A solution $V(\mathbf{x},t) \in C^2(\mathbb{R}^3 \times \mathbb{R}_+)$ to a certain linear hyperbolic partial differential equation can be expressed as:
$$V({\mathbf x}, t)= \frac{1}{4 \pi}\int_0^t\frac{1}{\tau}\int_{S^2(c \tau)}U(\mathbf{x-\xi},t-\tau)d\sigma_{c \tau}(\mathbf{\xi})d\tau$$
for some sufficiently smooth function $U$ on $\mathbb{R}^3 \times \mathbb{R}_+$. Here $$S^2(\tau):=\{\xi \in \mathbb{R}^3:||\xi||=\tau\}$$ is the sphere in $\mathbb{R}^3$ and $\sigma_{\tau}$ is the surface measure on $S^2(\tau)$. What I am having trouble with is the justification for converting the above iterated integral into a single volume integral though a change of variable, $r := c \tau$.
$$ \begin{aligned} V({\mathbf x}, t)= & \frac{1}{4 \pi}\int_0^{ct}\int_{S^2(r)}\frac{U(\mathbf{x-\xi},t-r/c)}{r}d\sigma_{r}(\mathbf{\xi})dr\\ =& \frac{1}{4 \pi}\int_{B^3(c \tau)}\frac{U(\mathbf{x-\xi},t-||\xi||/c)}{||\xi||}d\xi \end{aligned} $$
Where $B^3(\tau)$ is the ball at the of radius $\tau$ in Euclidean space at the origin. I know its a simple manipulation, but I am a bit confused. Is the divergence theorem being employed here?