Missing factor of 1/2 in a multivariable change of variables problem

Question

I am interested in evaluating the integral

$$ \iint_{E} xy dA $$ where $E$ is the region bounded by $xy = 3$, $xy = 1$, $y = 3x$, and $y = x$.

This gives a region of integration that looks like this:

Using the substitution $x = u/v$ and $y = v$, we obtain a Jacobian of

$$\frac{\partial(x,y)}{\partial(u,v)} = 1 \cdot \frac{1}{v} - 0 \cdot \frac{-u}{v^2} = \frac{1}{v} > 0$$

which I think should turn our integral into:

$$ \int_1^3\int_{1}^3 u \cdot \frac{1}{v} du dv = 4 \ln(3). $$

However, the answer is supposed to be $2 \ln (3)$, so I seem to be missing a factor of $1/2$ somewhere.

I changed my bounds by noting that $v = y$, and $y$ goes from $1$ to $3$ in the graph above. Also, $x = u/v$ implies $u = xv = xy$, and $u = xy$ goes from $1$ to $3$ as well.

So where is my mistake exactly?

Answer 1

The problem is that "$y$ goes from $1$ to $3$ in the graph above" doesn't mean that you should integrate from $v=1$ to $v=3$ after the change of variables - you would then effectively be integrating over this region:

Indeed we can kind of smell that something should be off about that step, because there are many different regions where the range of values of $y$ is $[1, 3]$, but they don't all have the same area.

The problem is that for different values of $u$, the range of values $y$ can take in your region is different. For a fixed $u$, we have $u \le y^2 \le 3u$, so you should integrate from $v = \sqrt u$ to $v = \sqrt{3u}$:

\begin{equation*} \int_1^3 \int_{v=\sqrt u}^{v=\sqrt{3u}} \frac uv \,\mathrm dv \,\mathrm du \end{equation*}

which does give the correct answer (it is a pretty confusing coincidence that the answer happens to be half of what you got!). As mentioned in the comments, it's perhaps a bit easier to choose a change of variables which gives a nicer boundary, like $xy$ and $x/y$.

Answer 2

Here is my modification of the given Image :

The Image by user "Izaak van Dongen" is better , I just modified OP Image & highlighted the Issue with little a bit of text.

Over-view :
OP is calculating with limits $y=1$ & $y=3$ though that will give a larger area. It is $y/x$ which has limits $1$ & $3$.
The Area gives DOUBLE Integral value , $4 \ln 3$ rather than $2 \ln 3$.

Hence the Better Way is the Substitution :
$u=xy$
$v=y/x$
Limits are $1$ & $3$.

UPDATE :

To make this complete :

Jacobian should be :

$$\frac{\partial(x,y)}{\partial(u,v)} = \cdots = \frac{1}{2v}$$

In a "Sense" , that is the missing factor of $1/2$ ...

Integral should be :

$$\int_1^3\int_{1}^3 u \cdot \frac{1}{2v} du dv = \cdots = 2 \ln(3)$$