# Curvature/surface area of a tractrix

1. Feb 18, 2017

### WendysRules

1. The problem statement, all variables and given/known data
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

2. Relevant 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.

3. 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!

2. Feb 23, 2017