Sketch the planar region D which is bounded by the following four curves and located in the positive quadrant: $y = x^3$, $y = 10x^3$, $y = x$ and $y = 2x$. Use transformation given by $u=\frac{y}{x^3}$ and $v=\frac{y}{x}$.
Evaluate: $\iint_D \frac{2y}{x^5} \,dx \,dy $
So far I have $x = \frac{u^\frac{1}{4}}{v^\frac{1}{4}}$ and $y = v\frac{u^\frac{1}{4}}{v^\frac{1}{4}}$. Using this to calcuate the Jacobian matrix:
$$\begin{equation} \begin{vmatrix} \frac{1}{4v^\frac{1}{4}u^\frac{3}{4}} & \frac{u\frac{1}{4}}{4v^\frac{3}{4}} \\ \frac{v\frac{3}{4}}{4u^\frac{3}{4}u^\frac{3}{4}} & \frac{3u\frac{1}{4}}{4v^\frac{3}{4}} \\ \end{vmatrix} \end{equation} = \frac{3-v^\frac{1}{2}}{4v^\frac{1}{2}u^\frac{1}{2}}$$
I used my Jacobian to calculate the integral and I also changed the region D with variables $u$ and $v$ however I ended up with a negative answer. Can anyone check if my Jacobian matrix is correct? I think this is where I have gone wrong.
Solving wrt to $u,v$ we get $$\left\{x=\sqrt{\frac{v}{u}},y=v \sqrt{\frac{v}{u}}\right\}$$ Jacobian is $$J=\left( \begin{array}{cc} -\frac{\sqrt{\frac{v}{u}}}{2 u} & -\frac{1}{2} \left(\frac{v}{u}\right)^{3/2} \\ \frac{\sqrt{\frac{v}{u}}}{2 v} & \frac{3 \sqrt{\frac{v}{u}}}{2} \\ \end{array} \right)$$ and $\det J=-\frac{v}{2 u^2}$
Substituting in the function we get $$\iint\limits_{D'} \frac{2 u^2}{v}\cdot \left(-\frac{v}{2 u^2}\right)\,dudv=\\= -\iint\limits_{D'}\,dudv$$