a derivation of the formula for arc length is simple enough:(adsbygoogle = window.adsbygoogle || []).push({});

given a function f[x], find the length of the arc from x0 to x1.

lim(x_{1}-x_{0})/n=dx

n->inf

x_{1}

[tex]S=^{i=n-1}_{i=0}\sum\sqrt{(x+(i+1)dx-(x+idx))^2+f(x+(i+1)dx)-f(x+dx))^2}[/tex]

x_{o}

[tex]S=^{i=n-1}_{i=0}\sum\sqrt{(dx)^2+f(x+(i+1)dx)-f(x+idx))^2}[/tex]

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\sqrt{dx^2+(f'(x+idx)*dx)^2}[/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,

D_______C

|...........|

|...........|

|_______|

A...........B

[tex]A=x+idx,y+jdy[/tex]

[tex]B=x+(i+1)dx,y+jdy[/tex]

[tex]C=x+(i+1)dx,y+(j+1)dy[/tex]

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

where

lim(x_{1}-x_{0})/n_{x}=dx

n_{x}->inf

lim(y_{1}-y_{0})/n_{y}=dy

n_{y}->inf

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?

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

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