[tex]\displaystyle\iint_R8xy\,\mathrm dA[/tex]
Assuming [tex]R[/tex] is the region bounded by the curves,
[tex]\begin{cases}y=\frac23x\\y=\frac32x\\xy=\frac23\\xy=\frac32\end{cases}[/tex]
then substituting [tex]x=\dfrac uv[/tex] and [tex]y=v[/tex] changes the boundaries of [tex]R[/tex] in the [tex]u[/tex]-[tex]v[/tex] plane to
[tex]\begin{cases}v=\frac23\frac uv\\v=\frac32\frac uv\\\frac uvv=\frac23\\\frac uvv=\frac32\end{cases}\iff\begin{cases}v=\sqrt{\frac23u}\\v=\sqrt{\frac32u}\\u=\frac23\\v=\frac32\end{cases}[/tex]
so that the region [tex]R[/tex] is given by all the points in the set [tex]\right\{(u,v)~:~\sqrt{\dfrac23u}\le v\le\sqrt{\dfrac32u},\,\dfrac23\le u\le\dfrac32\right\}[/tex].
Now the area element of the integral over [tex]R[/tex] in the [tex]u[/tex]-[tex]v[/tex] plane is given by the determinant of the following Jacobian matrix:
[tex]J=\dfrac{\partial(x,y)}{\partial(u,v)}=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}[/tex]
[tex]\det J=\begin{vmatrix}\frac1v&-\frac u{v^2}\\0&1\end{vmatrix}=\dfrac1v[/tex]
[tex]|\det J|=\mathrm dA=\mathrm dx\,\mathrm dy=\dfrac1v\,\mathrm du\,\mathrm dv[/tex]
So the integral is equivalent to
[tex]\displaystyle\iint_R8xy\,\mathrm dA=\int_{u=2/3}^{u=3/2}\int_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}8\frac uv v\frac1v\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle8\int_{u=2/3}^{u=3/2}\int_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}\frac uv\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle8\int_{u=2/3}^{u=3/2}u\ln v\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}\,\mathrm du[/tex]
[tex]=\displaystyle8\ln\frac32\int_{u=2/3}^{u=3/2}u\,\mathrm du[/tex]
[tex]=\displaystyle4\ln\frac32u^2\bigg|_{u=2/3}^{u=3/2}[/tex]
[tex]=\dfrac{65}{18}\ln\dfrac32[/tex]
