Paint volume for hemispherical dome using differentials

[buffgilville]

A hemispherical dome with radius 48 feet will be given a coat of paint 0.07 inches thick. Use differentials to estimate the number of gallons of paint that will be needed to paint the dome.

Here's what I did:
f(x+h) = f '(x)h + f(x) and v=4/3pi*r^3; r=48ft.

volume of the dome:
V=4/3pi*r^3 = f(r)

I set (r+h) to be the radius of the paint volume
so, V=4/3pi*(r+h)^3 = f(r+h)

then, f(r+h) = f '(r)h + f(r)

The question just want the volume of paint needed, so I subtracted f(r)
f(r+h)=f '(r)h + f(r) - f(r), simplify

then, f(r+h)=f'(r)h

derivative, V(paint) = 4/3pi*r^3, then dv/dr(paint)=4pi*r^2

given: r=48 feet, h=0.07inches = 0.0058333feet

f(r+h)=f'(r)h => f(r+h) = 4pi*(48)^2*0.0058333

I got approximately 168.891056 gallons, but the correct answer is 631.7000268 gallons.

What did I go wrong?

 

[Hurkyl]

Have you checked all of your work? (even the problem and your starting points)


Oh, and BTW,

The question just want the volume of paint needed, so I subtracted f(r)
f(r+h)=f '(r)h + f(r) - f(r), simplify

then, f(r+h)=f'(r)h

You know full well what you have written is wrong: the left hand side of this last equation should not be f(r+h).

 

[buffgilville]

I just realized the question asks for a hemispherical dome, not a sphere. Can someone please help me with this question? Thanks

 

[Hurkyl]

Doesn't this realization suggest a change to your attempt at solving it?

 

[buffgilville]

I keep getting the wrong answer.

 

[Hurkyl]

Have you checked your arithmetic and conversions as well? For instance, I entered this into google:

4 * pi * (48 feet)^2 * (0.07 inches) in gallons

and the result was 1,263 US gallons, not then 168 you got.

 

[photon_mass]

Find the volume of the outer sphere and the inner sphere. What is the volume between the spheres? How can differentials be used to get a volume? Also, the choice of a good (bad) coordinate system will make the problem much easier (harder).
/s