Quick math help.

[Enigma]

Let's say there's a sphere that has a diameter of 20,000km. I know the total surface area and the volume but I would like to know the surface area of a two thousand kilometer strip around its equator. How do I go about it in simple math equations?

Thanks.

[StarSword]

Clarify: Is the strip 2,000 km wide?

[Kuroneko]

That's actually one of the most surprising results of solid geometry. If you cut a sphere with two parallel planes separated by a distance h (i.e., length of segment perpendicular to them both), then the area is completely independent of where the cuts are, just on h:
[1] S = 2πRh
(Note h = 2R is a full sphere.)

So if the strip is 2000km between its edges as measured in space with a straight line, then just take h = 2000km. But I suspect you mean w = 2000km wide as measured along the surface of the sphere (e.g., by an inhabitant walking on it). Then let's find the angle θ the equator to the northern edge of the strip:
[2] Rθ = w/2.
Trig tells us that the height of this northern half-strip is:
[3] h = R(1-sin θ)
And therefore the area of the whole strip is then:
[4] S = 4πR²(1-sin(w/2R))
(Note w = πR is a full sphere.)

[Enigma]

Clarify: Is the strip 2,000 km wide?

The strip wraps around the circumference and the width of the strip is 2,000km along the surface.

[Feil]

In simple English and simpler math:

Calculate the surface area of the sphere
Calculate the ratio of the swept-out section of the circumference to the circumference (2000/[pi*diameter]).
Multiply the numbers together for the surface area of the strip.

This will work for any strip measured along the surface for which all lines that bisect it also bisect the sphere and vice versa, of which the equator-centered loop you have described is one of an infinite number of possibilities.

[Feil]

I made a stupid misspeaking in my post. Ignore the terminal paragraph. Instead of that, just use:

"This will work for a strip centered at the equator, like the one you described, or a strip that you could get by rotating that strip around."

Also note that my method is useful only if you're only interested in knowing about strips at the equator. Otherwise you will need to use Kuroneko's method. His is the general case; mine is just a special case that falls out of the calculus.

[Enigma]

I made a stupid misspeaking in my post. Ignore the terminal paragraph. Instead of that, just use:

"This will work for a strip centered at the equator, like the one you described, or a strip that you could get by rotating that strip around."

Also note that my method is useful only if you're only interested in knowing about strips at the equator. Otherwise you will need to use Kuroneko's method. His is the general case; mine is just a special case that falls out of the calculus.

Well the sphere is more or less perfect (as in as spherical as you can get). You could wrap that strip along the north-south region and the numbers would be the same. I hope I am not confusing you?

[Enigma]

In simple English and simpler math:

Calculate the surface area of the sphere
Calculate the ratio of the swept-out section of the circumference to the circumference (2000/[pi*diameter]).
Multiply the numbers together for the surface area of the strip.

This will work for any strip measured along the surface for which all lines that bisect it also bisect the sphere and vice versa, of which the equator-centered loop you have described is one of an infinite number of possibilities.

Let's see if I got this right. A 20,000km diameter sphere has a SA roughly around 1,256,000,000 km\sq. (I calculated pi at around 3.14 for simplicity's sake.).

So, #1 would be 1,256,000,000 km\sq.
#2, 2,000/3.14*20,000 = 2,000/62,800 = 0.0318 (rounded off)
#3 1,256,000,000*0.0318 = 39,940,800 km\sq?

Did I get it right?

[Enigma]

That's actually one of the most surprising results of solid geometry. If you cut a sphere with two parallel planes separated by a distance h (i.e., length of segment perpendicular to them both), then the area is completely independent of where the cuts are, just on h:
[1] S = 2πRh
(Note h = 2R is a full sphere.)

So if the strip is 2000km between its edges as measured in space with a straight line, then just take h = 2000km. But I suspect you mean w = 2000km wide as measured along the surface of the sphere (e.g., by an inhabitant walking on it). Then let's find the angle θ the equator to the northern edge of the strip:
[2] Rθ = w/2.
Trig tells us that the height of this northern half-strip is:
[3] h = R(1-sin θ)
And therefore the area of the whole strip is then:
[4] S = 4πR²(1-sin(w/2R))
(Note w = πR is a full sphere.)

That ain't simple at all.

Break it down for me as much as possible and use these numbers if it helps you.

The sphere has a diameter of 20,000 kilometers. It has a surface area of roughly 1,256,000,000 km\sq (rounding off pi at 3.14).

