metric of a $D-1-d$ dimensional compact manifold $\mathcal{M}_{\textrm{c}}$. The boundary is at large $r$ and we assume the occurrence of a horizon at $r=r_h$.\
Let us take a strip specified by the function $r=r(x_d)$. Then, the metric induced on it reads $$ds^2_{\textrm{ind}}\,=\,\left(B(r)^2 (dx_1^2+\dots dx_{d-1}^2)
+\left[\,\frac{(r')^2}{A(r)^2}+B(r)^2\right]dx^2_d  \right)
e^{-\frac{D-1-d}{d}\chi(r)}
+R^2 e^{\chi(r)} \gamma^{\textrm{(c)}}_{ij}(r) d\theta^i \theta^j\,.$$ To compute the area, we have to integrate $\sqrt{\textrm{det}(G_{\textrm{ind}})}$ over the strip $A$. In such determinant the dependence on $\chi(r)$ simplifies and therefore it does not occur anymore. If $\textrm{det}(\gamma^{\textrm{(c)}})$ does not depend on $r$, then the area of the surface is given by $$\label{area r}
\textrm{Area}(\gamma_A)\,=\,
\big[R^{D-1-d} \,\textrm{Vol}(\mathcal{M}_{\textrm{c}}) \big]\,
2L_\perp^{d-1}
\int_0^{L/2} B(r)^{d}\,
\sqrt{1+\frac{(r')^2}{A(r)^2 B(r)^2}}\,dx_d$$ where $L_\perp$ is the width of the strip along the directions $x_1 , \dots , x_{d-1}$ and $\textrm{Vol}(\mathcal{M}_{\textrm{c}})$.\
As done above, we take as Lagrangian density $\mathcal{L}_{\textrm{strip}}[r(x_d)]$ the integrand of (\[area r\]) and compute the momentum $p_r \equiv \partial\mathcal{L}_{\textrm{strip}}/\partial r'$. Then, being $\mathcal{L}_{\textrm{strip}}$ independent of $x_d$, we can employ the conserved quantity $\mathcal{H}_{\textrm{strip}} \equiv p_r r' -\mathcal{L}_{\textrm{strip}}$. Since at the minimum value of $r$ we have $r'(0)=0$, we set $\mathcal{H}_{\textrm{strip}}^2 \equiv B(r_{\textrm{min}})^{2d}$. This allows us to write (\[area r\]) as follows $$\label{area r infinity}
\textrm{Area}(\gamma_A)\,=\,
\big[R^{D-1-d} \,\textrm{Vol}(\mathcal{M}_{\textrm{c}}) \big]\,
2L_\perp^{d-1}
\int_{r_{\textrm{min}}}^{\infty} 
\frac{B(r)^{2d-1}}{A(r)\sqrt{B(r)^{2d}-B(r_{\textrm{min}})^{2d}}}\,dr\;.$$ It is 