# how to find a good change of variables to solve these kinds of multi-integral questions?

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.

• Your edit is correct. Commented Jul 22 at 18:01
• I guess you edited your post as I was posting my explanation. Yes, this is fine. (You could just as well omit the factor of $2$ in your definition of $v$.) Commented Jul 22 at 18:01

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 $$1/v$$.

When it comes to changes of variables, one possible approach is to determine formulas for $$x(u,v)$$, $$y(u,v)$$, compute the Jacobian, and apply the change of variables theorem as you outline. In some situations, you can actually get away with using the implicit formulas $$u(x,y)$$ and $$v(x,y)$$ instead. Let's see if that works in this case.

For my own benefit, I'll recall the change of variables theorem. Suppose $$(x,y) = T(u,v)$$ is a transformation taking the region $$D^*$$ in the $$uv$$ plane to the region $$D$$ in the $$xy$$ plane. Then:

$$\iint_D f(x,y)\, dx\,dy = \iint_{D^*}f\big(T(u,v)\big)\, \left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv$$

In our case, the transformation $$T$$ is given implicitly by $$u = x+y$$ and $$v = \frac{y}{2x^3}$$. Then we can compute the inverse Jacobian as follows:

$$\frac{\partial(u,v)}{\partial(x,y)} = \det \begin{pmatrix} 1 & \frac{-3y}{2x^4} \\ 1 & \frac{1}{2x^3} \end{pmatrix} = \frac{1}{2x^3} + \frac{3y}{2x^4} = \frac{x+3y}{2x^4}$$

Notice the $$x+3y$$ in the numerator matches with the $$x+3y$$ in the integrand. We can then exploit the fact that the Jacobian of a transformation is the (multiplicative) inverse of the Jacobian of the inverse transformation, or in other words $$\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{\frac{\partial(u,v)}{\partial(x,y)}}$$. Our integral then becomes:

$$\iint_{D^*} \frac{x+3y}{xy}\cdot\left|\frac{2x^4}{x+3y}\right|du\,dv$$

which can be simplified nicely, and the substitution $$(x,y) \to (u,v)$$ can be carried out implicitly as well, exploiting the structure of the integrand.

Edit: Well done with your solution! The implicit approach we have utilized here is strange at first glance, perhaps suspicious, but it is perfectly rigorous and formally valid.

Typically, we think of a substitution like $$(x,y) \to (u,v)$$ as exchanging one set of variables for another, and never the twain shall meet. In fact, we merely have four different variables which happen to be related in a particular way, namely that $$u = x+y$$ and $$v = y/(2x^3)$$. This relationship goes both ways, and it is always possible to express either integrand on either side of the change of variables theorem in terms of either $$(x,y)$$ or $$(u,v)$$, we just have to be careful that in the end, we integrate with respect to the correct variables and apply the correct bounds to those variables. You can think of our calculations here as merely "abbreviating" the terms of the integrand using the variables $$x$$ and $$y$$, which secretly stand in for some functions of $$u$$ and $$v$$ all along.