We are asked to evaluate $\displaystyle \int_A x^{-1}dV(A)$, with $A=\{ (x,y):2<x+y<4,y>0,x-y>0\}$. From the solutions we know that$\displaystyle \int_A x^{-1}dV(A)=2Log(2)$. The point is to use the change of coordinates theorem to evaluate the integral.

After proving that $\mathbf{g}(s,t)=(s^2+t^2,s^2-t^2)$ is regular (it's univalent only in each quadrant), I've tried computing $\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)$, with $B=\{ (s,t):2<s^2<4,\text{ }s^2>t^2,\text{ and } |s|>|t| \}$.

Well, when trying to compute $\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)$,

$\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)= \int^2_\sqrt{2}\int^s_0\frac{8st}{s^2+t^2}+\int^2_\sqrt{2}\int^0_{-s}\frac{-8st}{s^2+t^2}+\int^{-\sqrt{2}}_{-2}\int^0_s\frac{8st}{s^2+t^2}+\int^{-\sqrt{2}}_{-2}\int^{-s}_0\frac{-8st}{s^2+t^2}=4Log(16)$

Where did I go wrong? Any help will be appreciated.

  • $\begingroup$ It seems that A should also have $s,t>0$ for the exercise to be correct. $\endgroup$ Commented Jan 2, 2014 at 21:08

2 Answers 2


Method A: With Change of Variable as chosen by the OP

The OP changed the region A into B with the following variables: $$s^2+t^2=x\ \text{and} \ s^2-t^2=y$$ These equations define a transformation $T^{-1}$ from the $x\text{-}y$ plane to the $s\text{-}t$ plane. The Jacobian of T is:

$$\begin{vmatrix} 2s & 2t \\[0.3em] 2s & -2t \end{vmatrix}=-8st$$ To find the region A in the $s\text{-}t$ plane we look at the sides of the image in Method B.

$$y=0,\ \ x=y,\ \ x+y=4,\ \ x+y=2$$ This transforms to the image lines of B $$s^2=t^2, \ \ t=0, \ \ s^2=1, \ \ s^2=\sqrt{2} $$ Notice that this mapping is not $1\text{-}1$, hence its not possible to directly use the change of variable theorem. If we limit the region to only $s,t>0$ then we would have a one-to-one mapping. So an added constraint is $s,t>0$

The region B should be divided into symmetric parts $B1$ and $B2$ as shown in the image below. However, to insure $1\text{-}1$ mapping the relevant region is only $B1$

enter image description here

Below is the integral to find $B1$

$$\int \limits_{B1} \frac{|-8st|}{s^2+t^2} dtds=\int_{1}^{\sqrt{2}} \int_0^{s} \frac{|-8st|}{s^2+t^2} dtds=\ln(4)$$

Method B: Without Change of Variable

The first thing to do in this exercise is draw the region A. Its a very easy region to draw. I graphed it and the image is shown below, enter image description here

We are trying to find the double integral over the enclosed region. It is clear that this region needs to be divided into two parts, $A_1$ and $A_2$ as shown in the image below with the red region and green region representing $A_1$ and $A_2$ respectively. enter image description here

The two integrals are then supposed to be expressed as such:

$$\underbrace{\int_1^2\int_{2-x}^{x}x^{-1}dydx}_{A_1}+\underbrace{\int_2^4\int_{0}^{4-x} x^{-1} dydx}_{A_2}$$ $$\left[2-\ln(4) \right] +\left[\ln(16)-2\right]=\ln(4)=2\ln(2)$$

  • $\begingroup$ @John, Thanks for your answer, but the point is to use the change of coordinates theorem to evaluate the integral. And, if not to much trouble, to understand where I got it wrong. $\endgroup$ Commented Jan 2, 2014 at 18:31
  • $\begingroup$ @Anoldmaninthesea. If the problem can be solved in an easier method, then it's most logically solved in that method.There is no reason why the region should be parametrized with your given change of variables. $\endgroup$
    – John
    Commented Jan 2, 2014 at 18:41
  • $\begingroup$ @John, some exercises have a pedagogical point. Your resolution is fine, from what I could understand, but it's not what I asked at the beginning. That's why I cannot accept it as an answer to my question. However, it doesn't mean that I'm not grateful to you, because I am. Thanks for your interest in this question. $\endgroup$ Commented Jan 2, 2014 at 18:59
  • $\begingroup$ @Anoldmaninthesea. After you did your change of variables, did you draw your new region. And another thing, you jacobian determinant is $-8st$ $\endgroup$
    – John
    Commented Jan 2, 2014 at 19:02
  • 1
    $\begingroup$ Jeez, this question was annoying. $\endgroup$
    – John
    Commented Jan 2, 2014 at 21:09

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{? \equiv \int_{A}{\dd V \over x}\quad\mbox{where}\quad A \equiv \braces{\pars{x,y}\quad \ni\quad 2 < x + y < 4,\quad y>0,\quad x - y > 0}}$

The use of the $\it Heaviside\ Step$ function $\Theta:{\mathbb R}\verb*\*\braces{0} \to {\mathbb R}$: $$ \Theta\pars{x} \equiv \left\lbrace \begin{array}{lcl} 0 & \mbox{if} & x < 0 \\[1mm] 1 & \mbox{if} & x > 0 \end{array}\right. $$ avoids any pictures.

\begin{align} \color{#0000ff}{\Large ?}& = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{x + y - 2}\Theta\pars{4 - x - y}\Theta\pars{y}\Theta\pars{x - y}\, {\dd x\,\dd y \over x} \\[3mm]&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{y - 2}\Theta\pars{4 - y}\Theta\pars{y - x}\Theta\pars{2x - y}\, {\dd x\,\dd y \over x} \\[3mm]&= \int_{-\infty}^{\infty}\dd y\,\Theta\pars{y - 2}\Theta\pars{4 - y} \int_{y/2}^{y}\Theta\pars{y}\,{\dd x \over x} = \int_{2}^{4}\dd y\,\ln\pars{y \over y/2} = \color{#0000ff}{\large 2\ln\pars{2}} \end{align}


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