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## Homework Statement

Let [itex]R = \{ (x,y) \in \mathbb{R^{2}}: 0<x<1, 0<y<1\}[/itex] be the unit square on the xy-plane. Use the change of variables [itex]x = \frac{{\sin u}}{{\cos v}}[/itex] and [itex]y = \frac{{\sin v}}{{\cos u}}[/itex] to evaluate the integral [itex]\iint_R {\frac{1}

{{1 - {{(xy)}^2}}}dxdy}[/itex]

## Homework Equations

## The Attempt at a Solution

I've already computed the Jacobian:

[itex]\frac{{\partial (x,y)}}

{{\partial (u,v)}} = \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 - {\left( {\frac{{\sin u\sin v}}

{{\cos u\cos v}}} \right)^2}[/itex]

Now I'm left with finding out how [itex]R[/itex] would look like in the uv-plane. Hope someone can shed some light on this, thanks!