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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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up vote 2 down vote accepted

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 $x_{0}\ \in\ \left(0, 1\right)$.

\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*}

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