What is the formula for the surface area and volume of a concave cylinder, with radial radius 'r' and longitudinal radius 'R'? To explain a bit more, the shape is bicurved, based on a cylinder with radius r but with a narrower mid-point, with a radius R of curvature to either end, . The shape is encountered in a liquid 'capillary bridge' in which a drop of liquid between two cylindrical surfaces is slightly stretched.

Any help greatly appreciated. Thanks Julian

  • $\begingroup$ By words is difficult to understand. Why don't you make a orthographic drawing (top, side view, + maybe a section) of what you are asking about ? $\endgroup$
    – G Cab
    Nov 3, 2016 at 0:15
  • $\begingroup$ Doesn't such a liquid column become a catenoid due to surface tension? $\endgroup$
    – user856
    Nov 3, 2016 at 2:01

1 Answer 1


If the cross section of the shape is always circular, then the area of each infinitesimally thin cross section is $$A(r) = \pi r^2$$ where $r = r(t)$ is measured perpendicular to the line between the axis endpoints at $t=0$ and $t=h$, so that the deflection of the centerpoint from that axis does not matter.

In other words, if we assume the shape to have its axis between $(0,0,0)$ and $(0,0,z)$, each infinitesimally thin slice is taken on the $xy$ plane, and the radius $r(z)$ measured in the $xy$ plane. It does not then matter how much the axis then deflects (in the $xy$ plane) in $0 \lt z \lt h$.

(For an analog, consider a 2D parallelogram. If $w$ is the length of two opposite sides, and $h$ is the distance between the two (height), the area is always $wh$ no matter how skewed the parallelogram is. Here, we measure the radius similar to measuring the height of the parallelogram, and axis deflection corresponds to skewing the parallelogram.)

The volume is then a simple integral: $$V = \int_{0}^{h} A(z) dz = \pi \int_{0}^{h} r(z)^2 dz$$

If the radius is symmetric with respect to the center of the bridge, and is described using function $r(t)$, with $r(0)$ being the radius at endpoints, and $r(1)$ the radius at center, the integral simplifies to $$V = \pi h \int_0^1 r(t)^2 dt$$

For example, if the radius follows a simple interpolating polynomial ($3 t^2 - 2 t^3$), $$r(t) = r_0 + 3 (r_1 - r_0) t^2 - 2 (r_1 - r_0) t^3 = 2 (r_0 - r_1) t^3 - 3 (r_0 - r_1) t^2 + r_0$$ then $$V = \frac{\pi h}{35}\left( 13 r_0^2 + 9 r_0 r_1 + 13 r_1^2 \right)$$

Another example: If the radius follows a cosine curve, then $$r(t) = r_0 + \frac{r_1 - r_0}{2} \left( 1 - \cos(\pi t) \right )$$ and $$V = \frac{\pi h}{8}\left( 3 r_0^2 + 2 r_0 r_1 + 3 r_1^2 \right)$$


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