Using linear approximation?

Question

Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.010000 cm thick to a hemisperical dome with a diameter of 45.000 meters.

Can anyone help with this. I started by finding the area of the hemisphere 4pi(r)^2 all divided by 2. Not sure where to go from here.
thanks

Answer 1

V =½ * 4/3 * πr³
=2/3 * πr³
V(paint) = 2/3 * πR³ - 2/3 * πr³ Where R = (r + 0.0001)m (.01 cm = 0.0001 m)ⁿ
= 2π/3 ((r + 0.0001)³ - r³)
2π/3((r³ + 3*0.0001*r²) - r³) using (r + δ)³ ≈ r³ + 3r²δ for small δ so δⁿ≈0 for n ≥ 2
= 0.0002πr² m³
= 0.0002π * 45² * 100³ cm³
≈ 28274 cm³

Answer 2

To do this properly, you'd need to find the volume of hemispheres with radii of 45m and 45.0001m, then take the difference.

To do the linear approximation, you just need to multiply your surface area by 0.01cm, to get a volume of a rectangular prism.

Answer 3

surface area multiplied by thickness of coat of paint will give you amount of paint needed. make sure you do unit convesion properly to convert m^3 into cm^3.

Answer 4

You are on the right track. Now for the linear approximation all you have to do is multiply your area by 0.01 cm and multiply by (100^2 cm^2/m^2) to convert your calculations into cm^3

The exact amount of paint needed is
(4/3)π[(4500 +0.10)^3 - 4500^3] =
(4/3)π[(4500^3 +0.01*3*4500^2 + 3*4500*0.01^2 +0.01^3 - 4500^3] =
(4/3)π[0.01*3*4500^2 + 3*4500*0.01^2 +0.01^3] =
4*π[0.01*4500^2 + 4500*0.01^2 +(1/3)*0.01^3] =
0.01*4*π*4500^2 + 4π(0.45 +0.00000033333)
which is your approximation plus a few milliliters