Area of surface between two lines of latitude? 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.

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?
 A: EDIT: modified as per comment. Angle now taken from equator rather than the pole. Please see if this helps.
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' d(\theta)\\
&= \int\limits_{\theta = lat_1}^{lat_2} 2 \pi r \cos(\theta) d(\theta)\\
&= 2 \pi r\int\limits_{\theta = lat_1}^{lat_2}  \cos(\theta) d(\theta)\\
&= 2 \pi r (-\sin(\theta)) \bigg|_{\theta=lat_1}^{lat_2}\\
&= 2 \pi r (sin(lat_1)-sin(lat_2))
\end{align*}
For example, for latitudes 12 and 13 degrees north, we have an area of $2 \pi r (sin(13^\circ)-sin(12^\circ)) = 2\pi r(0.2249-0.2079) = 0.034\times \pi r$
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 $r' = r cos(\theta)$
