# Surface area from indicator function

I know that the volume and the surface area of a sphere of radius $R$ are related by a derivative: $$V(R)=\frac{4}{3}\pi R^3$$ $$A(R)=4\pi R^2=\frac{\partial V(R)}{\partial R}$$ I am asking if an analogous relation, in the sense that it allows to know the value of the surface from the value of the volume, exists for the indicator functions. I know the indicator function of a set $\Omega\in\mathbb{R}^n$ and $\vec{x}\in\mathbb{R}^n$ is a generic point : $$\chi_{\Omega}(\vec{x})= \begin{cases} \hfill 1 \text{ if } \vec{x}\in \Omega \\ \hfill 0 \text{ if } \vec{x}\notin \Omega \\ \end{cases}$$ the volume of $\Omega$ is easily computed:

$$V(\Omega)=\iiint_{\mathbb{R}^n} \chi_{\Omega}(\vec{x})d\vec{x}$$

Is it possible to compute the value of the surface area $A(\Omega)$ from the knowledge of $\chi(\Omega)$? Taking the derivative of $\chi_{\Omega}(\vec{x})$ I expect to have something related to the delta function. From an intuitive point of view, I expect the integral: $$\iiint_{\mathbb{R}^3} ||\nabla \chi_{\Omega}(\vec{x})|| d\vec{x} \tag{*}\label{*}$$ to be related to the surface area and this makes me think about a certain relationship. I also had a look online and in the book ''Shapes and Geometries Metrics, Analysis, Differential Calculus, and Optimization'' but I have not find anithing which solves my problem directtly. I have also thougt to use the divergence theorem but that would mean to find a field $\vec{F}$ whose divergence is $\chi$ and this is the countrary of what I am looking for by analaogy (something which allows me to compute the area from the derivative (gradient) of the volume).

Is my "intuition correct" and if yes could you give me a detailed answer or/and a good book/reference which attacks that problem directly?

---------------EDIT---------------

I reasoned a bit more on my question and I think I have found something. In particular, https://en.wikipedia.org/wiki/Surface_area remembered me that ''While for piecewise smooth surfaces there is a unique natural notion of surface area, if a surface is very irregular, or rough, then it may not be possible to assign an area to it at all.''

Then assuming to deal with a voulume $\Omega \in \mathbb{R}^n$ whose boundary $\partial \Omega$ is regular enough to have a well defined surface area, I reasoned as follow: the indicartor function is used to compute approximatively the surface area by implicitly assuming it to be smooth and calculating its derivative (which are nonvanishing only on the smooth-assumed boundary). This post Smooth approximation of characteristic function of a bounded open set gave me the idea: By seeing the indicator function $\chi_{\Omega}(\vec{x})$ as the limit of the following succession of functions: $$f_n(\vec{x})=\frac{n^3}{\pi^{\frac{3}{2}}}e^{-(n{\vec{x}})^2}$$ which has integral $1$ and approaches the Dirac delta function as $n\to \infty$. The convolution $\chi_{\Omega}*f_n$ is smooth $\forall n$ since $f_n$ is smooth and it converges everywhere to $\chi_{\Omega}$: $$[\chi_{\Omega}*f_n](\vec{x})=\int_{\mathbb R^3}\chi_{\Omega}(\vec{y})f_n(\vec{x}-\vec{y})d\vec{y}$$ $$\nabla^k_{\vec{x}}[\chi_{\Omega}*f_n](\vec{x})=\int_{\mathbb R^3}\chi_{\Omega}(\vec{y})\nabla^k_{\vec{x}}f_n(\vec{x}-\vec{y})d\vec{y}$$ Therefore, using this formalism, we can define the implicit equation for the surface as: $$h_n(\vec{x})=[\chi_{\Omega}*f_n](\vec{x})-0.5$$

$$\chi_{\Omega}(\vec{x})=\theta(h_n(\vec{x})) \tag{**}\label{**}$$

Given a 3D surface defined implicitly by $h_n(x,y,z)=0$ the normal versor to it is defined by: $$\hat{N}_n=\frac{\nabla h_n}{||\nabla h_n||}$$

