also important to express $L=L(r_{\textrm{min}})$ and it reads $$\label{L ABansatz}
L\,=\,2
\int_{r_{\textrm{min}}}^{\infty} 
\frac{B(r_{\textrm{min}})^{d}}{A(r)B(r)\sqrt{B(r)^{2d}-B(r_{\textrm{min}})^{2d}}}\,dr\;.$$ We require to have $AdS_{d+2}$ at large $r$, which means to impose $$\label{AB large r}
A(r)^2\,=\,
\frac{r^2}{R^2}+O(1)
\hspace{1cm}
B(r)^2\,=\,
\frac{r^2}{R^2}+O(1)
\hspace{1cm}
\chi(r)\,\rightarrow\, 0
\hspace{1cm}
r\,\rightarrow\, + \infty\;.$$ Because of this asymptotic behavior, the integral in (\[area r infinity\]) is divergent. Thus, one introduces the cut off $\alpha$ at large $r$, obtaining for the regularized area $$\begin{aligned}
\label{area r reg}
\frac{\textrm{Area}(\gamma_A)}{R^{D-1-d} \,\textrm{Vol}(\mathcal{M}_{\textrm{c}})} & &
\hspace{-.5cm}=\;
2L_\perp^{d-1}
\int_{r_{\textrm{min}}}^{\alpha} 
\frac{B(r)^{2d-1}}{A(r)\sqrt{B(r)^{2d}-B(r_{\textrm{min}})^{2d}}}\,dr \\
\label{area r finite integ 1}
\rule{0pt}{.9cm}
& &
\hspace{-3.3cm}=\;\;
\frac{2L_\perp^{d-1} }{(d-1)R^{d-2}}\,\Big(\alpha^{d-1}-r_{\textrm{min}}^{d-1}\Big)
+
2L_\perp^{d-1}  \int_{r_{\textrm{min}}}^{\alpha} 
 \left(
\frac{B(r)^{2d-1}}{A(r)\sqrt{B(r)^{2d}-B(r_{\textrm{min}})^{2d}}}
-\left(\frac{r}{R}\right)^{d-2}
\,\right)dr
\nonumber\\
& &\end{aligned}$$ where the integral in (\[area r finite integ 1\]) is finite when $\alpha \rightarrow \infty$, once the asymptotic behavior (\[AB large r\]) has been assumed. At this point, the finite term of area integral we are interested in is given by the sum of the integral and of the term proportional to $r_{\textrm{min}}^{d-1}$ in (\[area r finite integ 1\]). We remark that (\[L(zmax)\]) and (\[strip area\]) are special cases of (\[L ABansatz\]) and (\[area r reg\]) respectively. Indeed they are recovered by choosing $$A(r)\,=\,
\frac{r}{R}\,\sqrt{f(r)}
\hspace{2.2cm}
B(r)\,=\, \frac{r}{R}$$ and adopting the variable $z \equiv R^2/r$. The formula for 