Area in the Hyperbolic Plane Let $D_2(0) = \{(x,y) \in \mathbb{R}^2 \ | x^2+y^2 \leq 4\}$ with the Riemannian metric $$\langle u,v\rangle_{(x,y)} = \frac{u\cdot v}{g^2(x,y)} \ \ \ \ \ g(x,y) = 1 - \frac{x^2+y^2}{4}$$
I want to prove that the area of the region $u^2+v^2 < r^2$ (with $r<2$) on $D_2(0)$ is given by $$A(r) = \frac{\pi r^2}{1 -\frac{r^2}{4}} $$
So I tried the obvious. Let $U_1 = g e_1$ and $U_2 = ge_2$ be a basis for $T_pD_2(0)$, I know that the area of a region $D$ can be calculated by 
$$A(D) = \iint_D \sqrt{EG-F^2}dxdy $$
(Because we can think of the parametrizations as the identity)
With $E = \langle U_1,U_1 \rangle = g^2 = G$ and $F = \langle U_1,U_2 \rangle = 0$, and so $$A(D) = \iint_D g^2 dxdy$$
But whenever I compute this integral (using polar coordinates) I have a completely wrong answer. I guess there is something wrong with the expression on $A(D)$, but I can't correct it. Any help or hint will be appreciated.


With $E = \langle U_1,U_1 \rangle = g^2 = G$

Mistake here: should be $g^{-2}$, not $g^2$. Otherwise correct. You should get
$$
A(r) = 2\pi \int_0^r \frac{16}{(4-\rho^2)^2}\rho\,d\rho
$$
which leads to the correct area.
