- #1
Eidos
- 108
- 1
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?