Fattened volume of a curve

Let $\gamma:[0,1] \rightarrow \mathbb{R}^3$ be a smooth curve with nonvanishing velocity and let $C_\gamma = \gamma([0,1])$ be the image. Denote by $B_r = \{ x : \|x\| < r \}$ the open ball of radius $r$ centered at the origin. Then I expect the following to hold: $$\lim_{r \rightarrow 0} \frac1{r^2} \mathcal{H}^3 (C_\gamma + B_r) = \pi \cdot \mathcal{H}^1(C_\gamma)$$ where $A + B = \{ x+y : x \in A, y \in B \}$ is the Minkowski sum of two sets $A, B$ and $\mathcal{H}^d(A)$ is the $d$-dimensional Hausdorff measure of $d$.

The above setup can be easily generalised by replacing a curve with a $d$-dimensional embedded submanifold $M$ with boundary and by replacing $\mathbb R^3$ by $\mathbb R^D$: $$\lim_{r \rightarrow 0} \frac1{r^{D-d}} \mathcal{H}^D (M + B_r) = \omega_{D-d} \cdot \mathcal{H}^d(M)$$ with $\omega_k = \pi^{k/2}/\Gamma(\frac k2 + 1)$ being the volume of the $k$-dimensional unit ball.

Does anyone know if this type of result was proven elsewhere before?

As Ted Shifrin kindly pointed out, this turns out to have been researched thoroughly, starting with Weyl. The proposed formula should be true possibly with some additional conditions.

In the beginning of this paper, we see the Weyl's formula: if $M$ is a $d$-dimensional compact, connected submanifold of $\mathbb{R}^D$ with boundary and if $M^\perp _r = \{ x + y | x \in M, y \in T_x^\perp M, \|x\| \le r \}$ denotes the normal thickening of $M$, then there exist constants $k_l(M)$ such that $$ \text{Vol}(M_r^\perp) = \omega_{D-d} r^{D-d} \sum_{\substack{0 \le l \le D \\ l \text{ even} }} \frac{k_{l}(M) r^{l} }{(D-d+2) \cdots (D-d+l) } $$ where $\text{Vol}(M_r)$ is the Euclidean volume (which I believe is the Lebesgue measure). In particular, $k_0(M)$ is the Riemannian volume of $M$. Therefore, if $M_r = \{ x+y | x \in M, \|y\| \le r \}$ is the full thickening of $M$, then we can sandwich $\text{Vol}(M_r)$ as follows: $$\text{Vol}(M_r^\perp) \le \text{Vol}(M_r) \le \text{Vol}(M_r^\perp) + \text{Vol}((\partial M)_r^\perp)$$ Here, to get the second inequality, we assume that for each point $z \in \mathbb{R}^D$ there is a point $z' \in M$ such that either (1) $z' \in M$ and $z - z' \in T_{z'}^\perp M$ or (2) $z' \in \partial M$ and $z - z' \in T_{z'}^\perp (\partial M)$.

Seeing that the leading term of $\text{Vol}((\partial M)_r^\perp)$ in the Weyl's formula is of order $r^{D-d+1}$, we then see by a sandwiching argument that: $$\lim_{r \rightarrow 0} \frac{\text{Vol}(M_r)}{\omega_{D-d} r^{D-d}} = k_0(M)$$