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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: \begin{equation} \iiint_{\mathbb{R}^3} ||\nabla \chi_{\Omega}(\vec{x})|| d\vec{x} \tag{*}\label{*} \end{equation} 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: \begin{equation} f_n(\vec{x})=\frac{n^3}{\pi^{\frac{3}{2}}}e^{-(n{\vec{x}})^2} \end{equation} 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 $\chi_{\Omega}$: \begin{equation} [\chi_{\Omega}*f_n](\vec{x})=\int_{\mathbb R^3}\chi_{\Omega}(\vec{y})f_n(\vec{x}-\vec{y})d\vec{y} \end{equation} \begin{equation} \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} \end{equation} Therefore, using this formalism, we can define the implicit equation for the surface as: \begin{equation} h_n(\vec{x})=[\chi_{\Omega}*f_n](\vec{x})-0.5 \end{equation}

\begin{equation} \chi_{\Omega}(\vec{x})=\theta(h_n(\vec{x})) \tag{**}\label{**} \end{equation}

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

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: \begin{equation} \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{***} \end{equation} 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{**}: \begin{equation} \nabla \chi_{\Omega}(\vec{x})=\delta(h_n(\vec{x}))\nabla h_n(\vec{x}) \end{equation} Hence \ref{*} becomes: \begin{equation} \iiint_{\mathbb{R}^3} \delta(h_n(\vec{x})) ||\nabla h_n(\vec{x})|| d\vec{x} \end{equation}

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: \begin{equation} \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 \end{equation}

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

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    $\begingroup$ Check en.wikipedia.org/wiki/Stokes'_theorem $\endgroup$ – Captain Lama Apr 12 '16 at 21:30
  • $\begingroup$ I think this theorem is not useful for what I am trying to do... $\endgroup$ – Caso Apr 14 '16 at 15:56
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    $\begingroup$ 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. $\endgroup$ – Duccio Papini Apr 15 '16 at 10:18
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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 $c$.

If I understand what you're getting at, my answer to Why is the derivative of a circle's area its perimeter (and similarly for spheres)? is related, and may be of interest.

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  • $\begingroup$ 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. $\endgroup$ – Alex M. May 25 '16 at 16:12
  • $\begingroup$ @AlexM.: I don't disagree, but OP has made at least five edits (and offered the bounty) since I answered. $\endgroup$ – Andrew D. Hwang May 26 '16 at 1:54
  • $\begingroup$ True that, but even the original version (the one that Duccio Papini answered to) conveys the meaning that I am talking about. The subsequent edits didn't change the part before "---EDIT---". $\endgroup$ – Alex M. May 26 '16 at 7:34
  • $\begingroup$ @Alex: Just wanted to be clear that 1. I posted this answer before the bounty was offered. 2. This was a good faith effort to interpret the question after the first of the recent edits (which bumped the question to the front page). Obviously the indicator function of a region determines the region, and therefore the boundary, and therefore the measure of the boundary. I understood the OP to be asking about algebraic formulas, but at this point gather OP agrees with you that this answer isn't useful. The community is, of course, at liberty to downvote if that seems fit. $\endgroup$ – Andrew D. Hwang May 26 '16 at 12:57

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