= Solution
Use <Green's second identity> in the source coordinate:
$$
\int_\Omega(u\Delta v-v\Delta u)dV'=\int_{\partial\Omega}(u\partial_n v-v\partial_n u)dS'.
$$
Take $u(\mathbf x')=\psi(\mathbf x')$ and $v(\mathbf x')=-1/[4\pi|\mathbf x-\mathbf x'|]$, the <fundamental solution of the Laplace equation> satisfying $\Delta'v=\delta(\mathbf x'-\mathbf x)$. With localized charge and the decaying potential, the boundary integral tends to zero on expanding spheres. The volume identity then gives $\psi(\mathbf x)+(1/\epsilon_0)\int vq\,dV'=0$, hence the <Coulomb potential>
$$
\boxed{\psi(\mathbf x)=\frac1{4\pi\epsilon_0}\int_{\mathbb R^3}\frac{q(\mathbf x')}{|\mathbf x-\mathbf x'|}dV'.}
$$
Convergence of this integral and a condition fixing the potential at infinity are needed: mere pointwise decay of $q$ is not enough, and a harmonic function could otherwise be added. A bounded, compactly supported charge distribution is a sufficient setting for this derivation.
For the disk put $\sigma=Q/(\pi a^2)$. On its axis, every ring of radius $\rho$ is at distance $\sqrt{\rho^2+z^2}$, so
$$
\boxed{\psi(0,z)=\frac{\sigma}{2\epsilon_0}\int_0^a\frac{\rho\,d\rho}{\sqrt{\rho^2+z^2}}=\frac{Q}{2\pi\epsilon_0a^2}(\sqrt{a^2+z^2}-|z|).}
$$
For $z>0$ subtract $z$; for $z<0$ add $z$. This is the <electrostatic potential of a uniformly charged disk> with zero potential at infinity. The absolute value is essential for reflection symmetry and the correct behavior on the lower axis.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2-charged-disk-axis.png]
{title=The potential is continuous across a charged disk while its normal field jumps}
{description=Original axis potential and normal electric field for a uniform positive surface charge. Open circles at the center show the two limiting fields, not a single field value assigned on the ideal sheet.}
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