What is the wave propagated away from an impulsively excited spherical shell? Consider a spherical shell of radius $R$ centered on the coordinate origin, and an impulsive excitation $\delta (t)$ distributed over its surface ('ie. a single layer'). Each point on the sphere’s surface becomes a point source, which radiates at the time zero a spherical wave given by $1/(4\pi r) \delta (t-r/c)$. $r$ is the radial distance from the point on the sphere to the point in space at which the resultant wave field $\phi$ will be considered at the time $t$. The propagation velocity is $c$. It is assumed that the sphere is transparent to radiation so that the whole sphere will contribute to the field at any given point in space.
Since the spherical source is symmetrical, at each point it suffices to compute the wave field along the radial line passing through the sphere center and the point. The wave field will be the same on any other radial line, so it can be considered as depending on only one spatial coordinate, and time, $\phi (z,t)$. The question is how to find $\phi (z,t)$.

At first sight it might seem that the impulse wave would be spherical, and centered on the origin of coordinates. However, the wave field at a point $z$ will get contributions from all the point sources lying on circular slices of the spherical shell beginning at the time $t = z/c$, see the figures, and will continue to contribute until $t = (z+2R)/c$. So it doesn't seem that one gets a 'cleanly propagating' impulsive spherical wave. The following shows how as time progresses (t0,t1,t2,t3,t4,t5,t6,...) corresponding circular slices of the impulsive spherical source contribute to the wave field at the stationary point z. So the field is stretched out in time.  Also note that with the foregoing approach the 'inward' propagating waves are included in the results at point z.

If this problem has been addressed in published papers or in books I would really appreciate the references.
 A: Update: This solves the wave equation for the wrong initial conditions.
One way of seeing the solution is to see that a spherical wave emanating from a point is a solution, and if you take a picture of that wave at any time, you will see exactly your initial condition. This means one solution is an outgoing spherical wave. Also, since the wave equation has time reversal symmetry, an ingoing spherical wave must also be a solution. Therefore any superposition is also a solution. By conformally mapping the sphere to a plane and using symmetry, you can find that the correct superposition is when the ingoing and outgoing components have the same size in terms of integrated amplitude.
A: $\newcommand{\ro}{\mathbf{r}_o}
\newcommand{\zo}{z_o}
\newcommand{\rs}{\mathbf{r}_s}
\newcommand{\zs}{z_s}$My first answer was wrong, but I think it was wrong in an instructive way so I will leave it. Here is my second attempt. I couldn't find a clever way to do it, so I did it the straightforward way. I hope it is right this time, but there may be algebra mistakes.
If the sphere of radius $R$ is $S$, then we know that the field $f(\ro)$ at some observation point $\ro$ is given by 
$$ \int_S \frac{1}{4 \pi D} \delta (D - ct) d \rs, $$
where $D=D(\ro,\rs)=|\ro-\rs|$.
To evaluate the integral, we will choose coordinates. By symmetry we may choose $\ro = (0,0,\zo)$ with $\zo>0$. 
We may also choose $\rs = (\sqrt{R^2-\zs^2} \cos \phi,\sqrt{R^2-\zs^2}\sin \phi, \zs)$, so that $d \rs = R d \phi d \zs $, and $D^2=R^2-\zs^2 + (\zo-\zs)^2$. Then the integral becomes 
$$ \int_S \frac{1}{4 \pi \sqrt{ R^2-\zs^2 + (\zo-\zs)^2} } \delta(\sqrt{ R^2-\zs^2 + (\zo-\zs)^2} -ct) R d \phi d \zs $$
Doing the $\phi$ integral gives a factor of $2 \pi$, and we can pull the $2 \pi R$ out of the integral. We are left with 
$$\frac{R}{2} \int_{-R}^{R} \frac{1}{ \sqrt{ R^2-\zs^2 + (\zo-\zs)^2} } \delta(\sqrt{ R^2-\zs^2 + (\zo-\zs)^2} -ct) d \zs $$
Next we can work on the argument of the delta function. We want to find for what values of $\zs$ we have 
\begin{equation}
\begin{aligned}
0&=\sqrt{ R^2-\zs^2 + (\zo-\zs)^2} -ct \\
0&= R^2-\zs^2 + (\zo-\zs)^2 -c^2t^2 \\
0&= R^2-\zs^2 + \zs^2-2\zs\zo+\zo^2 -c^2t^2 \\
 0&= R^2-2\zs\zo+\zo^2 -c^2t^2\\
  \zs &= \frac{R^2+\zo^2 -c^2t^2}{2\zo}.
\end{aligned}
\end{equation}
We also need to compute 
\begin{equation}
\begin{aligned}
\qquad \partial_{\zs} \left(\sqrt{ R^2-\zs^2 + (\zo-\zs)^2} -ct\right) &= \frac{-z_s + (z_s-z_0)}{\sqrt{ R^2-\zs^2 + (\zo-\zs)^2}}\\
&=\frac{-z_0}{\sqrt{ R^2-\zs^2 + (\zo-\zs)^2}}
\end{aligned}
\end{equation}
Then using $\delta(f(x)) = \delta(x)/|f'(x)|$, our integral becomes
$$ \frac{R}{2\zo} \int_{-R}^{R}  \delta\left(\zs-\frac{R^2+\zo^2 -c^2t^2}{2\zo}\right) d \zs $$
Now $\int_{-R}^{R}  \delta\left(\zs-\frac{R^2+\zo^2 -c^2t^2}{2\zo}\right) d \zs $ is either one or zero. We know from the way we picked the delta function that the integral is one when $|\zo-R|<ct<\zo+R$ and the integral is zero otherwise. Therefore $$f(z_o,t)=\begin{cases}
\frac{R}{2\zo}, & |\zo-R|<ct<\zo+R \\
0, & \textrm{otherwise.}
\end{cases}
$$
