Arclength, Surface Area and Volume

[Eidos]

I was just thinking:

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

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

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

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

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

Is this correct?

 

[HallsofIvy]

I was just thinking:

If \iint dS is the surface area of a level surface, S, and \iiint dV is the volume of an enclosed solid, V, shouldn't \int df 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 \int d\sigma be the arclength of the curve \sigma", then the answer is "yes, of course".

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

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

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

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

Is this correct?

?? if y= f(x), then \alpha= y- f(x) is identically 0 and so \nabla\alpha= 0.

 

[Eidos]

Sorry about my fumble, \alpha is an implicit function for the curve, similar to how \Phi is an implicit function for a surface. We let \alpha or \Phi 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 \Phi=z-z(x,y) so |\nabla \Phi|=\sqrt{1+z_x^2+z_y^2} which is the jacobian determinant for an explicit surface. I do see your point regarding the fact that \Phi 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 \Phi=z(x,y)-z so that \Phi=0 is the level surface z=z(x,y). The rest follows from there. Similarly for the implicit curve \alpha. Thank you for pointing out my error