91Ó°ÊÓ

The region $\mathrm{Î\copyright }$is bounded by,

$x=0,y^2-x^2=4,y=2x,y^2-x^2=1$

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

In the above region, the equations of boundary curves are,

$$\begin{gathered} AB:y^2-x^2=1 \\ BC:y=2x \\ CD:y^2-x^2=4 \\ DA:x=0 \end{gathered}$$

Consider the new set of variables defined as,

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

After solving we get that,

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

We have,

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

Use these equations to determine the equation of each boundary of the region.

$$\begin{gathered} AB:y^2-x^2=1⇒u=1 \\ BC:y=2x⇒v=\frac{1}{2} \\ CD:y^2-x^2=4⇒u=4 \\ DA:x=0⇒v=0 \end{gathered}$$

Plot these limits on u v plane.

Set up the double integral,

$$\begin{gathered} ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=\frac{1}{4}{∫}_{u=1}^{u=4}{∫}_{v=0}^{v=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{1}{1-v^2}-1\right)\left(2v\right)dvdu \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=\frac{1}{4}{∫}_{u=1}^{u=4}{∫}_{v=0}^{v=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\frac{2v}{1-v^2}-2v\right)dvdu \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=\frac{1}{4}{∫}_{u=1}^{u=4}\left({\left[-\ln (1-v^2)\right]}_0^{1/2}-{\left[v^2\right]}_0^{1/2}\right) \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=\left(\frac{1}{4}\ln \left(\frac{4}{3}\right)-\frac{1}{8}\right){∫}_{u=1}^{u=4}du \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=3×\left(\frac{1}{4}\ln \left(\frac{4}{3}\right)-\frac{1}{8}\right) \\ ∫{∫}_{\mathrm{Î\copyright }}\left(\frac{x^3}{y^3}\right)dA=\left(\frac{3}{4}\ln \left(\frac{4}{3}\right)-\frac{3}{8}\right) \end{gathered}$$