# Area of surface of revolution between planes/concentric cylinders of Sphere/Pseudosphere

If area of surface of revolution with maximum radius $R,$ between two concentric cylinders radii $a,b$ is

$$2 \pi R (a-b), \tag 1$$

then find equation of its meridian.

EDIT2:

i.e., find r(z) if

$$\int_b^a 2 \pi \, \frac{r(z)} { \sin \phi} \, d r = 2 \pi R (a-b) \tag 2$$

where $\phi$ is slope of tangent to z-axis of symmetry.

EDIT1:

But if that had been between two parallel planes perpendicular to the axis of symmetry, distance (a-b) apart it would be a circle, a truncated segment of a sphere $R$, as is well known using differential equal of circle as a condition of normalcy, $r = R \cos \phi.$

$$\int_b^a 2 \pi \, r(z)\, d s = \int_b^a 2 \pi \, \frac{r \,dz} { \cos \phi} \, = 2 \pi R (z_a - z_b) = 2 \pi R \Delta z = 2 \pi R (a-b) \tag 3$$

EDIT3:

Wanted to share a result I obtained earlier in hyperbolic geometry. It is seen (2) is satisfied for a tractrix meridian $r = R \sin \phi$ as a condition of tangency , one more reason why pseudosphere can be named in this manner, apart from full area and volume equality properties of these two basic surfaces. Thus

$$\int_b^a 2 \pi \, r(z)\, d s = \int_b^a 2 \pi \, \frac{r \,dr} { \sin\phi} \, = 2 \pi R (r_a - r_b) = 2 \pi R \Delta r = 2 \pi R (a-b) \tag 4$$

$$\boxed {A_{sphere} = 2 \pi R \Delta z ,\; A_{pseudosphere} = 2 \pi R \Delta r}$$

EDIT4:

Motivating question of this post : Archimedes derived the area of the spherical truncated segment as equaling the projected lateral surface area of its touching cylinder.

What analogy enables finding the pseudosphere area in a similar way? All that we notice is that the product of cuspidal circle perimeter $2 \pi R$ and $\Delta r ,$ which is width of ring projection in the cuspidal plane.

• May I ask downvoters what is amiss here? – Narasimham Jul 4 '16 at 21:17