# Linear approximation of paint

1. Mar 9, 2007

### Weave

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$$Surface Area of sphere=4\pi(r^2)$$
Since it is hemipshereical, the surface area will be half
$$Surface Area of hemispherical dome=2\pi(r^2)$$
$$dSA=4\pi(r)dr$$

3. The attempt at a solution
I converted 45m into 4500cm for the radius. I set dr=.1cm

Last edited: Mar 9, 2007
2. Mar 9, 2007

### nrqed

The question says that 45 m is the diameter, not the radius.

3. Mar 9, 2007

### Weave

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

Last edited: Mar 9, 2007
4. Mar 9, 2007

### Weave

Am I approaching this the right way?

5. Mar 9, 2007

### 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.

6. Mar 9, 2007

### HallsofIvy

As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
but then say
Shouldn't it be
$$dSA= 2\pi r^2 dr$$?

7. Mar 9, 2007

### HallsofIvy

As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
but then say
You don't want Surface area, you want VOLUME. The volume of a sphere is $\frac{4}{3}\pi r^3$. The differential is $dV= \frac{4}r^2 dr$ 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!