In the proof for the mean value property for harmonic functions we start with the integral
$\frac{1}{2\pi r}\int_{\partial B(x,r)}u(y)dS(y)$
where $\partial B(x,r)$ is the boundary of the disk $B(x,r)$ with radius $r$. We're supposed to make the substitution $y=x+rz$ such that
$\frac{1}{2\pi r}\int_{\partial B(x,r)}u(y)dS(y)=\frac{1}{2\pi}\int_{\partial B(0,1)}u(x+rz)dS(z)$.
I understand most of the substitution, except for the part where $dS(y)=rdS(z)$. How do we achieve that result?
The proof is supposed to be a 2D version of the solution to exercise 1: https://www2.math.upenn.edu/~vedranso/Math425_Homework7_Solutions.pdf
In 2D, write the parametric equations for $\partial B(x,r)$ and $\partial B(0,1)$ as $$ \phi(t)=\big(x_1+r\cos(t),x_2+r\sin(t)\big),\quad t\in[0,2\pi] $$ $$ \psi(t)=\big(\cos(t),\sin(t)\big),\quad t\in[0,2\pi] $$ Then $$ \begin{align} \int_{\partial B(x,r)}u(y)\;dS(y) =&\int_0^{2\pi} u(\phi(t))|\phi'(t)|dt\\ =&\int_0^{2\pi} u(x_1+r\cos(t),x_2+r\sin(t))rdt\\ =&\int_0^{2\pi} u(x+r\psi(t))r|\psi'(t)|dt \\ =&\int_{\partial B(0,1)}u(x+rz)rdS(z)\\ \end{align} $$