with length $L$ along one direction, that we call $x$, and $L_\perp$ along the other orthogonal ones. Choosing the origin in the center of this strip, the symmetry of the problem allows us to restrict to surfaces described by the even function $z=z(x)$. Then, the area functional that we have to minimize to compute the holographic entanglement entropy reads $$\label{strip area0}
\textrm{Area}(\gamma_A)
=
2R^d L_\perp^{d-1} \int_{0}^{\frac{L}{2}} 
dx \, \frac{1}{z^d}\,\sqrt{1+\frac{(z')^2}{f(z)}}\:.$$ Considering as a Lagrangian density $\mathcal{L}_{\textrm{strip}}[z(x)]$ the integrand in (\[strip area0\]), one notices that it does not depend explicitly on $x$. This is the main simplification that makes the case of a rectangular region $A$ easier to solve than the case of a circular region. Indeed, the independence of $\mathcal{L}_{\textrm{strip}}$ on $x$ leads to the conserved quantity $\mathcal{H}_{\textrm{strip}}\equiv p_z z'-\mathcal{L}_{\textrm{strip}}$, where $p_z \equiv \partial \mathcal{L}_{\textrm{strip}}/\partial z'$. In particular, one gets $\mathcal{H}_{\textrm{strip}} = z^{-d}[1+(z')^2/f(z)]^{-1/2}$. By introducing $z_{\textrm{max}}^{2d}\equiv 1/\mathcal{H}_{\textrm{strip}}^2$, the constancy of $\mathcal{H}_{\textrm{strip}}$ reads $$\label{zprime^2 strip}
z'\,=\,- \sqrt{f(z)}\,\frac{\sqrt{z_{\textrm{max}}^{2d}-z^{2d}}}{z^{d}}$$ where we have used that $z' <0$. This equation tells us that $z_{\textrm{max}}$ is the turning point, namely $z'=0$ when $z=z_{\textrm{max}}$. Notice that, from (\[zprime\^2 strip\]), also at the horizon $z_0$ we could have $z'=0$ because $f(z_0)=0$, but we never reach it because $z_0 > 