Painting a cylinder with differentials

[mathtwit]

The interior of an open cylindrical tank is 12ft in diameter and 8 ft deep. Use differentials to find about how many gallons of waterproofing paint are needed to apply a 0.05" coat to the sides (not top or bottom) inside the tank. (1gallon is about 231 in^3)

I don't even know where to begin. I've messed with the surface area formula a little to try to derive something meaningful but I'm not getting anywhere.

 

[galactus]

Convert the ft dimensions to inches.

12 ft=144 in. , therefore radius = 72 inches

8 ft=96 in.

The surface formula=\(\displaystyle 2{\pi}rh=2{\pi}(72)(96)\)

Multiply by 1/20:

\(\displaystyle 2{\pi}(72)(96)\frac{1}{20}=\frac{3456{\pi}}{5}\;\ in^{3}\)

Dividing by 231, we get \(\displaystyle \frac{1152{\pi}}{385}\approx{9.4}\;\ gallon\)

 

[mathtwit]

Ok, I'm trying to work this using the the difference between the volume prior to painting and the volume after painting considering that dr is -0.05. (finding dV)

I should get the same answer in the end that you did though right?

I'm getting something close to 46 gal using the differential method.

 

[galactus]

The volume(in^3) before painting is \(\displaystyle {\pi}(72)^{2}(144)=746496{\pi}\)

The volume after painting is \(\displaystyle {\pi}(\frac{1439}{20})^{2}(144)=\frac

{18636489{\pi}}{25}\)

Subtract the two and get: \(\displaystyle \frac{25911{\pi}}{25}\approx{14.1}\;\ gallons\)