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

[okkvlt]

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.

lim(x₁-x₀)/n=dx
n->inf

x₁
$$S=^{i=n-1}_{i=0}\sum\sqrt{(x+(i+1)dx-(x+idx))^2+f(x+(i+1)dx)-f(x+dx))^2}$$
x_o
$$S=^{i=n-1}_{i=0}\sum\sqrt{(dx)^2+f(x+(i+1)dx)-f(x+idx))^2}$$
by the definition of the derivative, $$f(x+(i+1)dx)-f(x+idx)=f'(x+idx)*dx$$
$$S=^{i=n-1}_{i=0}\sum\sqrt{dx^2+(f'(x+idx)*dx)^2}$$
$$S=^{i=n-1}_{i=0}\sum dx*\sqrt{1+f'(x+idx)^2}$$
and by the definition of the integral
$$^{i=n-1}_{i=0}\sum dx*\sqrt{1+f'(x+idx)^2}=\int\sqrt{1+f'(x)^2}dx$$

(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 here's 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

$$A=x+idx,y+jdy$$
$$B=x+(i+1)dx,y+jdy$$
$$C=x+(i+1)dx,y+(j+1)dy$$
$$D=x+idx,y+(j+1)dy$$
where
lim(x₁-x₀)/n_x=dx
n_x->inf
lim(y₁-y₀)/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
$$\sqrt{1+(\partial f/\partial x)^2+(\partial f/\partial y)^2+(\partial f/\partial y)(\partial f/\partial x)}$$

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

 

[fresh_42]

This might be helpful:
https://www.khanacademy.org/math/mu...urface-integrals-articles/a/surface-integrals