Surface Area of ball floating in water

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SUMMARY

The discussion focuses on deriving a formula for the exposed surface area of a ball floating in water using calculus. The surface area of a sphere is defined as \(A = 4\pi r^2\), but the exposed area above water varies based on the sphere's density and water oscillations. The conversation highlights that while oscillations may affect the ball's position, they do not change the surface area above water. The challenge lies in accounting for scenarios where waves may fully submerge the ball, necessitating a consideration of the spherical cap's surface area.

PREREQUISITES
  • Understanding of calculus, particularly surface area derivations
  • Familiarity with the concept of spherical caps in geometry
  • Knowledge of fluid dynamics, specifically the effects of density on buoyancy
  • Basic principles of wave mechanics and their interaction with floating objects
NEXT STEPS
  • Research the derivation of surface area formulas for spherical caps
  • Study the principles of buoyancy and Archimedes' principle
  • Explore the effects of oscillating water surfaces on floating bodies
  • Learn about wave mechanics and their impact on submerged objects
USEFUL FOR

Students and professionals in mathematics, physics, and engineering, particularly those interested in fluid dynamics and geometric applications of calculus.

Dustinsfl
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Using calculus, how would I derive a formula for the exposed surface area of a ball floating in water?

For such a formula to be a good candidate, it would have to consider oscillations of the water and placid water.

The surface area of a sphere is \(A = 4\pi r^2\).
 
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dwsmith said:
Using calculus, how would I derive a formula for the exposed surface area of a ball floating in water?

For such a formula to be a good candidate, it would have to consider oscillations of the water and placid water.

The surface area of a sphere is \(A = 4\pi r^2\).
That will depend upon the density of the sphere. And if the surface of the water oscillates, the ball will oscillate with it but I don't think that will change the amount of surface area above the water.
 
HallsofIvy said:
That will depend upon the density of the sphere. And if the surface of the water oscillates, the ball will oscillate with it but I don't think that will change the amount of surface area above the water.

Without knowing density, how could this be done?

If we think of waves, there are waves that entirely overtake objects floating in the water even if it is just for a moment. Therefore, there can be times that the ball is fully submerged. So the the amount above can be anywhere from the max (placid) to 0 rogue waves.
 
If I were to use the calculus, I would treat the problem as a surface of rotation, at least to find a formula for the surface of a spherical cap.
 

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