Quarter circle surface opening outward

I have a Graphics3D issue, I have a cylinder within a cylinder. I want to generate a surface from a quarter circle, but the quarter circle should be "curving outward". The normal surface generated for a quarter circle "curving inward" is a hemisphere. I do not know if this surface has a specific name but it is similar to a wormhole, although the wormhole is somehow like a hyperboloid. The surface should start from the inner cylinder and ending perpendicularly on the outer cylinder, that is why I'm not using a hyperboloid for this surface. I have drawn a quarter circle curve below, if you rotate it by $$2 \pi$$ then that is the surface I'm looking for.

rbelow = 0.5;
dbelow = {0, 0, -1};
rabove = 0.5;
dabove = {0, 0, 1};
HemisphereBelow = ParametricPlot3D[rbelow {Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]} + dbelow, {u, -\[Pi]/2, \[Pi]/2}, {v, -\[Pi]/2, \[Pi]/2}, PlotStyle -> Blue, Mesh -> None, Boxed -> False, Axes -> None][];
HemisphereAbove = ParametricPlot3D[rabove {Cos[u] Sin[v], Sin[u] Sin[v], Cos[v]} + dabove, {u, -\[Pi]/2, \[Pi]/2}, {v, \[Pi]/2, 3 \[Pi]/2}, PlotStyle -> Blue, Mesh -> None, Boxed -> False, Axes -> None][];
Graphics3D[{Opacity[0.05, Black], Cylinder[], Opacity[0.2, Green], Cylinder[{{0, 0, -1}, {0, 0, 1}}, 0.5], Opacity[0.5, Blue], HemisphereBelow, Opacity[0.5, Blue], HemisphereAbove}, Boxed -> False]  This works because the equation for the upper half of a circle of radius $$r$$ centered at $$x_0$$ is $$y = \sqrt{r^2 - (x-x_0)^2}$$. With $$r= 1/2$$ & $$x_0 = 1$$, the quarter-circle you want can be obtained by plotting this from $$1/2$$ to $$1$$. RevolutionPlot3D then takes this curve and rotates it about the $$x$$-axis.