Computing the area and length of a curve

Question

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 $\partial D_p$.

Answer 1

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,

$$\sqrt{\det (g_{ij})} dudv = \frac{4}{(1-(u^2 + v^2))^2} du dv$$

Answer 2

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 $$l=\int_{\partial D}\sqrt{g_{11}\Big(\frac{du}{dt}\Big)^2+g_{22}\Big(\frac{dv}{dt}\Big)^2}dt=\int_0^{2\pi}\frac{2}{1-p^2}p~d\theta=\frac{4\pi p}{1-p^2}$$

Answer 3

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 $${\rm area}_g(D_p)=4\int\nolimits_{D_p}{1\over(1-|w|^2)^2}\ d{\rm area}(w)= 8\pi\int_0^p{r\over(1-r^2)^2}\ dr={4\pi p^2\over 1-p^2}\ .$$

Answer 4

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 $$ \frac{\partial\ l}{\partial\ u}=\frac{2}{1-u^2},\ {\rm Area}\ (\partial D_p)=\frac{1-p^2}{2}\frac{\partial }{\partial\ p}{\rm Area}\ (D_p)=\frac{4\pi p}{1-p^2} $$