- #1
okkvlt
- 53
- 0
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
[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]
xo
[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 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
[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(x1-x0)/nx=dx
nx->inf
lim(y1-y0)/ny=dy
ny->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?
given a function f[x], find the length of the arc from x0 to x1.
lim(x1-x0)/n=dx
n->inf
x1
[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]
xo
[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 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
[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(x1-x0)/nx=dx
nx->inf
lim(y1-y0)/ny=dy
ny->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?