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

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Homework Statement



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.

Homework 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
 
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Answers and Replies

  • #2
ppy
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its sorted no worries :)
 
  • #3
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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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