Answer
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 - {x^2}{y^2}}} = \frac{{{\pi ^2}}}{8}$
Work Step by Step
Write the mapping: $G\left( {u,v} \right) = \left( {\frac{{\sin u}}{{\cos v}},\frac{{\sin v}}{{\cos u}}} \right)$.
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{{\cos u}}{{\cos v}}}&{\frac{{\sin u\sin v}}{{{{\cos }^2}v}}}\\
{\frac{{\sin u\sin v}}{{{{\cos }^2}u}}}&{\frac{{\cos v}}{{\cos u}}}
\end{array}} \right|$
$ = 1 - \frac{{{{\sin }^2}u{{\sin }^2}v}}{{{{\cos }^2}u{{\cos }^2}v}}$
So, ${\rm{Jac}}\left( G \right) = 1 - {\tan ^2}u{\tan ^2}v$.
The domain of integration in $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 - {x^2}{y^2}}}$ implies that
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 1} \right\}$
Thus, ${\cal D}$ is the interior of the rectangle in the $xy$-plane with vertices $\left( {0,0} \right)$, $\left( {1,0} \right)$, $\left( {1,1} \right)$, $\left( {0,1} \right)$.
Using $x = \frac{{\sin u}}{{\cos v}}$ and $y = \frac{{\sin v}}{{\cos u}}$, we find the corresponding points in the $uv$-plane and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Vertex{\ }in}}{\ }{\cal D}}&{{\rm{Point{\ }in}}{\ }{\cal R}}\\
{\left( {x,y} \right)}&{\left( {u,v} \right)}\\
{\left( {0,0} \right)}&{\left( {0,0} \right)}\\
{\left( {1,0} \right)}&{\left( {\frac{\pi }{2},0} \right)}\\
{\left( {1,1} \right)}&{\left( {\frac{\pi }{4},\frac{\pi }{4}} \right)}\\
{\left( {0,1} \right)}&{\left( {0,\frac{\pi }{2}} \right)}
\end{array}$
We plot these points in the $uv$-plane and see that there is a triangle bounded by these points. So, we can describe the domain ${\cal R}$ as a vertically simple region:
${\cal R} = \left\{ {\left( {u,v} \right)|0 \le u \le \frac{\pi }{2},0 \le v \le \frac{\pi }{2} - u} \right\}$
Write $f\left( {x,y} \right) = \frac{1}{{1 - {x^2}{y^2}}}$
Using $x = \frac{{\sin u}}{{\cos v}}$ and $y = \frac{{\sin v}}{{\cos u}}$, we get
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = \frac{1}{{1 - {{\tan }^2}u{{\tan }^2}y}}$
Using the Change of Variables Formula, we evaluate the integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
Since $f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right| = 1$, so
$\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 - {x^2}{y^2}}} = \mathop \smallint \limits_{u = 0}^{\pi /2} \mathop \smallint \limits_{v = 0}^{\left( {\pi /2} \right) - u} {\rm{d}}v{\rm{d}}u$
$ = \mathop \smallint \limits_{u = 0}^{\pi /2} \left( {\frac{\pi }{2} - u} \right){\rm{d}}u$
$ = \left( {\frac{\pi }{2}u - \frac{1}{2}{u^2}} \right)|_0^{\pi /2} = \frac{{{\pi ^2}}}{4} - \frac{{{\pi ^2}}}{8} = \frac{{{\pi ^2}}}{8}$
So, $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 \frac{{{\rm{d}}x{\rm{d}}y}}{{1 - {x^2}{y^2}}} = \frac{{{\pi ^2}}}{8}$.
