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Start with a sphere of radius R, then take the part of the surface that exists between any two lines of latitude. I drew a pic with two examples, one near the equator and one near the pole.

enter image description here

What is the area of this blue section of surface? I don't know what it's proper name is, maybe a band or zone.

I did search around for this but I found a site that has a suspicious answer. There, they say it is simply 2piR*h, where h is the altitude between layers.

First of all, I don't know how to determine altitude except at the equator where I can use a chord function C = 2R*sin(a/2), but you have to use twice the angle there and then take half the chord to get the altitude.

Second of all, I don't understand how it can be purely based on h and not the cosine of latitude. A band near the pole will wrap around a much smaller part of the sphere. So the position of the band must be taken into account, not just its altitude (aka thickness in the projected cylinder).

So what is a good formula to find the area of a band between two lines of latitude? And if it really is only based on altitude, how and why?

DrZ214
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    Are you comfortable with integration? – Dhanvi Sreenivasan Apr 15 '21 at 05:31
  • @DhanviSreenivasan A single integral, probably. A double integral or elliptical integral, probably not. – DrZ214 Apr 15 '21 at 05:38
  • The name is spherical zone, as the site you found says, and the result on that site is correct. Compare Area of Spherical Zone – David K Apr 15 '21 at 05:40
  • Note https://math.stackexchange.com/questions/971321/showing-that-the-surface-area-of-a-zone-of-a-sphere-that-lies-between-two-parall – copper.hat Apr 15 '21 at 06:00
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    For the "altitude" of the zone, if you can get the altitude $h_1$ of one line of latitude above the equator, and the altitude $h_2$ of another line of latitude above the equator, the altitude of the zone between those lines of altitude is $|h_1 - h_2|$. You should similarly be able to work out the altitude between lines of latitude on opposite sides of the equator or both below the equator. The "altitude" is just the distance between two parallel planes. – David K Apr 15 '21 at 06:04

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EDIT: modified as per comment. (1) Angle now taken from equator rather than the pole. (2) Modified $d(\theta)$ to $rd(\theta)$.

Let $\theta$ be the angle from equator (as per your image). You can use integration as follows:

Let $r'$ be the radius of the circle at latitude $\theta$

\begin{align*} Area &= \int\limits_{\theta = lat_1}^{lat_2} 2 \pi r' (rd(\theta))\\ &= \int\limits_{\theta = lat_1}^{lat_2} 2 \pi r \cos(\theta) (rd(\theta))\\ &= 2 \pi r^2 \int\limits_{\theta = lat_1}^{lat_2} \cos(\theta) d(\theta)\\ &= 2 \pi r^2 (-\sin(\theta)) \bigg|_{\theta=lat_1}^{lat_2}\\ &= 2 \pi r^2 (sin(lat_1)-sin(lat_2)) \end{align*} For example, for latitudes 12 and 13 degrees north, we have an area of $2 \pi r^2 (sin(13^\circ)-sin(12^\circ)) = 2\pi r^2(0.2249-0.2079) = 0.034\times \pi r^2$

The reason for the $r' = r cos(\theta)$ is the following. The circle at a latitude of $\theta$ (where angle is computed from the equator), makes an angle $\frac{\pi}{2} - \theta$ from the pole. Now consider the triangle between the center of the sphere(O), the point on the radius of the circle (R), and the point towards the pole on the circle (P). In this $\Delta OPR$, $OR$ is the radius and we want $|PR|$, where angle $POR$ is $\frac{\pi}{2} - \theta$. Notice that this is a right triangle. Therefore $\sin(\frac{\pi}{2} - \theta) = \frac{|PR|}{|OR|} = \frac{r'}{r}$. Therefore $r' = r cos(\theta)$

Rahul Madhavan
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  • I don't think this is what i want. I want a band between any two lines of latitude in one hemisphere. E.g., between 60 and 61 degrees south, or 12 and 13 degrees north. So any formula will need to have two angles. It could start at exactly 90 south, but not necessarily. Edit unless thats what you mean by "You can switch the limits of integration depending on the orientation of the surface area." I thought that was talking about some change in the cylindrical projection. – DrZ214 Apr 15 '21 at 05:47
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    @DrZ214 Note that $|\sin(lat_2)-\sin(lat_1)|$ in this answer is the same as $h$ on the page you found. – David K Apr 15 '21 at 06:07
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    The integral is $\left. 2 \pi r (\sin(\theta)) \right|_{\theta=lat_1}^{lat_2}$. – David K Apr 15 '21 at 06:08
  • @DavidK Yeah, and as you said, finding the altitude of one point is just R*sin(lat). I think i see my previous confusion too, because i was thinking in terms of degrees. 12 to 13 degrees will not have the same area as 60 to 61 degrees for the reason i said, arctic circle being much smaller than equatorial circle. But I see now their altitudes will be different anyway, even tho they are both one degree apart. – DrZ214 Apr 15 '21 at 06:38
  • The units do not match those of an area. Quite sure r has to be squared. – Dusch Jan 17 '24 at 16:40
  • Thanks Dusch. That was an oversight. – Rahul Madhavan Feb 06 '24 at 05:03