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\begin{document} \huge

\noindent 1. Calculate
\begin{eqnarray} \nonumber
  I &=& \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
  dx\,dy\,e^{-(x+y)^2-(x-y)^2}\,.
\end{eqnarray} 
Solution:
\begin{eqnarray} \nonumber
  \xi\ =\ x+y\,, &~~~& \eta\ =\ x-y\,, \\ \nonumber
  x\ =\ \frac{1}{2}\,\left(\xi+\eta\right)\,, &~~~&
  y\ =\ \frac{1}{2}\,\left(\xi-\eta\right)\,,\\ \nonumber
  \left| \begin{matrix} 
  \frac{\partial x}{\partial\xi} & \frac{\partial x}{\partial\eta}\cr
  \frac{\partial y}{\partial\xi} & \frac{\partial y}{\partial\eta}\cr
  \end{matrix} \right|\ =\
  \left|\begin{matrix} \frac{1}{2} &   \frac{1}{2}\cr
                       \frac{1}{2} & - \frac{1}{2}\end{matrix}\right| 
  &=& - \frac{1}{2}\,.
\end{eqnarray} 
The minus comes because the transformation switches the righ-handed
system 
$$ d\vec{x}=\hat{x}\,dx\,,~~~d\vec{y}=\hat{y}\,dy $$ 
into the left-handed system
$$ d\vec{\xi}= \left( \begin{matrix} 0.5 \cr 0.5 \end{matrix} \right)\,
   d\xi\,,~~~ d\vec{\eta}= \left(\begin{matrix} 0.5\cr -0.5\end{matrix} 
   \right)\,d\eta\,. $$ 
Interchanging the definitions of $\xi$ and $\eta$ to $\xi=x-y$ and 
$\eta=x+y$ gives a right-handed system and Jacobian $+1/2$. Therefore,
\begin{eqnarray} \nonumber
  I &=& \frac{1}{2} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
  d\xi\,d\eta\,e^{-\xi^2-\eta^2}\ =\ \frac{\pi}{2}\,.
\end{eqnarray} 

\bigskip
\noindent 2. Calculate
\begin{eqnarray} \nonumber
  I &=& \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
  \int_{-\infty}^{+\infty} dx_1\,dx_2\,dx_3\,\\ \nonumber
  &~& \\ \nonumber
  &~& e^{-(x_1+x_2+x_3)^2-(x_1+x_2-x_3)^2-(x_1-x_2-x_3)^2}\,.
\end{eqnarray} 
Solution:
\begin{eqnarray} \nonumber
  \xi_1&=&x_1+x_2+x_3\,,\\ \nonumber
  \xi_2&=&x_1+x_2-x_3\,,\\ \nonumber
  \xi_3&=&x_1-x_2-x_3\,,
\end{eqnarray} 
\begin{eqnarray} \nonumber
  x_1&=&\frac{1}{2}\,\left(\xi_1+\xi_3\right)\,,\\ \nonumber
  x_2&=&\frac{1}{2}\,\left(\xi_2-\xi_3\right)\,,\\ \nonumber
  x_3&=&\frac{1}{2}\,\left(\xi_1-\xi_2\right)\,.
\end{eqnarray} 
Therefore,
\begin{eqnarray} \nonumber
  \left| \begin{matrix} \frac{\partial x_1}{\partial\xi_1} 
                      & \frac{\partial x_1}{\partial\xi_2} 
                      & \frac{\partial x_1}{\partial\xi_3} 
                    \cr \frac{\partial x_2}{\partial\xi_1} 
                      & \frac{\partial x_2}{\partial\xi_2} 
                      & \frac{\partial x_2}{\partial\xi_3} 
                    \cr \frac{\partial x_3}{\partial\xi_1} 
                      & \frac{\partial x_3}{\partial\xi_2} 
                      & \frac{\partial x_3}{\partial\xi_3} 
  \end{matrix} \right| &=&
  \left| \begin{matrix} 
              \frac{1}{2} &        0     &  \frac{1}{2} 
         \cr        0     &  \frac{1}{2} & -\frac{1}{2} 
         \cr  \frac{1}{2} & -\frac{1}{2} &        0
  \end{matrix} \right|\ =\ - \frac{1}{4}\,.
\end{eqnarray} 
Again the minus sign is due to a switch from a right-handed
to a left-handed system and the integral becomes 
\begin{eqnarray} \nonumber
  I &=& \frac{1}{4} \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
  \int_{-\infty}^{+\infty} d\xi_1\,d\xi_2\,d\xi_3\, \\ \nonumber &~& 
  \\ \nonumber &~& e^{-(\xi_1)^2-(\xi_2)^2-(\xi_3)^2}\ =\ 
  \frac{\pi^{3/2}}{4}\,.
\end{eqnarray} 

\vfil

\end{document}
