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. 
 A: 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 $u=0$).

A: Given $x_{0}\ \ni\ x_{0}^{2} + 1 = 1/x_{0}
       \quad\Longrightarrow\quad
       x_{0}^{3} + x_{0} - 1 = 0$ and $x_{0}\ \in\ \left(0, 1\right)$.
\begin{eqnarray*}
&&
\int_{0}^{x_{0}}{\rm d}x\int_{0}^{x^{2} + 1}\left(2x^{2} + y\right)\,{\rm d}y
+
\int_{x_{0}}^{1}{\rm d}x\int_{0}^{1/x}\left(2x^{2} + y\right)\,{\rm d}y
\\&&=
\int_{0}^{x_{0}}
\left\lbrack
2x^{2}\left(x^{2} + 1\right) + {\left(x^{2} + 1\right)^{2} \over 2}
\right\rbrack\,{\rm d}x
+
\int_{x_{0}}^{1}
\left\lbrack
2x^{2}\,{1 \over x} + {\left(1/x\right)^{2} \over 2}
\right\rbrack\,{\rm d}x
\\&&=
\int_{0}^{x_{0}}\left(%
{5 \over 2}\,x^{4} + 3x^{2} + {1 \over 2}\,
\right)\,{\rm d}x
+
\int_{x_{0}}^{1}\left(2x + {1 \over 2x^{2}}\right)\,{\rm d}x
\\&&=
\left({1 \over 2}\,x_{0}^{5} + x_{0}^{3} + {1 \over 2}\,x_{0}\right)
+
\left(1 - x_{0}^{2} - {1 \over 2} + {1 \over 2x_{0}}\right)
=
{x_{0}^{6} + 2x_{0}^{4} - 2x_{0}^{3} + x_{0}^{2} + x_{0} + 1
 \over
2x_{0}}
\\&&=
{\left(1 - x_{0}\right)^{2} + 2x_{0}\left(1 - x_{0}\right) - 2\left(1 - x_{0}\right)
 +
 x_{0}^{2} + x_{0} + 1
 \over
2x_{0}}
=
{3x_{0}
 \over
2x_{0}}
=
{3 \over 2}
\end{eqnarray*}
