I was just thinking:(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Arclength, Surface Area and Volume

**Physics Forums | Science Articles, Homework Help, Discussion**