How Much Paint Is Needed for a Hemispherical Dome?

[Weave]

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.

 

[nrqed]

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.

 

[Weave]

Oops. Well I inputed 2250cm for the radius and it is still wrong

 

[Weave]

Am I approaching this the right way?

 

[Dick]

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.

 

[HallsofIvy]

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]?

 

[HallsofIvy]

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!