Evaluate $\int_A x^{-1}$ 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.

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$\ds{? \equiv \int_{A}{\dd V \over x}\quad\mbox{where}\quad
A \equiv \braces{\pars{x,y}\quad \ni\quad 2 < x + y < 4,\quad y>0,\quad x - y > 0}}$

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}
