7
    OK, I think I got it, but that was ridiculously complicated.
     
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