Computing the area and length of a curve 
Using the Riemannian hyperbolic metric $$g = \frac{4}{(1-(u^2+v^2))^2}\pmatrix{ 1 & 0 \\ 0 & 1 \\}$$ on the disk $D_p = \{(u,v)\ | \ u^2 +v^2 \le p^2\}$ compute the area of $D_p$ and the length of the curve $\partial D_p$.

 A: You don't need to come up with a map $f: D_p \to \mathbb S^2$. "Normally", if you are given  a regular patch $f : U\to \mathbb R^3$, you have the first fundamental form 
$$g = \pmatrix{ f_u\cdot f_u & f_u\cdot f_v \\ f_v\cdot f_u & f_v\cdot f_v \\}$$
Using the first fundamental form alone (without using the second fundamental form), you can calculate the length and area. 
Now we take an abstract approach: instead of specifying a map $f$, we define directly on $U$ a first fundamental form 
$$\frac{4}{(1-(u^2 + v^2))^2}\pmatrix{ 1 & 0 \\ 0 & 1 \\}$$ 
Then one can use this to calculate length and area. For example, 
$$\sqrt{\det (g_{ij})} dudv = \frac{4}{(1-(u^2 + v^2))^2} du dv$$
A: Well, the question doesn't mention the dimension of space in which the surface is embedded. So strictly speaking, there's no reason to write down $f$. Computing the area of the region is straightforward,
$$A=\iint_D\frac{du\wedge dv}{||du\wedge dv||}=\iint_D\sqrt{|\det g|}~dx_1dx_2=\int_0^{2\pi}\int_0^p\frac{4}{(1-r^2)^2}rdrd\theta=\frac{4\pi p^2}{1-p^2}$$
And the length of the boundary
$$l=\int_{\partial D}\sqrt{g_{11}\Big(\frac{du}{dt}\Big)^2+g_{22}\Big(\frac{dv}{dt}\Big)^2}dt=\int_0^{2\pi}\frac{2}{1-p^2}p~d\theta=\frac{4\pi p}{1-p^2}$$
A: This is just the hyperbolic metric
$$ds={2|dw|\over 1-|w|^2}, \qquad w:=u+iv,\tag{1}$$
on the unit disk in the $(u+iv)$-plane. Therefore the length of $\partial D_p$ computes to
$$L(\partial D_p)=\int\nolimits_{\partial D_p} ds={2\over 1-p^2}\int\nolimits_{\partial D_p} |dw|={4\pi p\over1-p^2}\ .$$
For any conformal metric, i.e., a metric of the form $ds=g(w)|dw|$, one has
$$d{\rm area}_g(w)=g^2(w)\>d{\rm area}(w)\ ,$$
whereby $d{\rm area}$ denotes the euclidean area in the $w$-plane. Therefore the $g$-area of $D_p$ for the metric $(1)$ computes to
$${\rm area}_g(D_p)=4\int\nolimits_{D_p}{1\over(1-|w|^2)^2}\ d{\rm area}(w)=
8\pi\int_0^p{r\over(1-r^2)^2}\ dr={4\pi p^2\over 1-p^2}\ .$$
A: By Zhang's answer,
 $$ {\rm Area} (D_p=\{ (u,v)|\ u^2+v^2 \leq p^2 \}) = \frac{4\pi p^2}{1-p^2} $$
Consider length of a curve $c(u)=(u,0) $ So
 $$ c'(u)=(1,0),\ |c'(u)|=\frac{2}{1-u^2},\ l(u) = \int_0^u \frac{2\ dt}{1-t^2}=\ln\ \frac{1+u}{1-u} $$
So $$ \frac{\partial\ l}{\partial\ u}=\frac{2}{1-u^2},\ {\rm Area}\ (\partial D_p)=\frac{1-p^2}{2}\frac{\partial }{\partial\ p}{\rm Area}\ (D_p)=\frac{4\pi p}{1-p^2}  $$
