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.


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.


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.$

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


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

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

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


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.

PS & SPH area segments

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

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