How Much Paint Is Needed for a Hemispherical Dome?

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Homework Help Overview

The discussion revolves around estimating the amount of paint needed for a hemispherical dome using linear approximation. The problem specifies a dome with a diameter of 45 meters and a paint thickness of 0.1 cm, requiring participants to apply concepts from geometry and calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conversion of the diameter to radius and the implications for calculations. There are attempts to clarify the correct formulas for surface area and volume, with some questioning the use of differentials in the context of the problem.

Discussion Status

There is ongoing exploration of the correct approach to the problem, with participants providing insights into the necessary formulas and questioning assumptions about the dimensions and the type of calculation required (surface area vs. volume). Some guidance has been offered regarding the correct interpretation of the problem's parameters.

Contextual Notes

Participants note confusion regarding the diameter and radius, as well as the distinction between surface area and volume in the context of applying paint. The problem's requirements for linear approximation and the thickness of paint are also under discussion.

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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