Set 7 Problem number 1

How much paint is applied per square meter if 3 gallons of paint are uniformly spread out over the surface of a sphere of radius 2.1 meters?

By what factor does the amount per square meter change if the paint is applied over a sphere of double the radius, a sphere of quadruple the radius, and a sphere of radius 12 meters?

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 3 gallons.
The area of a sphere of radius r is 4 `pi r ^ 2, so the known radius 2.1 meters implies area area of first sphere = 55.417 m ^ 2.
The number of gallons per square meter is therefore area density of paint = 3 gallons / 55.417 m ^ 2 = .0541 gal/m ^ 2.

When the radius is doubled the same amount of paint is spread over a greater area. As shown below, a doubled radius implies a quadrupled area and therefore 1/4 the density.

We could follow the previous 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 by a factor of 4, since the area 4 `pi r^2 is proportional to the square of the radius r.
The paint is therefore spread out over 4 times the area, and the amount per unit of area (the density), which is inversely proportional to the area, therefore changes by a factor of 1/4 = .25.

The amount per unit area will therefore become

density on second sphere = .25 ( .0541 gal/m ^ 2) = .01353 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 be

density on third sphere = 1/16 ( .0541) gal/m^2 = .00338 gal/m^2.

Finally, if the radius is 12 meters, this radius will be in a ratio ( 12/ 2.1) with the first sphere, so that the square of the radius will be ( 12/ 2.1) ^ 2 times that of the first sphere.
It follows that the amount per square meter will be 1/( 12/ 2.1) ^ 2 = .03 times as great as on the first sphere.

This results in density on final sphere = .0541/( 12/ 2.1) ^ 2 gal/m ^ 2 = .00165 gal/m ^ 2.

This density is .03 times the original amount per square meter.

Generalized Solution

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. The symbol for `sigma is s (may appear as small s on browsers which lack character set).

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.
This ratio (r1 / r2) ^ 2 is the reciprocal square of the ratio of the radii.
We therefore say that the surface density is inversely proportional to the radius of the sphere.

Explanation in terms of Figure(s), Extension

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.