For finite $n$, the vector field $\hat{N}_n$ defined here is continuous and differentiable, hence we can apply the divergence theorem using $\hat{N}_n$ as a vector field: $$\iiint_V( \nabla\cdot\hat{N_n}) \;\text{d}\tau=\iint_{\partial V} (\hat{N_n}\cdot\hat{N_n})\;\text{dS}=\iint_{\partial V} \text{dS}= A \tag{***}\label{***}$$ Therefore we are able to compute the surface area integrating over the volume the divergence of the vector field defined by the normal to the surface.

The vector field $\hat{N}_n$ defined here is continuous and differentiable in the region around the border of V for finite $n$, but as $n\to\infty$ it becomes ill defined Therefore, up to now I think that my method allows to have an approximate estimation of the area of the surface for $n$ finite, but in the limir $n\to\infty$ we have that the vector field $\hat{N}_n$ becomes ill defined and so I cannot say anything about the convergence of the area to the real value...

I am now trying to show that \ref{***} becomes \ref{*} in the limit $n\to\infty$...intuitively this seems possible...

Recalling \ref{*}, we have that, using \ref{**}: $$\nabla \chi_{\Omega}(\vec{x})=\delta(h_n(\vec{x}))\nabla h_n(\vec{x})$$ Hence \ref{*} becomes: $$\iiint_{\mathbb{R}^3} \delta(h_n(\vec{x})) ||\nabla h_n(\vec{x})|| d\vec{x}$$

Now, using the coarea formula from geometric measure theory (https://en.wikipedia.org/wiki/Dirac_delta_function): $$\int_{\mathbf{R}^n} f(\mathbf{x}) \, \delta(g(\mathbf{x})) \, d\mathbf{x} = \int_{g^{-1}(0)}\frac{f(\mathbf{x})}{|\mathbf{\nabla}g|}\,d\sigma(\mathbf{x})$$ we have: $$\iiint_{\mathbb{R}^3} \delta(h_n(\vec{x})) ||\nabla h_n(\vec{x})|| d\vec{x}=\iint_{h_n^{-1}(0)} \frac{||\nabla h_n(\vec{x})||}{||\nabla h_n(\vec{x})||}dS=\iint_{h_n^{-1}(0)} dS$$

Therefore I have proved that \ref{*} is a good definition of the surface area. Now the question is how well \ref{***} approximates the area

• – Captain Lama Apr 12 '16 at 21:30
• I think this theorem is not useful for what I am trying to do... – Caso Apr 14 '16 at 15:56
• Check Chapter 5 about functions of bounded variation in "Measure theory and fine properties of functions" by Evans & Gariepy. The function $\chi_{\Omega}$ is a function of bounded variation if and only if the set $\Omega$ has finite perimeter, which means that the distributional gradient $\nabla \chi_{\Omega}$ of $\chi_{\Omega}$ is actually a (vector) measure concentrated on $\partial \Omega$. It can be shown that $\|\nabla \chi_{\Omega}\|$ coincides with the $(n-1)$-dimensional Hausdorff measure restricted to $\partial \Omega$ when $\partial \Omega$ is smooth enough. – Duccio Papini Apr 15 '16 at 10:18

It's always risky to answer "no" to open-ended "is it possible"-type questions. That said, in the case of using the volume formula for a family of regions to deduce surface area (the way the area of a sphere of radius $r$ is the derivative with respect to $r$ of the volume of a ball of radius $r$), the answer is probably "no": Think, for example, of a non-spheroidal ellipsoid with semi-axes $a$, $b$, and $c$. Its volume is $\frac{4}{3}\pi abc$, but its surface area is a non-elementary function of $a$, $b$, and $c$.
• This does not seem to address the OP question - at least not in my reading of the question. The OP asks whether one may compute the measure of $\partial \Omega$ by some integral involving $\chi_\Omega$ (and expressions derived from it, of course). I believe that @DuccioPapini's comment answers the question very nicely. – Alex M. May 25 '16 at 16:12