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Trigonometric substitution in double integral

  1. Mar 29, 2013 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant equations

    3. 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!
  2. jcsd
  3. Mar 30, 2013 #2


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    I presume that the image of R, the unit square in x & y, is a subset of the set [itex]\displaystyle \ \left\{ (u,\,v)\left|\, 0<u<\frac{\pi}{2}\,,\ 0<v<\frac{\pi}{2}\right.\right\}\ . [/itex]

    If x = 0, then u = 0 and v ≠ π/2 .

    If x = 1, then u = sin-1(cos(v)) = sin-1(sin(π/2 - v)) = π/2 - v .

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