I saw on this wikipedia page the following expression:
$$-\int_{\mathbb{R}^{n}} g(\mathbf{x}) \frac{\partial 1_{D}(\mathbf{x})}{\partial n} d \mathbf{x}=\int_{S} g(\mathbf{s}) d \sigma(\mathbf{s})$$
where $\sigma(s)$ is the Minkowski content, $n$ is the outward normal, $g$ is a smooth function, $D$ is a domain in $\mathbb{R}^N$, $S$ is the boundary of $D$ and $1_{D}$ is the indicator function on $D$.
I have an integral of the form:
$$\int_{f^{-1}(0)} \frac{1}{|\nabla f(\mathbf{s})|} d \sigma(\mathbf{s})$$
I want to write my integral above as with respect to $dx$ rather than $d\sigma(s)$ and thought I could do this with the wikipedia expression. By plugging my integral into the wikipedia expression I got this:
$$\int_{f^{-1}(0)} \frac{1}{|\nabla f(\mathbf{s})|} d \sigma(\mathbf{s})=-\int_{\mathbb{R}^{n}} \frac{1}{|\nabla f(\mathbf{x})|} \frac{\partial 1[f(\mathbf{x})>0]}{\partial n} d \mathbf{x}$$
I got this via chain rule:
$$=-\int_{\mathbb{R}^{n}} \frac{1}{|\nabla f(\mathbf{x})|} \frac{\partial 1[f(x)>0]}{\partial f(x)}\frac{\partial f(x)}{\partial n} d \mathbf{x}$$
The first derivative became the dirac delta function:
$$=-\int_{\mathbb{R}^{n}} \frac{\delta(f(x))}{|\nabla f(\mathbf{x})|} \frac{\partial f(x)}{\partial n} d \mathbf{x}$$
By the definition of gradient I got this:
$$=-\int_{\mathbb{R}^{n}} \frac{\delta(f(x))}{|\nabla f(\mathbf{x})|} (\nabla f(x)\cdot n) d \mathbf{x}$$
As $n$ is the outward normal vector (I assume normalized? I couldn't tell from wikipedia), I got this:
$$=-\int_{\mathbb{R}^{n}} \frac{\delta(f(x))}{|\nabla f(\mathbf{x})|} \bigg(\nabla f\cdot \frac{-\nabla f(x)}{|\nabla f(x)|}\bigg) d \mathbf{x}$$
Simplifying, I got this:
$$=\int_{\mathbb{R}^{n}} \delta(f(x))\frac{|\nabla f(\mathbf{x})|^2}{|\nabla f(\mathbf{x})|^2} d \mathbf{x}$$
And then finally this:
$$=\int_{f^{-1}(0)} \delta(f(x)) d \mathbf{x}$$
It seems to me like something has gone wrong here. Please could someone tell me if my working is correct? I don't fully understand what the vector $n$ is in this article and guessed its meaning above.