It seems in your definition of $A$ you are counting the area of the caps (if they exist).
This should probably lead to surfaces which are CMC and have certain contact angle at the two planes which explains your solutions. I also don't think your intuition is wrong and that the solutions should look like $$ r(x)=\sqrt{(a+b)^2-x^2} $$ for $b\geq 0$ (unless somehow Delanauy surfaces are better which I would find surprising -- this should in any case be a simple, though necessary, comparison argument). This is consistent with them being pieces of spheres.

As a (somewhat lengthy) aside, the more standard way to consider this sort of problem is to not count the area of the "caps". In this case the question has been completely answered (at least for $\mathbb{R}^3$). According to page 6 of this nice survey by Ritore and Ros the relevant people are Athanassenas, Vogel, and Pedrosa and Ritore.

In this case the answer is either an appropriate region of the sphere or of the cylinder. Essentially, your conditions are of free boundary type and so the surfaces all have to meet the two planes orthogonally (as pointed out by Will). This means that you actually have a hemi-sphere up to some critical volume where there is a cylinder with the same area and enclosing the same volume and for larger volumes the cylinder is optimal.

According to the survey article this continues to hold in $\mathbb{R}^{n+1}$ for $n\leq 7$. For $n\geq 9$ this is no longer true and there are certain types of rotationally symmetric CMC surfaces (called unduloids) which are optimal for some intermediary values of the volume (i.e. small volumes are still hemi-spheres and large volumes are still cylinders). The argument for this is actually quite beautiful. Essentially the hemisphere large enough to touch one plane tangentially actually has less area than the corresponding cylinder. The regularity of the problem then implies that there has to be a better surface which must be an unduloid.

Surprisingly, this question seems to be open for $n=8$.

It seems in your definition of $A$ you are counting the area of the caps (if they exist).
This should probably lead to surfaces which are CMC and have certain contact angle at the two planes which explains your solutions. I also don't think your intuition is wrong but that the solutions should look like $$ r(x)=\sqrt{(a+b)^2-x^2} $$ for $b\geq 0$ (unless somehow Delanauy surfaces are better which I would find surprising -- this should in any case be a simple, though necessary, comparison argument). This is consistent with them being pieces of spheres.

As a (somewhat lengthy) aside, the more standard way to consider this sort of problem is to not count the area of the "caps". In this case the question has been completely answered (at least for $\mathbb{R}^3$). According to page 6 of this nice survey by Ritore and Ros the relevant people are Athanassenas, Vogel, and Pedrosa and Ritore.

In this case the answer is either an appropriate region of the sphere or of the cylinder. Essentially, your conditions are of free boundary type and so the surfaces all have to meet the two planes orthogonally (as pointed out by Will). This means that you actually have a hemi-sphere up to some critical volume where there is a cylinder with the same area and enclosing the same volume and for larger volumes the cylinder is optimal.

According to the survey article this continues to hold in $\mathbb{R}^{n+1}$ for $n\leq 7$. For $n\geq 9$ this is no longer true and there are certain types of rotationally symmetric CMC surfaces (called unduloids) which are optimal for some intermediary values of the volume (i.e. small volumes are still hemi-spheres and large volumes are still cylinders). The argument for this is actually quite beautiful. Essentially the hemisphere large enough to touch one plane tangentially actually has less area than the corresponding cylinder. The regularity of the problem then implies that there has to be a better surface which must be an unduloid.

Surprisingly, this question seems to be open for $n=8$.