How Much Paint Is Needed for a Hemispherical Dome?

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SUMMARY

The discussion focuses on calculating the volume of paint required for a hemispherical dome with a diameter of 45 meters, using linear approximation. The correct radius for calculations is 2250 cm, derived from the diameter. The surface area formula for a hemispherical dome is confirmed as 2π(r²), while the volume differential is expressed as dV = (4/3)πr³. Participants clarify that the volume change should be calculated, not the surface area, emphasizing the importance of using the correct formulas for accurate results.

PREREQUISITES
  • Understanding of linear approximation in calculus
  • Familiarity with the formulas for surface area and volume of spheres
  • Ability to convert units from meters to centimeters
  • Knowledge of differential calculus concepts
NEXT STEPS
  • Review the concept of linear approximation in calculus
  • Study the formulas for the volume of a hemisphere and its differential
  • Practice unit conversion between metric measurements
  • Explore applications of differential calculus in real-world problems
USEFUL FOR

Students in mathematics or engineering fields, particularly those studying calculus and geometry, as well as professionals involved in construction or design of hemispherical structures.

Weave
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Homework Statement


Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.100000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

Homework Equations


Surface Area of sphere=4\pi(r^2)
Since it is hemipshereical, the surface area will be half
Surface Area of hemispherical dome=2\pi(r^2)
dSA=4\pi(r)dr

The Attempt at a Solution


I converted 45m into 4500cm for the radius. I set dr=.1cm
and the radius to 4500cm.
 
Last edited:
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Weave said:

Homework Statement


Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.100000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

Homework Equations


Surface Area of sphere=4\pi(r^2)
Since it is hemipshereical, the surface area will be half
Surface Area of hemispherical dome=2\pi(r^2)
dSA=4\pi(r)dr


The Attempt at a Solution


I converted 45m into 4500cm for the radius. I set dr=.1cm
and the radius to 4500cm.

The question says that 45 m is the diameter, not the radius.
 
Oops. Well I inputed 2250cm for the radius and it is still wrong
 
Last edited:
Am I approaching this the right way?
 
This time the problem is to determine a volume. So you want to estimate the change in volume if a hemisphere grows from diameter 45m to 45.002m.
 
Weave said:

Homework Statement


Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.100000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

Homework Equations


Surface Area of sphere=4\pi(r^2)
Since it is hemipshereical, the surface area will be half
Surface Area of hemispherical dome=2\pi(r^2)
dSA=4\pi(r)dr


The Attempt at a Solution


I converted 45m into 4500cm for the radius. I set dr=.1cm
and the radius to 4500cm.
As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
Since it is hemipshereical, the surface area willbe half
Surface Area of hemispherical dome=2\pi(r^2)
but then say
dSA=4\pi(r)dr
Shouldn't it be
dSA= 2\pi r^2 dr?
 
Weave said:

Homework Statement


Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.100000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

Homework Equations


Surface Area of sphere=4\pi(r^2)
Since it is hemipshereical, the surface area will be half
Surface Area of hemispherical dome=2\pi(r^2)
dSA=4\pi(r)dr


The Attempt at a Solution


I converted 45m into 4500cm for the radius. I set dr=.1cm
and the radius to 4500cm.
As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
Since it is hemipshereical, the surface area willbe half
Surface Area of hemispherical dome=2\pi(r^2)
but then say
dSA=4\pi(r)dr
You don't want Surface area, you want VOLUME. The volume of a sphere is \frac{4}{3}\pi r^3. The differential is dV= \frac{4}r^2 dr which is exactly the same as the surface area times the "thickness" dr. I thought that was what you were doing when you quoted the formula for surface area!
 

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