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

[ppy]

Homework Statement

Let S be the surface z = 1/(x$^{2}$ + y$^{2}$)$^{1/2}$, 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$^{2}$+y$^{2}$)$^{1/2}$ and for x,y in the disk D which is the circle seen when the surface is viewed from the top given by x$^{2}$+y$^{2}$≤ 1 z=0. Then the surface area of S is ∫∫$_{S}$dS=∫∫$_{D}$(1+(∂f/∂x)$^{2}$+(∂f/∂y)$^{2}$)$^{1/2}$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

 

[ppy]

its sorted no worries :)

 

[jedishrfu]

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