I'd like to know the surface area of a 2,000 kilometer strip along the circumference of the sphere (I chose the equator for simplicity's sake since it is more or less a perfect shape and not slightly egg shaped and whatnot. The strip could wrap around the north-south region and the surface area would be the same. ).

Can you show me step by step using these numbers as how to reach the answer? Thank you. It'll help me greatly this way so I'll be able to get a better grasp.

[Darth Holbytlan]

In simple English and simpler math:

Calculate the surface area of the sphere
Calculate the ratio of the swept-out section of the circumference to the circumference (2000/[pi*diameter]).
Multiply the numbers together for the surface area of the strip.

This will work for any strip measured along the surface for which all lines that bisect it also bisect the sphere and vice versa, of which the equator-centered loop you have described is one of an infinite number of possibilities.

Let's see if I got this right. A 20,000km diameter sphere has a SA roughly around 1,256,000,000 km\sq. (I calculated pi at around 3.14 for simplicity's sake.).

So, #1 would be 1,256,000,000 km\sq.
#2, 2,000/3.14*20,000 = 2,000/62,800 = 0.0318 (rounded off)
#3 1,256,000,000*0.0318 = 39,940,800 km\sq?

Did I get it right?

Your calculations are fine. Unfortunately, Feil's formula is wrong, so you didn't get anything close to the correct answer. Kuroneko's is better, but also has an error:

That's actually one of the most surprising results of solid geometry. If you cut a sphere with two parallel planes separated by a distance h (i.e., length of segment perpendicular to them both), then the area is completely independent of where the cuts are, just on h:
[1] S = 2πRh
(Note h = 2R is a full sphere.)

So if the strip is 2000km between its edges as measured in space with a straight line, then just take h = 2000km. But I suspect you mean w = 2000km wide as measured along the surface of the sphere (e.g., by an inhabitant walking on it). Then let's find the angle θ the equator to the northern edge of the strip:
[2] Rθ = w/2.
Trig tells us that the height of this northern half-strip is:
[3] h = R(1-sin θ)

This step is wrong. The correct trig formula is:
[3] h/2 = R sin θ.

(Note that the error is substituting (1-sin θ) for sin θ. The h/2 is for clarity: We're ultimately interested in the height of the whole strip, which was h in [1]. Here we calculated half that height for just the northern half of the strip.)

And therefore the area of the whole strip is then:
[4] S = 4πR²(1-sin(w/2R))
(Note w = πR is a full sphere.)

Corrected, the formula becomes:
[4] S = 4πR²sin(w/2R).

That ain't simple at all.

Break it down for me as much as possible and use these numbers if it helps you.

The sphere has a diameter of 20,000 kilometers. It has a surface area of roughly 1,256,000,000 km\sq (rounding off pi at 3.14).

I'd like to know the surface area of a 2,000 kilometer strip along the circumference of the sphere (I chose the equator for simplicity's sake since it is more or less a perfect shape and not slightly egg shaped and whatnot. The strip could wrap around the north-south region and the surface area would be the same. ).

Can you show me step by step using these numbers as how to reach the answer? Thank you. It'll help me greatly this way so I'll be able to get a better grasp.

2R is twice the radius of the sphere. Also known as the diameter.
w is the width of the strip along the surface of the sphere.
4πR² is the area of the sphere, equal to π(2R)².

In your example, 2R = 20000 and w = 2000.

The formula is S = 4πR²sin(w/2R), or the area of the sphere times the proportion of the sphere that the strip takes up.

The full area of the sphere is about 1256M km², as you calculated earlier.
The proportion of the area the strip takes up is sin(w/2R) = sin(2000/20000) = sin(1/10) ≈ 0.099833, or about 1/10. (For |x| ≪ 1, sin x ≈ x.)

So the approximate surface area is 125M km².

It's hard to clarify the proportional calculation without a diagram, but this may help with the concept: As the width of the strip increases linearly, more area is swept out at first than later, as the lines of latitude at the cut shrink in diameter. For example, at 30° N and S of the equator, only 1/3 of the total width is covered, but 1/2 the area of the earth is taken up.

The key to understanding this formula is to understand the Lambert cylindrical equal-area projection. It's based on the insight that if you take any sphere, poke holes at the poles, and stretch out those holes while keeping each line of latitude at a fixed height until the sphere is converted into a cylinder, the area of the sphere and any shape on its surface will remain unchanged. This means that you can project the band around the equator to a cylinder and calculate the area there.

[Kuroneko]

You're right. I don't know why I was calculating the distance to the north pole's plane rather than the equator; complete brain fart.