# Arclength, Surface Area and Volume

## Main Question or Discussion Point

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
Homework Helper
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 $\nabla\alpha= 0$.

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 Last edited: