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## Main Question or Discussion Point

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?

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?