Arclength, Surface Area and Volume

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I was just thinking:

If [tex]\iint dS[/tex] is the surface area of a level surface, S, and [tex]\iiint dV[/tex] is the volume of an enclosed solid, V, shouldn't [tex]\int df[/tex] be the arclength of a function f(x)?

Lets say that our surface is given implicitly by [tex]\Phi[/tex]
For the surface area we get:
[tex]\iint dS[/tex] = [tex]\int_{y_0}^{y_1}\int_{x_0}^{x_1}|\nabla\Phi|dxdy[/tex]
where [tex]|\nabla\Phi|[/tex] is the Jacobian determinant.

Now if we define function f implicitly by [tex]\alpha[/tex]
[tex]\int df[/tex] = [tex]\int_{x_0}^{x_1} |\nabla \alpha| dx[/tex]

Arclength is usually given by [tex]\int_{x_0}^{x_1} \sqrt{1+(\frac{dy}{dx})^2} dx[/tex].

This works out the same;
say our function is y=f(x) then [tex]\alpha=y-f(x)[/tex],
[tex]\nabla\alpha=(-f'(x),1,0)[/tex] whose modulus is exactly [tex]\sqrt{1+(\frac{dy}{dx})^2}[/tex]

Is this correct?
 

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  • #2
HallsofIvy
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I was just thinking:

If [tex]\iint dS[/tex] is the surface area of a level surface, S, and [tex]\iiint dV[/tex] is the volume of an enclosed solid, V, shouldn't [tex]\int df[/tex] be the arclength of a function f(x)?
Why should it? A surface has an area and a solid has a volume: both are geometric figures. A function is not a geometric figure and does not have an "arclength". If you meant "shouldn't [itex]\int d\sigma[/itex] be the arclength of the curve [itex]\sigma[/itex]", then the answer is "yes, of course".

Lets say that our surface is given implicitly by [tex]\Phi[/tex]
For the surface area we get:
[tex]\iint dS[/tex] = [tex]\int_{y_0}^{y_1}\int_{x_0}^{x_1}|\nabla\Phi|dxdy[/tex]
where [tex]|\nabla\Phi|[/tex] is the Jacobian determinant.

Now if we define function f implicitly by [tex]\alpha[/tex]
[tex]\int df[/tex] = [tex]\int_{x_0}^{x_1} |\nabla \alpha| dx[/tex]

Arclength is usually given by [tex]\int_{x_0}^{x_1} \sqrt{1+(\frac{dy}{dx})^2} dx[/tex].

This works out the same;
say our function is y=f(x) then [tex]\alpha=y-f(x)[/tex],
[tex]\nabla\alpha=(-f'(x),1,0)[/tex] whose modulus is exactly [tex]\sqrt{1+(\frac{dy}{dx})^2}[/tex]

Is this correct?
?? if y= f(x), then [itex]\alpha= y- f(x)[/itex] is identically 0 and so [itex]\nabla\alpha= 0[/itex].
 
  • #3
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Sorry about my fumble, [tex]\alpha[/tex] is an implicit function for the curve, similar to how [tex]\Phi[/tex] is an implicit function for a surface. We let [tex]\alpha[/tex] or [tex]\Phi[/tex] be equal to a constant and it kicks out a curve or surface respectively.

I was using a process analogous to:
z=z(x,y)
let [tex]\Phi=z-z(x,y)[/tex] so [tex]|\nabla \Phi|=\sqrt{1+z_x^2+z_y^2}[/tex] which is the jacobian determinant for an explicit surface. I do see your point regarding the fact that [tex]\Phi[/tex] is identically zero.

I should have called f a curve in the beginning, I must learn to be more precise.

Edit: Indeed I have put the cart before the horse, the trick is to define [tex]\Phi=z(x,y)-z[/tex] so that [tex]\Phi=0[/tex] is the level surface z=z(x,y). The rest follows from there. Similarly for the implicit curve [tex]\alpha[/tex]. Thank you for pointing out my error :smile:
 
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