# Painting a large sphere

An artist duo (Artist A and Artist B) were hired to make a large blue sphere resembling the earth. They decided to make it out of three sections, two spherical caps and one middle ring with the same width when seen with a 2D perspective (see picture).

When they were deciding how to split up the work, Artist A said he wanted to paint the two spherical caps, leaving Artist B with the middle section.

Who is getting the better deal (i.e. who gets to use less paint)? If one is painting more, how much more paint will that artist have to use?

Part 2: If they had split up the sphere into 5 parts of equal width and Artist A still got the two spherical caps while Artist B got the middle strip, then what would the answer be? (The remaining two strips would be outsourced)

Note: This will probably be answered within minutes, but for future viewers, try answering this without looking it up before checking the answer.

For part 1, I used a sphere with a diameter of 1 (so $h = \frac{1}{3}$)

Artist B (middle section) gets the better deal (paints half as much).
All 3 section areas are the same:
Cap Area is $2\pi rh = 2\pi \frac{1}{2}\frac{1}{3} = \frac{\pi}{3}$
Total Area of sphere is $4\pi r^2 = 4\pi \frac{1}{2}^2 = \pi$
So, each cap is $\frac{\pi}{3}$, meaning the middle section is $\frac{\pi}{3}$ also.

For part 2, I used a sphere with a diameter of 1 (so $h = \frac{1}{5}$)

Artist B (middle section) gets the better deal (paints half as much).
All 5 section areas are the same:
Cap Area of 2 end sections is $2\pi rh = 2\pi \frac{1}{2}\frac{2}{5} = \frac{2\pi}{5}$
Cap Area of end section is $2\pi rh = 2\pi \frac{1}{2}\frac{1}{5} = \frac{\pi}{5}$
Total Area of sphere is $4\pi r^2 = 4\pi \frac{1}{2}^2 = \pi$
So, each of the 5 parts is $\frac{\pi}{5}$, meaning the middle section is $\frac{\pi}{5}$ also.

• How is r=1/2 in the first part? – CodeNewbie Jul 10 '15 at 14:44
• @JLee Take a look at the MathJax Guide on Meta.Math – Engineer Toast Jul 10 '15 at 14:53
• Wouldn't the radius of the caps be smaller than the radius of the middle section, though? – Bailey M Jul 10 '15 at 14:54
• I think r = $\sqrt2\div3$ – CodeNewbie Jul 10 '15 at 14:55
• @BaileyM I used this link. Does it look correct to you? – JLee Jul 10 '15 at 14:55