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Vector calculus question on showing the area of a surface is infinite

  1. Oct 15, 2013 #1

    ppy

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    1. The problem statement, all variables and given/known data

    Let S be the surface z = 1/(x[itex]^{2}[/itex] + y[itex]^{2}[/itex])[itex]^{1/2}[/itex], 1 ≤ z < ∞.
    Show that the area of S is infinite.

    2. Relevant equations
    the surface S is given by z=f(x,y) with f(x,y)=1/(x[itex]^{2}[/itex]+y[itex]^{2}[/itex])[itex]^{1/2}[/itex] and for x,y in the disk D which is the circle seen when the surface is viewed from the top given by x[itex]^{2}[/itex]+y[itex]^{2}[/itex]≤ 1 z=0. Then the surface area of S is ∫∫[itex]_{S}[/itex]dS=∫∫[itex]_{D}[/itex](1+(∂f/∂x)[itex]^{2}[/itex]+(∂f/∂y)[itex]^{2}[/itex])[itex]^{1/2}[/itex]dxdy. Where has the last line come from. an explanation would be great as I cannot see where this is coming from. This is an example I have found. I am only stuck on this line.

    Thanks
     
    Last edited: Oct 15, 2013
  2. jcsd
  3. Oct 15, 2013 #2

    ppy

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    its sorted no worries :)
     
  4. Oct 15, 2013 #3

    jedishrfu

    Staff: Mentor

    If you're referring to the square root factor it comes from differential surface element's area.

    Check the wikipedia article:

    http://en.wikipedia.org/wiki/Surface_integral
     
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