How Heavy Is a Drilled Gold Wedding Band?

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SUMMARY

The weight of a drilled gold wedding band, with a width of 1 cm and a hole diameter of 2 cm, is calculated to be 2.529 grams. The calculation involves determining the volume using the formula for the volume of revolution, specifically integrating the function that represents the band’s shape. The density of gold, specified as 19.32 g/cm³, is used to convert the volume into mass. The final mass calculation confirms that the weight of the wedding band is indeed 2.529 grams.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of volume of revolution.
  • Knowledge of density and its application in mass calculations.
  • Basic geometry related to circles and cylindrical shapes.
NEXT STEPS
  • Study integration techniques for calculating volumes of solids of revolution.
  • Learn about the properties and applications of gold, including its density and uses in jewelry.
  • Explore geometric shapes and their volume formulas, particularly for cylindrical objects.
  • Investigate the relationship between mass, volume, and density in various materials.
USEFUL FOR

Mathematicians, goldsmiths, jewelry designers, and anyone interested in the physics of materials and their applications in crafting jewelry.

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A goldsmith makes a wedding band by taking a gold ball and drilling a wide hole through it. The wedding band has to be 1 cm wide and the hole has to have a diameter of 2 cm.

Given that gold weighs 19.32g/cm^3, how heavy will the wedding band be?

does 2.529 g seem like the correct answer?
 
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Haven't the faintest idea. Show your work, and we'll be the judges.

However:
For goodness sake, spare us the figures, set up and solve the problem IN SYMBOLS!
 
x^2 + y^2 = (1/2)^2 + (1)^2 = 5/4
x^2 = 5/4 - y^2

The volume of wedding band

= pi integral{-1/2 to 1/2} [(5/4 - y^2) - (1^2)] dy

= pi integral{-1/2 to 1/2} (1/4 - y^2) dy

= pi [y/4 - (1/3)y^3]

= pi /24 cm^3

mass=(pi/24)(19.32) = 2.529 g

is this correct?
 

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