Element of area: mistake in the book? In the book of Miles Reid and Balazs Szendroi "Geometry and Topology" authors consider the hyperbolic geometry on the hyperboloid $x^2+y^2-t^2=-1$ and claim in Ex. 3.22 (b) that an element of area is $\frac{dxdy}{t}$. They also compute it in polar coordinates (basically just multiplying by Jacobian). However a well-known formula for area element
$$\sqrt{1+f_x^2+f_y^2}dxdy$$
with $f(x,y)=t=\sqrt{1+x^2+y^2}$ clearly gives completely different result. Is it a mistake?
Thank you.
 A: In Euclidean space the hyperboloid is not a surface of constant (negative) curvature. You need to look at the hyperboloid in Minkowski space, instead, and so the change in the inner product will change your formula. 
P.S. I have in fact verified that their formula is correct. (Heuristically, we can tell it's right because that is the area form for the unit sphere $x^2+y^2+t^2=1$ in $\Bbb R^3$ and the hyperboloid in Minkowski space is the dual symmetric space to the sphere.)
EDIT: The usual formula is derived by noting that the surface area element $dS$ on $t=f(x,y)$ satisfies
$$dS = \frac1{|\cos\gamma|}dx\,dy,$$
where $\gamma$ is the angle between the $xy$-plane and the tangent plane of the surface. This angle is in turn the angle between the unit normal vectors, namely $(0,0,1)$ and $\dfrac{(-f_x,-f_y,1)}{\sqrt{1+f_x^2+f_y^2}}$, and the usual dot product formula gives us $1/|\cos\gamma| = \sqrt{1+f_x^2+f_y^2}$. 
The problem in our situation is that because both the $xy$-plane and our hyperboloid are spacelike surfaces, the normal vectors to them are timelike (i.e., have negative length squared). But since the square length of the normal vector $(-f_x,-f_y,1)$ is now $f_x^2+f_y^2-1$ (which, again, is negative), you might guess that the appropriate factor will instead be $\sqrt{1-f_x^2-f_y^2}$. (The way I first derived this used differential forms and the interior product with the normal vector.) Here's a straightforward way to find this, and I'll leave you to work out the details. Instead of using the inner product of the unit normal vectors, let's instead use the inner product of two spacelike vectors $v,w$ in the respective planes: The two planes have a line in common, and let's take $v$ and $w$ to be vectors in the planes both orthogonal to that line. Take $v=\big(f_x,f_y,0\big)$ and $w=\big(f_x,f_y,-(f_x^2+f_y^2)\big)$ (I'm not bothering to normalize yet), and, abracadabra, $$\dfrac1{\cos\gamma} = \dfrac{\sqrt{v\cdot v}\sqrt{w\cdot w}}{v\cdot w} = \sqrt{1-f_x^2-f_y^2}.$$
(Here $\cdot$ represents the inner product $v\cdot w = v_1w_1+v_2w_2-v_3w_3$.)
