# Estimate Paint for Hemispherical Dome - Linear Approximation

• MHB
• josesalazmat
In summary, the conversation is about using linear approximation to estimate the amount of paint needed to apply a coat of paint 0.040000 cm thick to a hemispherical dome with a diameter of 45.000 meters. The formula used is V=2/3pir^3 and the result should be 127.17 cm ^3, but the result is showing as wrong. The speaker asks for advice and suggests using 3.14 for pi or an exact solution of $\dfrac{81\pi}{4} \, cm^3$.
josesalazmat
Hello
I have tried to resolve the problem below

Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.040000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

My procedure was:

the volume of the sphere is $$\displaystyle V=4/3 pi r^3$$ but this is a hemispherical dome, so the formula should be $$\displaystyle V=2/3pir^3$$

I derived it, so
$$\displaystyle dv=2pir^2 * dr$$

dr is 0.04000 cm
Then, the result should be 127.17 cm ^3 but this result is showed as wrong
I don't know where I am making the mistake

Thanks

josesalazmat said:
Hello
I have tried to resolve the problem below

Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.040000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

My procedure was:

the volume of the sphere is $$\displaystyle V=4/3 pi r^3$$ but this is a hemispherical dome, so the formula should be $$\displaystyle V=2/3pir^3$$

I derived it, so
$$\displaystyle dv=2pir^2 * dr$$

dr is 0.04000 cm
Then, the result should be 127.17 cm ^3 but this result is showed as wrong
I don't know where I am making the mistake

Thanks

I get 127.23 $cm^3$ ... did you use 3.14 for pi? I used my calculator's approximation for pi.

Maybe an exact solution? ... $\dfrac{81\pi}{4} \, cm^3$

Last edited by a moderator:

1)

## What is the purpose of estimating paint for a hemispherical dome using linear approximation?

The purpose of estimating paint for a hemispherical dome using linear approximation is to determine the approximate amount of paint needed to cover the surface area of the dome. This can help with planning and budgeting for a painting project.

2)

## How is linear approximation used to estimate paint for a hemispherical dome?

Linear approximation is used by dividing the dome into smaller sections, such as triangles or rectangles, and then calculating the surface area of each section. The total surface area is then estimated by adding together the surface areas of all the sections. This provides a close approximation of the actual surface area of the dome.

3)

## What factors can affect the accuracy of the estimated paint for a hemispherical dome using linear approximation?

The accuracy of the estimated paint for a hemispherical dome using linear approximation can be affected by factors such as the complexity of the dome's shape, the precision of the measurements used, and any irregularities on the surface of the dome.

4)

## Are there any limitations to using linear approximation for estimating paint for a hemispherical dome?

Yes, there are limitations to using linear approximation for estimating paint for a hemispherical dome. Linear approximation assumes that the dome can be divided into smaller, easily calculable sections, which may not always be the case. Additionally, the accuracy of the estimation may be affected by the curvature and slope of the dome.

5)

## Can linear approximation be used for any type of dome, or just hemispherical domes?

Linear approximation can be used for any type of dome, including hemispherical domes, as long as the dome's surface can be divided into smaller sections. However, the accuracy of the estimation may vary depending on the complexity of the dome's shape.

• Calculus
Replies
2
Views
4K
• Calculus
Replies
4
Views
31K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
19
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
11K
• Classical Physics
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
11K
• Calculus and Beyond Homework Help
Replies
7
Views
8K