Let $\mu(x)dx$ be a measure in $\mathbb{R}^{2n-2}$, where $\mu$ (a $C^\infty$ and positive function) is the density of the volume in the sense that $\DeclareMathOperator{\Vol}{\mathrm{Vol}} \Vol_\mu(B_r)=\int_{B_r}\mu(x)dx$, where $B_r$ be the ball of radius $r$ on $\mathbb{R}^{2n-2}$. Now suppose that $\mu(x)dx=d\Gamma$: by Stokes's theorem we get $$\Vol_\mu(B_r)=\int_{B_r}\mu(x)dx=\int_{B_r}d\Gamma=\int_{S_r}\Gamma$$ where $S_t$ is the sphere in $\mathbb{R}^{2n-3}$. The question is: can we express the area of $S_r$ as a function of the density $\mu$?
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$\begingroup$ Do you want to express the Euclidean area of $S_r$ in terms of $\mu$ or do you want to define an area density in terms of $\mu$? The former, I think it can't be done in general, the latter can be done in a multitude of ways. For instance, take the derivative of the $\mu$ volume of $B_r$. $\endgroup$– alvarezpaivaCommented Jun 20, 2021 at 18:38
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Denote by $A_R$ the area of the ball $B_R$ in $\mathbb{R}^d$ of radius $R$, and denote by $r$ the radial coordinate, then $$A_R=\int_{\mathbb{R}^{d}}\delta(r-R)\mu(\mathbf{x})\,d\mathbf{x},$$ with $\delta(r)$ the Dirac delta function.
Example in three dimensions, $d=3$, with measure $\mu(\mathbf{x})\,d\mathbf{x}=r^2 \sin\theta\, drd\theta d\phi$ in spherical coordinates,
$$A_R=\int_0^\infty \delta(r-R) r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin \theta d\theta =4\pi R^2.$$
answered Jan 20, 2021 at 14:06