I wonder if there's a method/strategy to find a good change of variables for double/triple integrals over some bounds.
In my problem, I'm supposed to find the integral of $f(x,y)=\frac{x+3y}{xy}$ over the bounded region which is given by $$D=\begin{cases}
-x+1\leq y\leq-x+4\\
2x^{3}\leq y\leq8x^{3}
\end{cases}$$
giving us
$$\iint_D f(x,y)$$
what first screamed to me is to set $u=y+x$ which would cause the first area to become $1\leq u\leq 4$ which seems elegant, but then what screams to me in the second area is to set $v=\frac{y}{2x^3}$ collapsing the second area into $1\leq v\leq 4$.
so far so good but then I'm supposed to find $x(u,v)$ and $y(u,v) $ but with the variables I chose it becomes a mess and I can't separate the variables, so that's why I'm asking this question.
Btw in this problem I was asked to specifically solve it with change of variables.
EDIT:
after many trials, I did the following and I want to know if it's alright or not rigorous enough (feels kinda cheaty)
as I've previously stated $$\begin{cases}
1\leq u\leq4 & | y+x=u\\
1\leq v\leq4 & | \frac{y}{2x^{3}}=v
\end{cases}$$
now I've decided to calculate the inverse of the Jacobian in the following way:
$$J^{-1}=\begin{vmatrix}u_{x} & u_{y}\\
v_{x} & v_{y}
\end{vmatrix}=\begin{vmatrix}1 & 1\\
-\frac{3y}{2x^{4}} & \frac{1}{2x^{3}}
\end{vmatrix}=\frac{1}{2x^{3}}\left(1+\frac{3y}{x}\right)=\frac{1}{2x^{3}}\left(\frac{x+3y}{x}\right)$$
which means the Jacobian is:
$$J=2x^{3}\left(\frac{x}{x+3y}\right)$$
and then the integral will become:
$$\begin{align*}
\iint_{E}f\left(x,y\right)\cdot J & =\intop_{1}^{4}\intop_{1}^{4}\frac{x+3y}{xy}\cdot2x^{3}\left(\frac{x}{x+3y}\right)dudv\\
& =\intop_{1}^{4}\intop_{1}^{4}\frac{2x^{3}}{y}dudv\\
& =\intop_{1}^{4}\intop_{1}^{4}\frac{1}{v}dudv\\
& =3\cdot\ln4
\end{align*}$$
My question now is, is this right? in this way I never defined $x,y$ as functions of $u,v$ but it seems like it works out anyway.
Best Answer
You do not need to solve for $x$ and $y$. These problems are usually constructed very carefully (and artificially) to work out just right. Remember that the Jacobian you need is the reciprocal of the Jacobian $\dfrac{\partial (u,v)}{\partial(x,y)}$, but in the end everything must be writteni in terms of $u,v$. In your case you will end up with a simple integrand of $1/v$.