= Solution
On the boundary, integrate the expansion from part (a) over one angular period. Every nonconstant <sine> and <cosine> has zero average, so
$$
\int_0^{2\pi}f(\phi)\,d\phi=2\pi A_0.
$$
All positive-power radial terms vanish at $r=0$, leaving $\Psi(0)=A_0$. Hence
$$
\boxed{\Psi(0)=\frac1{2\pi}\int_0^{2\pi}f(\phi)\,d\phi
=\frac1{2\pi R}\int_{r=R}f\,ds}.
$$
This is the <mean value property for harmonic functions>. It is an average over arclength as well as over angle, since $ds=R\,d\phi$. The usual continuity or integrable-boundary-data hypotheses justify taking this angular average.
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