Understanding a step in Kirchoff's formula. Why does $\int_{B(x,t)} g(y) dS(y) = \int_{B(0,1)} g(x+tz) dS(z)$ ? In some book the solution to the I.V.P for a wave equation in 3D can be represented by the Kirchoff's formula as: 
$\newcommand{\avint}{⨍}$
$ u(x,t) = \frac{\partial}{\partial t} \big[ t \avint_{\partial B(x,t)} g(y) dS(y) \big] + t\avint_{\partial B(x,t)} h(y) dS(y) $
where dS(y) is the surface measure of the sphere $B(x,t)$. 
However, this could also be written as: 
$ u(x,t) = \frac{\partial}{\partial t} \big[ t \avint_{\partial B(0,1)} g(x+tz) dS(z) \big] + t\avint_{\partial B(0,1)} h(x+tz) dS(z) $ 
My question is how is it okay to change from $\partial B(x,t)$ to $\partial B(0,1)$ ? 
Any clarification would be greatly appreciated. 
 A: When you write$\newcommand{\avint}{⨍}$
$$u(x,t) = \frac{\partial}{\partial t} \big[ t \avint_{\partial B(x,t)} g(y) dS(y) \big] + t\avint_{\partial B(x,t)} h(y) dS(y)$$
this $dS$ is the measure on the sphere with center in $x$ and radius $t$.
When you write 
$$u(x,t) = \frac{\partial}{\partial t} \big[ t \avint_{\partial B(0,1)} g(x+tz) dS(z) \big] + t\avint_{\partial B(0,1)} h(x+tz) dS(z)$$
the $dS$ is the measure on the sphere with center in $0$ and radius $1$.
Consider the integral the integral
$$\int_{\partial B(x,t)} f(y) dS_{\partial B(x,t)}(y).$$
Make $y=z+x$, where $z\in \partial B(0,t)$. We obtain $dS_{\partial B(x,t)}(y) = dS_{\partial B(0,t)}(z)$. Therefore
$$\int_{\partial B(x,t)} f(y) dS_{\partial B(x,t)}(y)=\int_{\partial B(0,t)} f(z+x)dS_{\partial B(0,t)}(z).$$
On the other hand, make $z=tw$, where $w\in \partial B(0,1)$, we obtain $dS_{\partial B(0,t)}(z) =t^{n-1}dS_{\partial B(0,1)}(w)$ and therefore 
$$\int_{\partial B(0,t)} f(z+x)dS_{\partial B(0,t)}(z) = \int_{\partial B(0,1)} f(tw+x)t^{n-1}dS_{\partial B(0,1)}(w).$$
So, we conclude that 
$$\int_{\partial B(x,t)} f(y) dS_{\partial B(x,t)}(y)= \int_{\partial B(0,1)} f(tw+x)t^{n-1}dS_{\partial B(0,1)}(w).$$
Now from this identity we obtain
\begin{align*}\avint_{\partial B(x,t)}g(y) dS(y) &= \frac{1}{|\partial B(x,t)|}\int_{\partial B(x,t)}g(y) dS_{\partial B(x,t)}(y)
\\&= \frac{1}{t^{n-1}|\partial B(0,1)|}\int_{\partial B(x,t)}g(y) dS_{\partial B(x,t)}(y)
\\&=\frac{1}{t^{n-1}|\partial B(0,1)|}\int_{\partial B(0,1)}g(tw+x) t^{n-1} dS_{\partial B(0,1)}(w) 
\\&=\frac{1}{|\partial B(0,1)|}\int_{\partial B(0,1)}g(tw+x)  dS_{\partial B(0,1)}(w)
\\&= \avint_{\partial B(0,1)}g(tw+x) dS_{\partial B(0,1)}(w).
\end{align*}
With this is easy to see that the first formula implies the second one.
