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Given a surface $S$ in $\mathbb R^3$, what is the relationship between the gradient (when $S$ is defined as a level curve of function $F: \mathbb R^3 \to \mathbb R$) and surface area?

I noticed such a relationship when examining the surface area of the saddle surface $z = x^2 - y^2$. The surface area has integrand $\sqrt{1 + 4x^2 + 4y^2}$. And, when defined as a level curve $F(x,y,z) = x^2 - y^2 - z$, we get $\|\nabla F\| = \sqrt{1 + 4x^2 + 4y^2}$.

What is the relationship between surface area and the gradient? Does it hold for $\mathbb R^n$ or only $\mathbb R^3$?

My (incomplete) attempt is below.


Partial Solution

If $S$ is a surface defined as the level curve of $F(x,y,z)$, and $\frac {\partial F}{\partial z} = \pm 1$, then the magnitude of $\nabla F$ equals the ratio of surface area of a portion of $S$ to its projection on the $xy$ plane, and $$\text{Surface Area } = \iint \|\nabla F\| dx dy.$$

Given a surface $S$, point $v \in S$, and plane $P$, define surface amplification of $S$ over $P$ as $$\lim_{\delta \to 0} \frac{\text{surface area of }R}{\text{area of }R \text{ projected onto } P}$$ where $R$ is a region of $S$ including point $v$ with area $\delta$. (Formalizing this limit is hard.)

If $S$ is parameterized by $(x, y, Z(x,y))$, then surface amplification over the $xy$ plane equals $$\sqrt{1 + \frac{\partial Z}{\partial x}^2 + \frac{\partial Z}{\partial y}^2}$$ as can be easily seen from the standard surface area integral.

If $S$ is defined as the level surface of $F(x,y,z)$, and $\frac {\partial F}{\partial z} = \pm 1$, then surface amplification over the $xy$ plane equals $\| \nabla F \|$, which follows from the definition of gradient. (How do I show this clearly?)

I conjecture this applies for any $\mathbb R^n$. This post may be related.

Is this correct? How do we complete the proof? What is the geometric intuition behind this?

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Quoting from https://math.stackexchange.com/a/4651813/73934, the volume form of the ambient space $dx\wedge dy\wedge dz$, given a normal vector field to a surface, induces the following surface area form:

$ds=n_{x}\ dy\wedge dz+n_{y}\ dz\wedge dx+n_{z}\ dx\wedge dy=\dfrac{dy\wedge dz}{n_{x}}=\dfrac{dz\wedge dx}{n_{y}}=\dfrac{dx\wedge dy}{n_{z}}$

The last expression is particularly useful when $F(x,y,z) = z-f(x,y)$ because then $n_z\equiv\frac{1}{\|\nabla F\|}$ and $ds=\|\nabla F\|dx\wedge dy$

The geometric intuition may be that the gradient shows by how much a surface patch on a $F=const$ has been stretched compared to its $xy$ projection. Looking for intuition it helps to reduce the dimensionality. Let $F(x,y)=y-f(x)$. An infinitesimal slope length over $dx$ is $\sqrt{ dx^2 + (f_xdx)^2}=\|\nabla F\|dx.$

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