= Solution
For the disk, the hemisphere reaches
$$
z_{\max,D}=R.
$$
For half of the strip surface, invert the first-order equation and integrate from the midpoint to the boundary:
$$
R=\int_0^{z_{\max,T}}
\frac{z^2\,dz}{\sqrt{z_{\max,T}^4-z^4}}
=z_{\max,T}\int_0^1
\frac{u^2\,du}{\sqrt{1-u^4}}.
$$
The <beta function> integral is
$$
\int_0^1\frac{u^2\,du}{\sqrt{1-u^4}}
=\frac{\sqrt\pi\,\Gamma(3/4)}{\Gamma(1/4)}
\simeq0.59907.
$$
Therefore
$$
\boxed{z_{\max,T}
=\frac{\Gamma(1/4)}{\sqrt\pi\,\Gamma(3/4)}R
\simeq1.669R>R=z_{\max,D}}.
$$
The strip's holographic entropy surface lies deeper in the bulk than that of a disk with the same half-width or radius.
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