Mean value theorem for a disk (complex analysis)

Question

How would I start to prove $$ f(z_0) = \frac{1}{πr^2} \int\int_{|z-z_0|<r}f(x+iy)dxdy $$ for $0<r<R$, using the Mean Value Theorem $f(z_0) = \frac{1}{2π} \int_0^{2π} f(z_0 + re^{it})dt$?

Answer 1

Evaluate the double integral using polar coordinates. $\frac 1 {\pi r^{2}} \int_{|z-z_0| <r}f(x+iy)dxdy= \frac 1 {\pi r^{2}} \int _0 ^{2\pi} \int _0^r f(z_0+se^{i\theta}) sdsd\theta$. To evaluate this multiply the second equation you have by $s$ after changing $r$ to $s$ and then integrate.