A round-bottomed vessel is more than a cylinder. What area needs coating, and how much liquid fits? Build an ideal model, then check what changes when dimensions are known only within intervals.
Define what is being counted
Prerequisites: adding areas, disk and sector, heights and distances, volumes and sphere area.
Our open-top vessel joins a cylinder of radius r=4 cm and height h=7 cm to a lower hemisphere of the same radius. Total height is 11 cm. Ignore wall thickness for now, allowing capacity and one curved surface to use the same r and h.
The interiors do not overlap, so:
V = π(r)^(2)h + (2)/(3)π(r)^(3) = (464π)/(3) (cm)^(3).
This is about 485.90 cm³, or 0.48590 L. Coating one side of the curved wall requires S=2πrh+2πr²=88π≈276.46 cm². There is no top disk; the shared joining disk is internal and adds no exposed area. A thick shell needs separate inner and outer dimensions; coating both sides needs both surfaces.
Vessel: open top and internal join
Assumed bounds and nominal values
Separate: closed cylinders, V=128π cm³
Whether to add or subtract depends on the object. Drilling a cylindrical hole subtracts its volume from a block, but the hole’s wall adds surface area. Describe the boundary before choosing a formula.
Dimension bounds give answer bounds
Assume 3.9≤r≤4.1 cm and 6.8≤h≤7.2 cm. These are our chosen bounds, not an instrument specification or manufacturing standard. Both formulas increase with positive r and h, so both lower endpoints give minima and both upper endpoints give maxima.
Volume ranges approximately from 449.166 to 524.581 cm³. Round outward to preserve the model’s bounds: 449.16≤V≤524.59 cm³. Likewise 262.19≤S≤291.10 cm². Nominal 485.90 cm³ is one central substitution, not every vessel’s guaranteed capacity.
These are deterministic bounds under stated assumptions. Calling them a “95% confidence interval” would require a probability model. Shape error, wall thickness and uncertainty in the bounds themselves are excluded.
A reserve is a separate choice
If we choose an 8% allowance for this exercise, multiply nominal area by 1.08: about 298.58 cm². To add the same allowance above the upper model bound, use 291.10×1.08=314.388 cm² instead. These are different agreements; 8% is no recommended technical standard. Cutting layouts, seams and costs add constraints.
Optimize a specified class
Now a separate problem: a closed cylinder with two bases, fixed V=128π cm³. For numerical r,h in centimeters, h=128/r² and S/π=256/r+2r². Radii 2,4,8 give heights 32,8,2 and areas 136π,96π,160π cm². Among these three, r=4 wins; three examples do not prove a global minimum.
Branch · With formulasOptional · may be skipped
A minimum without derivatives
For every numerical r>0:
(S)/(2π) − 48 = (r)^(2) + (128)/(r) − 48 = ((r − 4)^(2)(r + 8))/(r) ≥ 0.
Equality holds only at r=4: h=8, S=96π cm². The right-hand product is nonnegative, proving the claim for all positive radii. The ratio h=2r agrees with the general closed-cylinder optimization problem.
This optimum applies to the closed cylinder, not our open hemispherical vessel. Geometry answers only after shape, surfaces, constraints and objective are specified.
Sources and limits
Volume addition and formulas: Merzliak and coauthors, 2019, sections22–23. University of Waterloo, CEMC: optimal closed cylinder; our 128π factorization is original. NIST uncertainty glossary: confidence requires distribution assumptions. The vessel, intervals and 8% allowance are chosen examples, not real manufacturing tolerances.