Statement regarding mean value theorem for harmonic functions

Question

Suppose $u$ is a function that is harmonic on a domain $D$. Could someone offer a proof of the following statement?

$$ u(z_0) = \frac{1}{\pi r^2}\iint_{\{z-z_0\}<r} u(x+iy) \, dx\, dy $$

for $z_0 \in D$.

Answer 1

Hint. Transform the integral to polar coordinates, i. e. \begin{align*} \frac 1{\pi r^2} \int_{\{z-z_0\} < r} u(x+iy)\, d(x,y) &= \frac 1{\pi r^2} \int_0^r \int_0^{2\pi} u(z_0+\rho e^{i\theta})\,d\theta\,\rho \,d\rho \end{align*} Now use what you know (see your comment above).