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.
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?
Note: I didn't check your computation of the metric components, but they seem reasonable.
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: $\text{Area}(\{ r \leq a \}) = \int_0^a \text{length}(r) \, dr = \int_0^a 2\pi r \, dr$. )