Trigonometric substitution in double integral

[drawar]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I've already computed the Jacobian:

\frac{{\partial (x,y)}}<br /> {{\partial (u,v)}} = \left| {\begin{array}{*{20}{c}}<br /> {\frac{{\partial x}}<br /> {{\partial u}}} &amp; {\frac{{\partial x}}<br /> {{\partial v}}} \\<br /> {\frac{{\partial y}}<br /> {{\partial u}}} &amp; {\frac{{\partial y}}<br /> {{\partial v}}} \\<br /> <br /> \end{array} } \right| = \left| {\begin{array}{*{20}{c}}<br /> {\frac{{\cos u}}<br /> {{\cos v}}} &amp; {\frac{{\sin u\sin v}}<br /> {{{{\cos }^2}v}}} \\<br /> {\frac{{\sin u\sin v}}<br /> {{{{\cos }^2}u}}} &amp; {\frac{{\cos v}}<br /> {{\cos u}}} \\<br /> <br /> \end{array} } \right| = 1 - {\left( {\frac{{\sin u\sin v}}<br /> {{\cos u\cos v}}} \right)^2}

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

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I've already computed the Jacobian:

\frac{{\partial (x,y)}}<br /> {{\partial (u,v)}} = \left| {\begin{array}{*{20}{c}}<br /> {\frac{{\partial x}}<br /> {{\partial u}}} &amp; {\frac{{\partial x}}<br /> {{\partial v}}} \\<br /> {\frac{{\partial y}}<br /> {{\partial u}}} &amp; {\frac{{\partial y}}<br /> {{\partial v}}} \\<br /> <br /> \end{array} } \right| = \left| {\begin{array}{*{20}{c}}<br /> {\frac{{\cos u}}<br /> {{\cos v}}} &amp; {\frac{{\sin u\sin v}}<br /> {{{{\cos }^2}v}}} \\<br /> {\frac{{\sin u\sin v}}<br /> {{{{\cos }^2}u}}} &amp; {\frac{{\cos v}}<br /> {{\cos u}}} \\<br /> <br /> \end{array} } \right| = 1 - {\left( {\frac{{\sin u\sin v}}<br /> {{\cos u\cos v}}} \right)^2}

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

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

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 .

etc.