91Ó°ÊÓ

The given region $\mathrm{Î\copyright }$is said to be bounded by,

$y^2-x^2=12,y^2-x^2=3,y=-2x,y=2x$

Plot the given points to form the trapezoid and name the vertices.

In the region $\mathrm{Î\copyright }$, the equations of boundary curves are,

role="math" $$\begin{gathered} AB:y=-2x \\ BC:y^2-x^2=12 \\ CD:y=2x \\ DA:y^2-x^2=3 \end{gathered}$$

Consider the new set of variables defined as

$$\begin{gathered} u=\frac{y}{x} \\ v=y^2-x^2 \end{gathered}$$

After solving we get that,

$$\begin{gathered} \sqrt{\frac{v}{u^2-1}}=x \\ y=\sqrt{\frac{v{u}^2}{u^2-1}} \end{gathered}$$

Use these equations to determine the equation of each boundary of the region in terms of u and v.

$$\begin{gathered} AB:y=-2x⇒u=-2 \\ BC:y^2-x^2=12⇒v=12 \\ CD:y=2x⇒u=2 \\ DA:y^2-x^2=3⇒v=3 \end{gathered}$$

Plot these limits on u v- plane.

Set up the double integral.

$$\begin{gathered} ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{∫}_{u=-2}^{u=2}{∫}_{v=3}^{v=12}\left(\frac{vu}{u^2-1}\right)dvdu \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{∫}_{-2}^2\left(\frac{u}{u^2-1}{∫}_3^{12}\left(v\right)dv\right)du \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=\frac{1}{2}{∫}_{-2}^2\left(\frac{u}{u^2-1}{\left[\frac{v^2}{2}\right]}_3^{12}\right)du \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=\frac{135}{4}{∫}_{-2}^2\left(\frac{u}{u^2-1}\right)du\left[\frac{u}{u^2-1}isanoddfunction\right] \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=\frac{135}{4}×0 \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{y^3}{x}-xy\right)dA=0 \end{gathered}$$