I have to calculate the value of an area in $\mathbb{R}^2$ given as $ax < y^2 < bx$, $cy < x^2 < dy$ with $0<a<b$ and $0<c<d$ by using a transformation which turns this area into a rectangle. But I can't even imagine what this area looks like.
Thanks for any hints.
Let $(x,y)=(u^2v,uv^2)$
\begin{array}{ccccc} ax<y^2<bx & \implies & au^2v<u^2v^4<bu^2v & \implies & a<v^3<b \\ cy<x^2<dy & \implies & cuv^2<u^4v^2<duv^2 & \implies & c<u^3<d \end{array}
\begin{align*} \iint_A dx\, dy &= \int_{\sqrt[3]{a}}^{\sqrt[3]{b}} \int_{\sqrt[3]{c}}^{\sqrt[3]{d}} \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix} du \, dv \\ &= \int_{\sqrt[3]{a}}^{\sqrt[3]{b}} \int_{\sqrt[3]{c}}^{\sqrt[3]{d}} \begin{vmatrix} 2uv & u^2 \\ v^2 & 2uv \end{vmatrix} du \, dv \\ &= \int_{\sqrt[3]{a}}^{\sqrt[3]{b}} \int_{\sqrt[3]{c}}^{\sqrt[3]{d}} 3u^2v^2 \, du \, dv \\ &= \frac{(b-a)(d-c)}{3} \end{align*}