= Solution
In the <upper half-plane model> $H=\{(x,y):y>0\}$ the <hyperbolic metric> is
$$
ds^2=\frac{dx^2+dy^2}{y^2}.
$$
The length of a differentiable curve is $\int\sqrt{\dot x^2+\dot y^2}\,dt/y$. The area element is the square root of the metric determinant times $dx\,dy$, so the <hyperbolic area> of a measurable region $S$ is $\int_S y^{-2}\,dx\,dy$.
The <Gauss-Bonnet theorem> for a <geodesic> <hyperbolic triangle> of angles $\alpha,\beta,\gamma$ in <curvature> $-1$ gives \b[area $\pi-\alpha-\beta-\gamma$]; an ideal vertex has angle zero. The horizontal edge in the present region is not a <geodesic>, so it is simpler to use the area element directly:
$$
\operatorname{Area}_H(R)=\int_0^{1/2}\int_{\sqrt{1-x^2}}^1\frac{dy\,dx}{y^2}
=\int_0^{1/2}\left(\frac1{\sqrt{1-x^2}}-1\right)dx
=\boxed{\frac\pi6-\frac12}.
$$
The open boundary does not affect the <integral> because its area is zero.
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