Surface area of segment of a sphere radius $a$ at the equator, between two parallels, is given by $ 2 \pi a (z_2-z_1) $,where $z_2, z_1$ are heights of spherical segment at radii of parallel circles $r_2, r_1$.
Find the corresponding formula for a tractricoid segment area if $a$ is radius at cuspidal equator .
Area= $ \int 2 \pi r ds = 2 \pi \int r . dr /sin(\phi) $
Due to tangent property: $ r/sin(\phi) = a, $
and evaluating between limits of radius $r_1$ and $r_2$
we have its area = $2 \pi a (r_2 - r_1)$
The area of tracticoid or pseudosphere is similar in appearance corresponding to that of
sphere $2 \pi a (z_2 - z_1)$ and for that reason easy to remember.
For sphere we reckon between two parallels $ \Delta z$ and for the pseudosphere,
between two concentric cylinders $ \Delta r$ .
I did not find it anywhere stated so simply in area computation of the segmented
pseudosphere, so now bringing it to your notice. Kind Regards.