Percent uncertainty in Volume of a beach ball?

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SUMMARY

The percent uncertainty in the volume of a spherical beach ball with a radius of r = 5.60 ± 0.05 m can be calculated using the formula for the volume of a sphere, V = (4/3)πr³. The percentage error in the volume is three times the percentage error in the radius. Given the radius uncertainty of ±0.05 m, the percent uncertainty in the volume is approximately 2.68%. This calculation is essential for understanding how measurement errors propagate in geometric contexts.

PREREQUISITES
  • Understanding of the formula for the volume of a sphere, V = (4/3)πr³
  • Basic knowledge of percentage calculations
  • Familiarity with uncertainty propagation in measurements
  • Introductory calculus concepts related to differentiation
NEXT STEPS
  • Study the concept of uncertainty propagation in physics
  • Learn how to apply calculus to derive volume formulas
  • Explore the implications of measurement errors in scientific experiments
  • Review the principles of dimensional analysis in physics
USEFUL FOR

Students in physics or engineering courses, educators teaching measurement and uncertainty, and anyone interested in the mathematical principles behind volume calculations and error analysis.

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Homework Statement


What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r=5.60±0.05m?


Homework Equations


∏r^2


The Attempt at a Solution

 
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Are you sure that pi*r^2 is the volume of a sphere?
 
I know that pi*r^2 isn't the volume of the sphere but I couldn't remember the formula for the volume of a sphere and regardless, I'm still not sure how to find the percent uncertainty, because my physics teacher isn't very clear when explaining things and it's really difficult to get the just of what he's saying.
 
As Yogi Berra would say, you could look it up.

Obviously, you could compare the volumes of a ball where the radii are 0.05 m greater or less than a radius of 5.60 m, which is more of a beach balloon. Are you sure the radius of the ball is 5.6 m?

Or, you could take the cool approach using calculus.
 
Since you say "roughly", there is a rule of thumb that when quantities multiply, their percentage errors add. Since this is a volume problem, it must involve a product of three distances (and since the radius, r, is the only distance involved, radius cubed). That is, the percentage error in the volume is three times the percentage error in the radius.
 
Accidental double post.
 

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