Let $P$ be a topologically embedded submanifold in a Riemannian manifold $M$. Then the tube $T(P, r)$ of radius $r \geq 0$ about $P$ is the set of all points $m \in M$ such that there exists a geodesic $\xi$ of length $\leq r$ from $m$ that meets $P$ orthogonally.


If $M = R^n$ or $M = S^n$, Weyl gave in "The volume of tubes", 1939 a formula for the volume of $T(P, r)$ for small $r$ (see also the book of Alfred Gray "Tubes" gives lots of background and formulas related to this problem). In particular, for $M = R^n$, the volume $V\_{P} ( r )$ turns out to be a polynomial in $r$, whose coefficients carry some interesting geometric information related to intrinsic curvature (e.g. the coefficient of $r^{n}$ turns out to be of the form $f(n) \chi(P)$, where $\chi(P)$ is the Euler characteristic of $P$).


I am actually interested in $\bar{V}\_P$ defined as the volume of the r-neighbourhood: $P + B\_r$ in $R^n$, which is different from the $T(P, r)$ if P has a boundary. See [this image](https://i.stack.imgur.com/FkX6b.png) for a visualization of the difference.


A rough approximation would suggest
$$
\bar{V}\_{P} (r) = V\_{P} (r) + \frac{1}{2} V\_{\partial P} (r) + \theta ( r )
$$
for some correction function $\theta$ that is small for small $r$.


For example, if $P$ is convex, then the Steiner formula gives $\theta ( r ) = \omega\_n r^n$, where $\omega\_n$ is the volume of the $n$-dimensional 1-ball. On the other hand, $\theta ( r ) = 0$ if $P$ is half a sphere.


Does there exist a formula for $\bar{V}\_{P}$?



Tube formula for r-neighbourhood of a manifold
Apparently, the boundary corrections were computed by D. Q. Naiman - see [this paper by Catherine Loader.](http://arxiv.org/pdf/math/0511502.pdf)


