find the area of the first-quadrant region bounded by the curves $y = x^3,y=2x^3,x=y^3,x=4y^3$ be the curves that bound the area:
now we can use the substitution:
$$y =ux^3$$ $$x =vy^3$$
in order to find the jacobian we have to express x and y in terms of u and v hence:
$$y = \frac{1}{u^{1/8}v^{3/8}}$$ $$x = \frac{1}{u^{3/8}v^{1/8}}$$
now we can find the Jacobian:
$$\frac{\partial (x,y)}{\partial(u,v)} = \frac{9}{64u^{3/4}v^{3/4}} - \frac{1}{64u^{5/4}v^{5/4}}$$
hence the area must be:
$$\int \int \Big|\frac{9}{64u^{3/4}v^{3/4}} - \frac{1}{64u^{5/4}v^{5/4}} \Big| dudv$$
Am I going on the right path?
I have trouble finding the limits of integration. by the way the book answer is:
$$\frac{1}{8}(2-\sqrt{2})$$
For the bounds we want $y^3\leq x\leq 4y^3$ and $x^3\leq y\leq 2x^3$ where $u=y/x^3$ and $v=x/y^3$, so $$1\leq v\leq 4\\1\leq u\leq 2$$
Also $$\begin{align}x&= u^{-3/8}v^{-1/8}\\y&=u^{-1/8}v^{-3/8}\\\mathrm J &= \begin{Vmatrix} -(3/8)u^{-11/8}v^{-1/8} &-(1/8)u^{-3/8}v^{-9/8}\\ -(1/8)u^{-9/8}v^{-3/8} & -(3/8)u^{-1/8}v^{-11/8}\end{Vmatrix}\\&=(1/64)\lvert (9-1) u^{-12/8} v^{-12/8} \rvert\\&=(1/8)\lvert u^{-3/2} v^{-3/2} \rvert \end{align}$$