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How do i generalize this result to higher dimensions? (arc length, surface area)

  1. Dec 31, 2009 #1
    a derivation of the formula for arc length is simple enough:
    given a function f[x], find the length of the arc from x0 to x1.


    by the definition of the derivative, [tex]f(x+(i+1)dx)-f(x+idx)=f'(x+idx)*dx[/tex]
    [tex]S=^{i=n-1}_{i=0}\sum dx*\sqrt{1+f'(x+idx)^2}[/tex]
    and by the definition of the integral
    [tex]^{i=n-1}_{i=0}\sum dx*\sqrt{1+f'(x+idx)^2}=\int\sqrt{1+f'(x)^2}dx[/tex]

    (the first equation uses the pythagorean theorem to estimate the length of the curve from x+idx,f(x+idx) to x+(i+1)dx,f(x+(i+1)dx).
    now heres where it gets messed up. suppose i want to find the surface area of the function f[x,y] by the same technique.
    i have a square,


    [tex]D=x+idx,y+(j+1)dy [/tex]

    anyway, to avoid a long drawn out thing that arrives to the wrong conclusion, i multiplied the distance A,f(A) to B,f(B) by the distance A,f(A) to D,f(D) and i came up with the integrand being
    [tex]\sqrt{1+(\partial f/\partial x)^2+(\partial f/\partial y)^2+(\partial f/\partial y)(\partial f/\partial x)}[/tex]

    which is wrong. How do i use the same method of finding the arc length formula to find the surface area formula?
  2. jcsd
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