If I have the integral $$ \int_{\partial B(x_0,r)}f(z)\,d\sigma(z) $$
over a open ball of center $x_0$ and radius $r$, how can 'move' this integral to the open ball $B(x,\tilde{r})$, where $\tilde{r}>0$, I mean
$$ \int_{\partial B(x_0,r)}f(z)\,d\sigma(z)\to\int_{\partial B(x,\tilde{r})}f(\tilde{z})\,d\sigma(\tilde{z}) $$
My approach was take the change of variable $z\to z-x_0+x$, then $$ \int_{\partial B(x_0,r)}f(z)\,d\sigma(z)\to\int_{\partial B(x,r)}f(z-x_0+x)\,d\sigma(z) $$ this change the center of the ball, but I don't know this is correct. Any hint will be appreciated. Thanks!