z_{\textrm{max}}\geqslant z(x) \geqslant 0$. The equation (\[zprime\^2 strip\]) provides the profile of the minimal surface we are looking for and, by construction, it satisfies $z(L/2)=0$ and $z(0)=z_{\textrm{max}}$. As a check of (\[zprime\^2 strip\]), one can write the equation of motion coming from $\mathcal{L}_{\textrm{strip}}$ $$\frac{z'' z}{f(z)}
+d \left[1+\frac{(z')^2}{f(z)}\right]
- \frac{(z')^2 z}{2 f(z)^2}\,f'(z)
\,=\,0$$ and verify that the same equation can be found by deriving the conservation law $\mathcal{H}_{\textrm{strip}}= \textrm{const}$ w.r.t. $x$. Then, separating the variables in (\[zprime\^2 strip\]), we find that the inverse function $x(z)$ is $$\label{x(z)}
x(z)\,=\,\int_0^x d\tilde{x}\,=\,- \int_{z_{\textrm{max}}}^{z}
\frac{w^d}{\sqrt{f(w)}\,\sqrt{z_{\textrm{max}}^{2d}-w^{2d}}}\,dw\:.$$ Imposing in (\[x(z)\]) the relation $x(0)=L/2$, one gets that $$\label{L(zmax)}
\frac{L}{2}\,=\,
\int_0^{z_{\textrm{max}}}
\frac{w^d}{\sqrt{f(w)}\,\sqrt{z_{\textrm{max}}^{2d}-w^{2d}}}\,dw$$ which provides $L=L(z_{\textrm{max}})$ and the correspondence between $z_{\textrm{max}}$ and $L$.\
As for the area of the minimal surface defined by (\[zprime\^2 strip\]), we can employ its definition to change integration variable in (\[strip area0\]), which therefore becomes $$\label{strip area1}
\textrm{Area}(\gamma_A)
= 2R^dL_\perp^{d-1} \int_{0}^{z_{\textrm{max}}} 
\frac{z_{\textrm{max}}^d}{z^d \sqrt{f(z)}\sqrt{z_{\textrm{max}}^{2d}-z^{2d}}}\,dz\:.$$ It is important to remark that, since $f(z) \rightarrow 1$ as $z \rightarrow 0$, the integral in (\[strip area1\]) diverges at $z=0$. This leads us to put a UV cutoff $z \geqslant a$ in the integration domain of (\[strip area1\]). Thus, the integral we have to compute reads $$\label{strip area}
\textrm{Area}(\gamma_A)
\,=\,2R^dL_\perp^{d-1} 