I appended this to a previous question of mine, but I don't think anyone saw it.
Let's say $R$ is the region defined by the curves:
$y=x$, $xy=2$, $xy=4$ and $y=\frac{1}{8}x^3$
If $y=v$ and $x=\frac{u}{v}$, then the new curves are:
$v=u^{1/2}$, $u=2$, $u=4$ and $v=\frac{1}{8^{1/4}}u^{3/4}$
Now, I have $\int\int_{R}^{}x^2 y^2 dA = \int_{2}^{4}\int_{\frac{1}{8^{1/4}}u^{3/4}}^{u^{1/2}}f(u,v)dvdu$ and am looking for $f (u,v)$. I think it's $\frac{u^2}{v}$
The equation for change of variable involving the Jacobian I learned about has the bounds in the opposite order, $du dv$ How do I do this in the $dvdu$ order?
In my opinion there is no need to change variables. We have that \begin{align*}\iint_{R}^{}x^2 y^2 dxdy&=\int_{x=2^{1/2}}^2x^2\left(\int_{y=2/x}^x y^2 dy\right)dx+\int_{x=2}^{2^{5/4}}x^2\left(\int_{y=x^3/8}^{4/x} y^2 dy\right)dx\\ &=\frac{1}{3}\int_{x=2^{1/2}}^2\left(x^5-\frac{2^3}{x}\right)dx +\frac{1}{3}\int_{x=2}^{2^{5/4}}\left(\frac{2^6}{x}-\frac{x^{11}}{2^9}\right)dx\\ &=\frac{1}{3}\left(\frac{28}{3}-4\ln(2)+16\ln(2)-\frac{14}{3}\right)=\frac{14}{9}+4\ln(2). \end{align*} Edited question (see comment below). We have that for $x=u/v$ and $y=v$, $$\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u}&\displaystyle\frac{\partial x}{\partial v}\\ \displaystyle\frac{\partial y}{\partial u}&\displaystyle\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}\displaystyle\frac{1}{v}&\displaystyle-\frac{u}{v^2}\\ \displaystyle0&\displaystyle1\end{vmatrix}=\frac{1}{v}.$$ Hence \begin{align*}\iint_{R}^{}x^2 y^5 dxdy&= \iint_Q\left(\frac{u}{v}\right)^2 v^5 \cdot \frac{dudv}{v} =\int_{2}^{4}u^2\left(\int_{\frac{u^{3/4}}{8^{1/4}}}^{u^{1/2}}v^2dv\right)du\\ &=\int_{2}^{4}u^2\left(\frac{u^{3/2}}{3}-\frac{8^{1/4}u^{9/4}}{24}\right) du =\frac{7264}{189}-\frac{32\sqrt{2}}{27}-\frac{1024\sqrt[4]{2}}{63} \end{align*} where $$Q=\left\{(u,v): 2\leq u\leq 4, \frac{u^{3/4}}{8^{1/4}}\leq v\leq u^{1/2}\right\}.$$ So in (c) the function $f(u,v)$ is $u^2v^2$.