= Solution
On the $t=0$ slice, rotational symmetry lets the surface be written $z=z(r)$. The inversion symmetry fixing the boundary circle extends to an isometry of the Poincare half-space whose fixed set is
$$
r^2+z^2=R^2.
$$
The unique extremal surface anchored at $r=R,z=0$ must be invariant under that isometry. Hence the <Ryu–Takayanagi surface for a disk> is the hemisphere
$$
\boxed{\gamma_D:\quad z(r)=\sqrt{R^2-r^2},
\qquad 0\leq r\leq R}.
$$
One may verify directly that it extremizes the area functional $2\pi\int r z^{-2}\sqrt{1+z'^2}\,dr$.
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