Finding the area between 4 curves by changing my basis to create a square and integrating

Question

The question number $3$ at hand: https://i.stack.imgur.com/WYHsh.jpg

My work on it: https://i.stack.imgur.com/yjw6L.jpg

I thought I was doing everything right but it seems to be wrong, the final answer should be $$\frac{6}{35}(b^{\frac{7}{6}}-a^{\frac{7}{6}})(d^{\frac{5}{6}}-c^{\frac{5}{6}})$$

Answer 1

Let $(x,y)=(u^3 v^3,u^2 v^4)$

\begin{array}{ccccc} ax^2<y^3<bx^2 & \implies & au^6v^6<u^6v^{12}<bu^6v^6 & \implies & a<v^6<b \\ cy^3<x^4<dy^3 & \implies & cu^6v^{12}<u^{12}v^{12}<du^6v^{12} & \implies & c<u^6<d \end{array}

\begin{align*} \iint_A dx\, dy &= \int_{\sqrt[6]{a}}^{\sqrt[6]{b}} \int_{\sqrt[6]{c}}^{\sqrt[6]{d}} \begin{vmatrix} x_u & x_v \\ y_u & y_v \end{vmatrix} du \, dv \\ &= \int_{\sqrt[6]{a}}^{\sqrt[6]{b}} \int_{\sqrt[6]{c}}^{\sqrt[6]{d}} \begin{vmatrix} 3u^2v^3 & 3u^3v^2 \\ 2uv^4 & 4u^2v^3 \end{vmatrix} du \, dv \\ &= \int_{\sqrt[6]{a}}^{\sqrt[6]{b}} \int_{\sqrt[6]{c}}^{\sqrt[6]{d}} 6u^4v^6 \, du \, dv \\ &= \frac{6}{35}(b^{7/6}-a^{7/6})(d^{5/6}-c^{5/6}) \end{align*}

See also another answer of mine here.