Problem: If 2 gallons of paint are uniformly spread out over the surface of a sphere of radius 10.1 meters, how much paint is applied per square meter? If the radius doubles, by what factor does the amount per square meter change? If the radius quadruples, what is the factor? If the radius is 6 meters, what is the factor?
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Solution: To find the number of gallons per square meter, we will divide the number of gallons by the number of square meters. We know that we have 2 gallons. The area of a sphere of radius r is 4 `pi r ^ 2, so the known radius 10.1 meters implies area
area of first sphere = 1281 m ^ 2.
The number of gallons per square meter is therefore
area density of paint = 2 gallons / 1281 m ^ 2 = .001561 gal/m ^ 2.
We could follow the same procedure, using the doubled radius then the quadrupled radius to obtain the gal/m ^ 2 amounts we need to calculate the desired factors. However, it is more instructive to follow the following line of reasoning:
When the radius is doubled, the square of the radius increases by a factor of 4. This increases the area 4 `pi r ^ 2 by a factor of 4. The paint is therefore spread out over 4 times the area, and the density therefore changes by a factor of .25. The amount per unit area will therefore become
density on second sphere = .25 ( .001561 gal/m ^ 2) = .0003903 gallons/m ^ 2.
Similarly, when the radius is quadrupled, the square of the radius will increase by a factor of 16. The amount per unit area will therefore be 1/16 of the original amount, which is a factor of 1/16 = .0625. The amount per unit area will in fact be
density on third sphere = 1/16 ( .001561) gal/m^2 = 9.760 * 10 ^ -5 gal/m^2.
Finally, if the radius is 6 meters, this radius will be in a ratio ( 6/ 10.1) with the first sphere, so that the square of the radius will be ( 6/ 10.1) ^ 2 times that of the first sphere. It follows that the amount per square meter will be 1/( 6/ 10.1) ^ 2 = ||value missing for vbl19|| times as great as on the first sphere. This results in
density on final sphere = .001561/( 6/ 10.1) ^ 2 gal/m ^ 2 = .004421 gal/m ^ 2.
This is 2.833 times the original amount per square meter.
Generalized Response: The surface density of any quantity Q spread over area A is Q / A. Therefore if a quantity Q is spread over the surface of a sphere of radius r, is is spread over area A = 4 `pi r^2 and therefore has surface density `sigma = Q / (4 `pi r^2). Note: `sigma is the Greek letter often used to designate surface density.
If the same quantity Q is spread over two spheres, one with radius r1 and the other with radius r2, the two surface densities are
`sigma1 = Q / (4 `pi r1^2)
and
`sigma2 = Q / (4 `pi r2^2).
The ratio of surface densities is therefore
`sigma2 / `sigma1 = [ Q / (4 `pi r2^2) ] / [ Q / (4 `pi r1^2) ] = (r1 / r2)^2.
The surface density is therefore said to be inversely proportional to the radius of the sphere.
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Figure Description: The figure below depicts two spheres with radii r1 and r2. Their areas are 4 `pi r1^2 and 4 `pi r2^2, so the area ratio is (r2 / r1) ^ 2, as indicated. Any quantity spread over the larger sphere will be spread more thinly than the same quantity over the smaller, with a area density ratio which is the inverse of the area ratio. The area density ratio will therefore be (r1 / r2) ^ 2.
