Positive Gaussian Curvature Surfaces of Revolution
📂Geometry Positive Gaussian Curvature Surfaces of Revolution Overview The surface of revolution created by the unit speed curve α ( s ) = ( r ( s ) , z ( s ) ) \boldsymbol{\alpha}(s) = \left( r(s), z(s) \right) α ( s ) = ( r ( s ) , z ( s ) ) is called M M M .
M = { ( r ( s ) cos θ , r ( s ) sin θ , z ( s ) ) : 0 ≤ θ ≤ 2 π , s ∈ ( s 0 , s 1 ) }
M = \left\{ \left( r(s)\cos\theta, r(s)\sin\theta, z(s) \right) : 0 \le \theta \le 2\pi, s \in (s_{0}, s_{1}) \right\}
M = { ( r ( s ) cos θ , r ( s ) sin θ , z ( s ) ) : 0 ≤ θ ≤ 2 π , s ∈ ( s 0 , s 1 ) }
The coordinate patch mapping x \mathbf{x} x of M M M is as follows.
x ( s , θ ) = ( r ( s ) cos θ , r ( s ) sin θ , z ( s ) )
\mathbf{x}(s, \theta) = \left( r(s)\cos\theta, r(s)\sin\theta, z(s) \right)
x ( s , θ ) = ( r ( s ) cos θ , r ( s ) sin θ , z ( s ) )
At this time, the Gaussian curvature of this surface of revolution is as follows .
K = − r ′ ′ r
K = -\dfrac{r^{\prime \prime}}{r}
K = − r r ′′
Describes the surface of revolution with the Gaussian curvature being K = − r ′ ′ r = a 2 K = -\dfrac{r^{\prime \prime}}{r} = a^{2} K = − r r ′′ = a 2 positive.
Description Based on the above assumptions, we obtain r ′ ′ + a 2 r = 0 r^{\prime \prime} + a^{2}r = 0 r ′′ + a 2 r = 0 . Since this is a second-order ordinary differential equation, the solution is as follows .
r ( s ) = a 1 cos ( a s ) + a 2 sin ( a s ) = A cos ( a s + b )
r(s) = a_{1}\cos(as) + a_{2}\sin(as) = A\cos(as + b)
r ( s ) = a 1 cos ( a s ) + a 2 sin ( a s ) = A cos ( a s + b )
Let’s denote this as b = 0 b=0 b = 0 . Then, assuming that there was r > 0 r>0 r > 0 , A > 0 A > 0 A > 0 , and therefore, ∣ s ∣ < π / 2 a \left| s \right| \lt \pi/2a ∣ s ∣ < π /2 a . Also, since z ′ = ± 1 − ( r ′ ) 2 z^{\prime} = \pm\sqrt{1 - (r^{\prime})^{2}} z ′ = ± 1 − ( r ′ ) 2 , the following theorem is obtained.
Theorem Let’s call M M M a surface of revolution created by the unit speed curve α ( s ) \boldsymbol{\alpha}(s) α ( s ) , and a constant K = a 2 K = a^{2} K = a 2 with positive Gaussian curvature . Then, α ( s ) = ( r ( s ) , z ( s ) ) \boldsymbol{\alpha}(s) = \left( r(s), z(s) \right) α ( s ) = ( r ( s ) , z ( s ) ) is given as follows.
r ( s ) = A cos ( a s ) , ∣ s ∣ < π / 2 a z ( s ) = ± ∫ 0 s 1 − a 2 A 2 sin 2 ( a t ) d t + C
\begin{align}
r(s) &= A\cos (as),\quad \left| s \right| \lt \pi/2a \nonumber \\
z(s) &= \pm \int_{0}^{s} \sqrt{1 - a^{2}A^{2} \sin^{2}(at)}dt + C
\end{align}
r ( s ) z ( s ) = A cos ( a s ) , ∣ s ∣ < π /2 a = ± ∫ 0 s 1 − a 2 A 2 sin 2 ( a t ) d t + C
Here, A > 0 , C A > 0, C A > 0 , C is a constant.
If A ≠ 1 a A \ne \dfrac{1}{a} A = a 1 , the given integral ( 1 ) (1) ( 1 ) does not appear as an elementary integral, which is called an elliptic integral .
If A = 1 a A = \dfrac{1}{a} A = a 1 , it’s simply z ( s ) = ± ∫ 0 s cos ( a t ) d t + C = ± 1 a sin ( a s ) + C z(s) = \pm {\displaystyle \int}_{0}^{s} \cos(at)dt + C = \pm\frac{1}{a}\sin (as) + C z ( s ) = ± ∫ 0 s cos ( a t ) d t + C = ± a 1 sin ( a s ) + C . Therefore, since r 2 + z 2 = 1 a 2 r^{2} + z^{2} = \frac{1}{a^{2}} r 2 + z 2 = a 2 1 , the revolution surface M M M is a sphere.
The shape of the surface of revolution depending on the value of A A A is as follows:
Also, since r = A cos ( a s ) r = A\cos(as) r = A cos ( a s ) , the matrix of metric coefficients is:
[ g i j ] = [ 1 0 0 A 2 cos 2 ( a s ) ]
\begin{bmatrix}
g_{ij}
\end{bmatrix}
= \begin{bmatrix} 1 & 0 \\ 0 & A^{2} \cos^{2}(as) \end{bmatrix}
[ g ij ] = [ 1 0 0 A 2 cos 2 ( a s ) ]