How Much Paint Is Needed for a Hemispherical Dome?

In summary, to estimate how much paint is needed to cover a 45m diameter hemispherical dome with a 0.100000 cm thickness, you would need to use the following linear approximation: 4500 cm * .100000 cm = 4500 cm3.
  • #1
Weave
143
0

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


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

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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  • #2
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


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


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.
 
  • #3
Oops. Well I inputed 2250cm for the radius and it is still wrong
 
Last edited:
  • #4
Am I approaching this the right way?
 
  • #5
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.
 
  • #6
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


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


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
[tex]Surface Area of hemispherical dome=2\pi(r^2)[/tex]
but then say
[tex]dSA=4\pi(r)dr[/tex]
Shouldn't it be
[tex]dSA= 2\pi r^2 dr[/tex]?
 
  • #7
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


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


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
[tex]Surface Area of hemispherical dome=2\pi(r^2)[/tex]
but then say
[tex]dSA=4\pi(r)dr[/tex]
You don't want Surface area, you want VOLUME. The volume of a sphere is [itex]\frac{4}{3}\pi r^3[/itex]. The differential is [itex]dV= \frac{4}r^2 dr[/itex] 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!
 

FAQ: How Much Paint Is Needed for a Hemispherical Dome?

What is linear approximation of paint?

The linear approximation of paint is a mathematical technique used to estimate the amount of paint needed for a surface or object based on its dimensions and the desired thickness of the paint layer. It assumes that the paint will be evenly distributed and that the surface is flat.

How is linear approximation of paint calculated?

To calculate the linear approximation of paint, you will need to know the area of the surface or object to be painted, the desired thickness of the paint layer, and the coverage rate of the paint. The formula is: Total paint needed = (Area * Desired thickness) / Coverage rate.

What are the limitations of linear approximation of paint?

Linear approximation of paint is not always accurate as it assumes a flat surface and even distribution of paint, which may not be the case in real life. It also does not account for wastage or loss of paint during application. Additionally, different surfaces may require different amounts of paint due to their porosity or texture.

Can linear approximation of paint be used for different types of paint?

Yes, linear approximation of paint can be used for different types of paint as long as the coverage rate for each type is known. However, different types of paint may have different coverage rates, so it is important to use the correct rate for accurate calculations.

Are there any alternative methods to linear approximation of paint?

Yes, there are other methods for estimating the amount of paint needed, such as the area-volume method which takes into account the shape of the surface or object. Some paint manufacturers also provide online calculators or charts to help estimate the amount of paint needed for a specific project.

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