= Solution
{c}
Use source <spherical coordinates> with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the <spherical shell> has $a\leq s\leq2a$, $\pi/2\leq\alpha\leq\pi$, and $0\leq\theta\leq2\pi$. Its <volume element> is $s^2\sin\alpha\,ds\,d\alpha\,d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos\alpha}$, so
$$
\phi(h)=-2\pi G\rho\int_a^{2a}s^2
\int_{\pi/2}^{\pi}\frac{\sin\alpha\,d\alpha}
{\sqrt{s^2+h^2-2sh\cos\alpha}}\,ds.
$$
The angular <integral> is elementary: put $q=\cos\alpha$ to obtain
$$
\int_{-1}^0\frac{dq}{\sqrt{s^2+h^2-2shq}}
=\frac{s+h-\sqrt{s^2+h^2}}{sh}.
$$
We used $\sqrt{(s+h)^2}=s+h$ since both $s,h$ are positive. Hence
$$
\begin{aligned}
\phi(h)
&=-\frac{2\pi G\rho}{h}\int_a^{2a}
\bigl(s^2+hs-s\sqrt{s^2+h^2}\bigr)\,ds\\
&=-\frac{2\pi G\rho}{h}
\left[\frac{s^3}{3}+\frac{hs^2}{2}
-\frac{(s^2+h^2)^{3/2}}{3}\right]_a^{2a}.
\end{aligned}
$$
Thus the <axial potential of a uniform hemispherical shell>, at every point of the positive $z$-axis, is
$$
\boxed{\phi(h)=-\frac{2\pi G\rho}{3h}
\left[7a^3+\frac92ha^2-(4a^2+h^2)^{3/2}+(a^2+h^2)^{3/2}\right],\qquad h>0.}
$$
There is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower <hemisphere> in all three cases. As checks, $\phi(h)\to-3\pi G\rho a^2$ as $h\downarrow0$, and at large $h$,
$$
\phi(h)\sim-\frac{G}{h}\,\frac{14\pi\rho a^3}{3}.
$$
The last factor is exactly the total <mass> of the lower half of the <spherical shell>.
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