Curvature/surface area of a tractrix

[WendysRules]

Homework Statement

Given $ds^2 = a^2cot^2 \alpha d\alpha^2 + a^2sin^2 \alpha d\phi^2$ where a is some constant. Find:
a) The gaussian curvature
b) the surface area of the upper half of the tractrix

Homework Equations

Stoke theorem: $\int_S dF = \int_{\partial S} F$
and for curvature, we know the $\Omega^1_2 = K\omega$ where K is the gaussian curvature, and omega is the basis.

The Attempt at a Solution

I'll start with the curvature: So to get $\Omega^1_2$ I need to first find $\omega^1_2$
So, given my line element, I see that my basis is $\{ a\cot \alpha d alpha, a \sin \alpha d\phi \}$

So, my torsion conditions are: 1) $0 = d\sigma^1+\omega^1_1 \wedge \sigma^1+\omega^1_2 \wedge \sigma^2$
2) $0 = d\sigma^2+ \omega^2_1 \wedge sigma^1 + \omega^2_2 \wedge sigma^2$
Plugging in our basis and simplifying, we see that: 1) $0 = 0 + 0 + \omega^1_2 \wedge a\sin\alpha d\phi$ which forces $\omega^1_2 = b d\phi$ for some b.
Thus, 2 becomes : $0 = a\cos\alpha d\alpha \wedge d\phi + \omega^2_1 \wedge a\cot\alpha d\alpha + 0$ which becomes $0 = a\cos\alpha d\alpha \wedge d\phi + ab\cot\alpha d\alpha \wedge d\phi$ which forces $b= -sin\alpha$. Thus, $\omega^1_2 = -sin\alpha d\phi$ this makes $\Omega^1_2 = d\omega^1_2 = d(-sin\alpha d\phi) = -cos \alpha d\alpha \wedge d\phi$ since $\omega = \sigma^1 \wedge \sigma^2$ we see that $\Omega^2_1 = \frac{-cos\alpha}{a^2cot\alpha sin\alpha} d\alpha \wedge d\phi$ which would make $K = \frac{-1}{a^2}$ giving us a constant negative curvature.

b) This one, in my opinion, is more difficult. My boundaries will be $\alpha \in \left[0, \frac{\pi}{2} \right]$ and $\phi \in \left[0, 2\pi \right )$ So, for my F, I know that $F= \vec{F} \cdot d\vec {r}$ where I can either convert to Cartesian, or try to just say that $\vec{F} = F_{\alpha} \hat {\alpha} + F_{\phi} \hat {\phi}$ and $d\vec {r} = a \cot \alpha d\alpha \hat {\alpha} + a \sin \alpha d\phi \hat {\phi}$ making $F = a F_{\alpha} \cot \alpha d\alpha + a F_{\phi} \sin \alpha d\phi$ which would make $dF = \left( a \frac{\partial F_\phi}{\partial \alpha}+a F_\phi \cos \phi-a \cot \alpha \frac{\partial F_\alpha}{\partial \phi} \right) d\alpha \wedge d\phi$ using stokes tells us that our surface integral is $\int_S \left( -a \cot \alpha \frac{\partial F_\alpha}{\partial \phi} + a \frac{\partial F_\phi}{\partial \alpha}+a F_\phi \cos \phi \right) d\alpha \wedge d\phi = \int^{\frac{\pi}{2}}_{0} F_{\alpha} a \cot \alpha d\alpha + \int^{\frac{\pi}{2}}_{0} F_{\phi} a \sin \alpha d\phi$
which if I do it, I get

however, there is a problem with this, I don't know my functions true identities? So must I really go through converting to Cartesian? Or is there something I'm missing? I know in Cartersian that $x = a\sin \alpha \cos \phi$ $y = a \sin \alpha \sin \alpha$ and $z = -a \cos \alpha - ln {\tan{\frac{\alpha}{2}}}$ so I could work it out if i really had to, but I think I might be missing something obvious!

 

[mighty2000]

Thank you for your post. I appreciate your attempt at solving the given problem and your use of relevant equations and concepts. However, I would like to offer some suggestions and corrections to your solution.

Firstly, in the torsion conditions, you have used the basis $\{a\cot\alpha d\alpha, a\sin\alpha d\phi\}$, which is incorrect. The correct basis would be $\{a\cot\alpha d\alpha, a\sin\alpha d\alpha\}$. This is because in the given line element, the variables are $\alpha$ and $\phi$, not $\alpha$ and $\alpha$.

Secondly, in your attempt at finding the gaussian curvature, you have correctly found the basis, but your calculation of $\Omega^1_2$ is incorrect. It should be $\Omega^1_2 = d\omega^1_2 = d(-\sin\alpha d\phi) = -\cos\alpha d\alpha\wedge d\phi$. This leads to the correct value of $K = \frac{-1}{a^2}$.

Moving on to the calculation of surface area, I would suggest using the first fundamental form, which is given by $E = a^2\cot^2\alpha, F = 0, G = a^2\sin^2\alpha$. Using this, we can calculate the surface area of the upper half of the tractrix as follows:

$$
\begin{align}
A &= \iint_S \sqrt{EG-F^2}d\alpha\wedge d\phi \\
&= \int_0^{2\pi}\int_0^{\pi/2}\sqrt{a^4\cot^2\alpha\sin^2\alpha}d\alpha d\phi \\
&= a^2\int_0^{2\pi}\int_0^{\pi/2}\cot\alpha\sin\alpha d\alpha d\phi \\
&= a^2\int_0^{2\pi}\left[-\cos\alpha\right]_0^{\pi/2}d\phi \\
&= 2\pi a^2
\end{align}
$$

I hope this helps you in solving the given problem. Let me