Answer
The Jacobian of the map:
${\rm{Jac}}\left( G \right) = {{\rm{e}}^{2r}}$
Work Step by Step
We have the map:
$G\left( {r,s} \right) = \left( {{{\rm{e}}^r}\cosh \left( s \right),{{\rm{e}}^r}\sinh \left( s \right)} \right)$
Write $x = {{\rm{e}}^r}\cosh \left( s \right)$, $y = {{\rm{e}}^r}\sinh \left( s \right)$. So,
$\frac{{\partial x}}{{\partial r}} = {{\rm{e}}^r}\cosh \left( s \right)$, ${\ \ \ \ }$ $\frac{{\partial x}}{{\partial s}} = {{\rm{e}}^r}\sinh \left( s \right)$
$\frac{{\partial y}}{{\partial r}} = {{\rm{e}}^r}\sinh \left( s \right)$, ${\ \ \ \ }$ $\frac{{\partial y}}{{\partial s}} = {{\rm{e}}^r}\cosh \left( s \right)$
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial r}}}&{\frac{{\partial x}}{{\partial s}}}\\
{\frac{{\partial y}}{{\partial r}}}&{\frac{{\partial y}}{{\partial s}}}
\end{array}} \right| = \frac{{\partial x}}{{\partial r}}\frac{{\partial y}}{{\partial s}} - \frac{{\partial x}}{{\partial s}}\frac{{\partial y}}{{\partial r}}$
${\rm{Jac}}\left( G \right) = {{\rm{e}}^{2r}}{\cosh ^2}\left( s \right) - {{\rm{e}}^{2r}}{\sinh ^2}\left( s \right)$
$ = {{\rm{e}}^{2r}}\left( {{{\cosh }^2}\left( s \right) - {{\sinh }^2}\left( s \right)} \right)$
Recall Eq. (1) in Section 7.9:
${\cosh ^2}\left( s \right) - {\sinh ^2}\left( s \right) = 1$
So, ${\rm{Jac}}\left( G \right) = {{\rm{e}}^{2r}}$.
