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 $\partial D_p$.


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