Element of area in hyperbolic plane with polar coordinates on hyperboloid model The hyperbolic plane in the hyperboloid model in polar coordinates is the set of points (t,x,y) with $$x = r \cos{\theta}, y = r \sin{\theta}, t = \sqrt{1+r^2}.$$  I want to prove that the area element is
$$ dA=\frac{r dr d\theta}{\sqrt{1+r^2}}.$$
I consider a small sector $[\theta,\theta + \delta \theta] \times [r,r+\delta]$ in the $(\theta,r)$-plane and wish to show that the sector $S$ in the upper hyperboloid lying over it is close to a hyperbolic rectangle with length and sides equal to $$r \delta \theta \mbox{ and }  \frac{\delta r}{\sqrt{1+r^2}} .$$
The sector $S$ has four vertices, two of which are:
$$(\sqrt{1+r^2} , r \cos(\theta) , r \sin(\theta)) $$
and
$$(\sqrt{1+(r+\delta r)^2)}, (r+\delta r)\cos(\theta), (r+\delta r)\sin(\theta)). $$
I think the distance between these points should be close to $$ \frac{\delta r}{\sqrt{1+r^2}}.$$
I use the distance formula $d(P,Q) = \cosh^{-1}(- P \cdot_L Q)$ where $\cdot_L$ is the Lorentz product $a \cdot_L b = -a_1 b_1 + a_2 b_2 + a_3 b_3$.
I have computed the hyperbolic distance between the points above and it doesn't match what it should be in order to give the area element above.
What am I missing?
 A: Here's a calculation in change-of-coordinates language. (The use of $r$ throughout may be simpler despite my comment; leaving that as an exercise. Also, it should be straightforward to use differentials and geometry to get the same result, though I haven't written up the details.)
In the parametrization $f(\rho, \theta) = (t, x, y)$ defined by
\begin{align*}
  t &= \cosh\rho, \\
  x &= \sinh\rho \cos\theta, \\
  y &= \sinh\rho \sin\theta,
\end{align*}
we have
\begin{align*}
  dt &= \sinh\rho\, d\rho, \\
  dx &= \cosh\rho \cos\theta\, d\rho - \sinh\rho \sin\theta\, d\theta, \\
  dy &= \cosh\rho \sin\theta\, d\rho + \sinh\rho \cos\theta\, d\theta.
\end{align*}
Consequently, the metric on the $(\rho, \theta)$ plane induced by the Lorentz/Minkowski metric $-dt^{2} + dx^{2} + dy^{2}$ is
\begin{align*}
  -(\sinh\rho\, d\rho)^{2}
  &+ (\cosh\rho \cos\theta\, d\rho - \sinh\rho \sin\theta\, d\theta)^{2} \\
  &+ (\cosh\rho \sin\theta\, d\rho + \sinh\rho \cos\theta\, d\theta)^{2} \\
  &= d\rho^{2} + \sinh^{2}\rho\, d\theta^{2}.
\end{align*}
If we set $r = \sinh\rho$, then $dr = \cosh\rho\, d\rho = \sqrt{1 + r^{2}}\, d\rho$, so
\begin{align*}
  d\rho^{2} + \sinh^{2}\rho\, d\theta^{2}
  &= \frac{dr^{2}}{1 + r^{2}} + r^{2}\, d\theta^{2} \\
  &= \biggl[\frac{dr}{\sqrt{1 + r^{2}}}\biggr]^{2} + [r\, d\theta]^{2}.
\end{align*}
The area element is the square root of the determinant, i.e.,
$$
dA = \frac{dr}{\sqrt{1 + r^{2}}} \times r\, d\theta
= \frac{r\, dr\, d\theta}{\sqrt{1 + r^{2}}}.
$$
