Metric Questions Using the First Fundamental Form Consider the Poincare Disk with local coordinates $\rho=\log\frac{1+r}{1-r}$ and $\theta$, where $r$ is the radius and $\theta$ is the angle with $x$ axis. After finding the metric tensors as $g_{11}=1$, $g_{12}=g_{21}=0$, and $g_{22}=\sinh^2 \rho$, now calculate the following quantities.


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*The length of the circle $\{\rho=a\}$; My attempt is as follows. The circle can be parameterized as $s(\theta)=\left(\frac{e^a-1}{e^a+1}\cos\theta, \frac{e^a-1}{e^a+1}\sin\theta\right)$. Hence, $s'(\theta)=\left(-\frac{e^a-1}{e^a+1}\sin\theta, \frac{e^a-1}{e^a+1}\cos\theta\right)$. To this end, $$\|s'(\theta)\|=\sqrt{1\times \left(-\frac{e^a-1}{e^a+1}\sin\theta\right)^2+\sinh^2(a)\times\left(\frac{e^a-1}{e^a+1}\cos\theta\right)^2}=\frac{e^a-1}{e^a+1}\sqrt{\left(\sin\theta\right)^2+\sinh^2(a)\left(\cos\theta\right)^2}.$$ The last step is to integrate the last formula with respect to $\theta$ over the interval $[0, 2\pi]$. I have two questions here. First, is my calculation so far correct? Second, if so, how do I do the integration?

*The area of the disk $\{\rho\leq a\}$. My attempt is as follows. By the formula for area, it is enough to calculate the following integral. $$\int\int_{0\leq\rho\leq a, 0\leq\theta\leq 2\pi}\sqrt{\sinh^2(\rho)}d\theta d\rho=\int\int_{0\leq\rho\leq a, 0\leq\theta\leq 2\pi}\sinh(\rho) d\theta d\rho=2\pi[\cosh(a)-1].$$ Is this correct?
 A: Note: I didn't check your computation of the metric components, but they seem reasonable.


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*Now that you have the coordinates $\rho$ and $\theta$, you should forget about the original coordinates on the disk (at least for the purposes of answering this question). The circle $\{ \rho = a \}$ is parametrized by the curve $\gamma(t)$ defined by $\rho(t) = a$ and $\theta(t) = t$ for $0 \leq t < 2\pi$. The velocity vector for this curve is $\gamma'(t) = \rho'(t) \frac{\partial}{\partial \rho} + \theta'(t) \frac{\partial}{\partial \theta} = \frac{\partial}{\partial \theta}$, so its length is $||\gamma'(t)|| = \sqrt{g_{\theta \theta}} = \sinh a$. Integrating this expression over $0 \leq \theta \leq 2\pi$ gives that the length of the circle is $2\pi \sinh a$. (As a reality check, this expression is increasing in $a$, is zero when $a=0$, and tends to $\infty$ as $a$ tends to $\infty$, which is what we expect for the Poincare disk.)

*Your computation of the area is correct. Note that the nature of the metric in this problem  means that the area of the disk $\{ \rho \leq a \}$ should be $\int_0^a \text{length}(\rho) \, d \rho$, where $\text{length}(\rho)$ denotes the length of the circle of radius $\rho$ (which is $2 \pi \sinh \rho$, as we computed in 1). This should have been a clue that you did something wrong in part 1. (This formula for area is not true in general; here, it works because $g_{\rho \rho} = 1$ and $g_{\rho \theta} = 0$. Of course, the formula is also true in $\mathbb{R}^2$ with the Euclidean metric: $\text{Area}(\{ r \leq a \}) = \int_0^a \text{length}(r) \, dr = \int_0^a 2\pi r \, dr$. )
