Finding the area between 4 curves by changing my basis to create a square and integrating The question number $3$ at hand: https://i.imgur.com/DL11Izh.jpg
My work on it: https://i.imgur.com/AzaQKHS.jpg?1
I thought I was doing everything right but it seems to be wrong, the final answer should be $$\frac{6}{35}(b^{\frac{7}{6}}-a^{\frac{7}{6}})(d^{\frac{5}{6}}-c^{\frac{5}{6}})$$

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.
