Help me understand the integral $\dfrac{1}{4\pi} \int_{t}^{\infty} {\dfrac{dr}{r} \int_{|x-y| = r} {\alpha(y)dS_y}}$ while studying wave equation

Question

I'm studying equations of mathematical physics, currently, the solution of Cauchy problem for wave equation using Fourier transform over $\{x \in \mathbb{R}^3,t>0\}$ with initial functions $\phi(x)$ and $\psi(x)$. The following is found. $u(x,t) = -\dfrac{\partial^2}{\partial t^2}u_{\phi}(x,t) - \dfrac{\partial}{\partial t}u_{\psi}(x,t)$, where $u_{\alpha}(x,t) \equiv \int_{\mathbb{R}^3}{K(x-y,t)\alpha(y)dy}$, where $K(z,t) = \dfrac{1}{4\pi |z|}$, if $|z| > t$; $0$, if $|z| < t$. The result for $K(x-y,t)$ is placed in $u_{\alpha}(x,t)$ and the latter is rewritten as $\dfrac{1}{4\pi} {\int_{|x-y| > t} {\dfrac{\alpha(y)}{|x-y|}dy}}$. Now, this is the step I don't understand. The latter is rewritten as $\dfrac{1}{4\pi} \int_{t}^{\infty} {\dfrac{dr}{r} \int_{|x-y| = r} {\alpha(y)dS_y}}$. I really don't understand what happened and what is $dS_y$. It's not defined in the textbook. Any help? After this Kirchhoff's formula is found.