# Double integral with messy variable substitution

I was looking through old multivariable analysis exams and found this double integral (with solution). My problem is I can't seem to understand how the transformation to the new area of integration is done. So here's the question:

Calculate $\int\int_D (2x^2+y)\,dx\,dy$ where $D$ is limited by the functions: $x = 0, x= 1, y= 0, y=\frac{1}{x}$ and $y = x^2 + 1$

$D$ looks somewhat like a house and the intersection $x^2 + 1=\frac{1}{x}$ gives a messy solution so that's why this substitution is used (in the supplied solution) instead:

$\begin{cases} u = xy \\ v = y -x^2 \end{cases}$

We get the new area $E$:

$\begin{cases} u-1 \leq v \leq 1 \\ 0 \leq u \leq 1 \end{cases}$

From here it's very easy to solve since:

$\dfrac{d(u,v)}{d(x,y)} = y+2x^2$ so we have $(y+2x^2)\,dx\,dy = du\,dv$

What I don't understand is how $v$ gets the lower limit $u-1$. How this way of solving the problem should become obvious at all is also still a mystery to me.

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Here is a picture of the 'house' and the different bounds :

The idea of the parameterization is :

• for $\,u:=xy\;$ to have a simple way to represent the hyperbola $y=\frac 1x$ for $u=1$ and, as $u$ decreases to $0$, a 'propagation' of hyperbolae symmetric around $y=x$ down to the asymptotic $x$ and $y$ axis for $u=0$.
• for $\,v:=y-x^2\;$ chosen too to get a simple upper bound $v=1$. Of course here $v=0$ is simply the parabola $y=x^2$ and won't give the 'wall' at the right.
To get the correct bound simply set $x=1$ to obtain $\;u=y\,$ and $\,v=y-1\;$ so that, eliminating the parameter $y$, we get indeed $\,\boxed{v=u-1}$ (starting with $x=1$ should be clearer that starting the other way : from $v=u-1$ get $\;y-x^2=xy-1\,$ rewritten as $\;1-x^2=y(x-1)\,$ that is $\,x=1\,$ or $\,y=-x-1\,$ not retained because under $u=0$).
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I don't think I've seen a more clear explanation of domain parameterization, albeit a special case! Thank you! –  A.E Aug 19 '13 at 5:03
Glad you liked it @AEdwards ! –  Raymond Manzoni Aug 19 '13 at 8:54
Given $x_{0}\ \ni\ x_{0}^{2} + 1 = 1/x_{0} \quad\Longrightarrow\quad x_{0}^{3} + x_{0} - 1 = 0$ and $x_{0}\ \in\ \left(0, 1\right)$.