<s>[INST] Double integral with a change of variables where $u = \frac{y}{x^3}$ and $v = \frac{y}{x}$. 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.
 [/INST] 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
$$\iint\limits_{D'} \frac{2 u^2}{v}\cdot \left(-\frac{v}{2 u^2}\right)\,dudv=\\=
-\iint\limits_{D'}\,dudv$$
</s>