Circle-Circle intersection area in the hyperbolic space Is there any closed formula for the area of the intersection of two circles  in the hyperbolic plane $\mathbb{H}^2$? The two circles have radii $R, R'$ and a distance of $d$ between centers. If possible, the curvature should be a parameter (though I am not sure if the curvature affects the area).

I have a version without any integrals: four acute angles related by four equations, but not an easy system to solve. Then your area is a simple function of those angles. If it will help, i can type it in. By the way, the calculation is intrinsic. The upper half plane or disc just makes everything worse. 
Meanwhile, looking at your hyperbolic network question, you should become aware that the hyperbolic plane is the natural setting for any extremely regular network, one where every edge is the same length, each node is connected to the same number of neighbors, spaced at equal angles. See material on Ford Circles, but especially  http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane 
EDIT, Sunday 23 Feb.
$$ \frac{\sin \alpha}{\sinh a} = \frac{\sin \beta}{\sinh b} =\frac{\sin \gamma}{\sinh c}  $$
$$  \cos \alpha = \frac{\cosh b \cosh c - \cosh a}{\sinh b \sinh c }  $$
$$  \cosh a = \frac{\cos \alpha + \cos \beta \cos \gamma}{ \sin \beta \sin \gamma}  $$
IF $\gamma = \pi / 2,$
$$ \cosh c = \cosh a \cosh b = \cot \alpha \cot \beta  $$
$$  \cosh a = \frac{\cos \alpha}{\sin \beta},  \; \;   \sin \alpha = \frac{\sinh a}{\sinh c},  $$
$$  \cos \alpha = \frac{\tanh b}{\tanh c},  \; \;   \tan \alpha = \frac{\tanh a}{\sinh b} $$
IF also two infinite edges, we have a finite edge $t$ and an acute angle $\Pi(t).$
$$  \tanh t =  \cos \Pi(t), \; \;  \sinh t = \cot \Pi(t), $$
$$  \cosh t =  \csc \Pi(t), \; \;  e^{-t}  = \tan \frac{ \Pi(t)}{2}. $$
Circle. Area
$$ A = 2 \pi (\cosh r - 1) = 4 \pi \sinh^2 (r/2)   $$
Circumference
$$ C = 2 \pi \sinh r.  $$
% 523
$$ \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y   $$
$$ \sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y   $$
% 559
If $ u = \tan (x/2), $
$$ \cos x = \frac{1 - u^2}{1 + u^2},  $$
$$ \sin x = \frac{2u}{1 + u^2},  $$
$$ \tan x = \frac{2u}{1 - u^2},  $$
$$   dx =   \frac{2 du}{1 + u^2}.  $$
Alright; your question, with distance scale set to $1,$ comes down naturally to four equations in four unknowns, which are angles. Area for triangles is most naturally stated in terms of angles, same on the surface of a sphere, but not in the Euclidean plane. For this material, including the distance scale $k,$ I recommend   George E. Martin, The Foundations of Geometry and the Non-Euclidean Plane. For example , Theorem 32.17 on page 434 includes $k.$
Anyway, I have always worked with $k=1.$ We get four acute angles $\alpha,\beta,\gamma, \delta,$ and three (given) intrinsic distances $R,R', d,$ with the supposition that $d < R + R'.$ The four equation system becomes
$$  \frac{\cos \alpha}{\sin \beta} =  \frac{\cos \gamma}{\sin \delta}   $$
$$  \cosh R = \cot \alpha \cot \beta   $$
$$  \cosh R' = \cot \gamma \cot \delta   $$
$$ \cosh d =  \frac{\cos \beta}{\sin \alpha}   \frac{\cos \delta}{\sin \gamma} + \sqrt{ \frac{\cos^2 \beta}{\sin^2 \alpha} -1 }  \sqrt{ \frac{\cos^2 \delta}{\sin^2 \gamma} -1 }    $$  
After solving for the four angles, your area is
$$ 2 \alpha \cosh R  +  2 \gamma \cosh R' + 2 \alpha + 2 \beta  + 2 \gamma + 2 \delta   - 2 \pi - 2.     $$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
