



 
Richard, I drew the crucible upright to allow me to use the standard form of the mathematical expression for a cylinder. The radius of the crucible is r units and the height h units. The plane that contains the surface of the liquid just before it is going to pour makes an angle of s degrees with the XY plane. A more natural angle fo you might be 90  s degrees. The equation of this plane is
The volume of liguid in the crucible is the volume inside the cylinder and between this plane and the XY plane. This volume can then be found using a double integral,
This expression is valid as long as the low side of the liquid surface meets the cylindrical side of the crucible, that is for 0 ≤ s ≤ tan^{1}(h/2r). For tan^{1}(h/2r) ≤ s ≤ 90^{o} the low side of the liquid surface meets the bottom of the crucible. The volume of liquid in the crucible in this case is volume = cubic units. This is not a particularly nice expression but for a specific crucible (known values of r and h) it is possible to construct a table that gives the volume for a range of values of the angle s. Harley  


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