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The question is motivated by a simple example: the area of a ring is $\pi(R^2-r^2)$, where $R$ and $r$ are the radii of the outer and inner circles respectively. Let $C$ be the 'middle circle' with radius $(R+r)/2$. The area can also be written as $\int_C (R-r) d\mathscr{H}$, where $\mathscr{H}$ is the 1-dim Hausdorff measure. The integrand $R-r$ is the distance between the outer and inner circles.

Can this be generalized in the following way: if in $\mathbb{R}^n$ we have two $(n-1)$ dimensional closed manifolds $C_1$ and $C_2$, where $C_1$ is enclosed by $C_2$, so that we have a ring-shaped region. Let's assume that $C_1$ and $C_2$ have positive reach and one manifold is within the enlargement of the other manifold by its reach. Is there a formula to express the volume of the region between two manifolds by using some Hausdorff integration on the manifolds? A naive idea: let $C$ be the 'middle manifold' between $C_1$ and $C_2$ (maybe created by medial surface?). For each point $x\in C$, let $v(x)$ be a vector orthogonal to $C$ at $x$ and $f(x)$ be the length of the part of $v(x)$ between $C_1$ and $C_2$. But in general we cannot write the volume of the region between $C_1$ and $C_2$ as $\int_C f(x) d\mathscr{H}(x)$...

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If $C_1$ and $C_2$ are the boundary components of an $\epsilon$-neighborhood of $C$, then the volume of the region is $2\epsilon$ times the volume of $C$. This is a special case of the Weyl tube formula, see e.g. the book "Tubes" by Alfred Gray.

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  • $\begingroup$ Thanks for pointing me to the right reference. I guess this should be enough for me for the question I really want to solve, even though a more general formula is also desirable, i.e. when the width of a tube is not a constant... $\endgroup$
    – Bill J
    Commented Jan 23, 2017 at 19:00

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