\documentclass[a4paper,11pt,twoside,reqno]{amsart}
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%\title{Nota_Problema_23f_Tema2_Integracio}
%\author{Juan R Pacha}
%\date{May 2014}

\textwidth 170mm %155%
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\usepackage{natbib}
\usepackage{graphicx}
\usepackage{subfigure}

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\usepackage{fancyhdr}
 
\pagestyle{fancy}
\fancyhf{}
\fancyhead[LO, RE]{Tema 2}
\fancyhead[LE, RO]{Integraci\'{o} de funcions de varies variables}
 

\newcommand{\rmd}{\ensuremath{\mathrm{d}}}

\begin{document}
%\thispagestyle{empty}

%\maketitle

%\section*{Nota}
\thispagestyle{fancy}

\section*{Problema 23. Apartat f}
\noindent
Es comprova que el canvi 
\begin{equation}\label{eq:xyx}
	T:\
u = \frac{y^{2}}{x},\qquad
v = \frac{z^{2}}{y},\qquad
w = \frac{x^{2}}{z},
\end{equation}
transforma el domini original $\mathcal{D}$, en
$\mathcal{D'}=[1,a]\times [1,a]\times [1,a]$. És a dir, en un cub
d'aresta $a-1$ (veure Figura~\ref{fig:1}). Aleshores
el Jacobià corresponent surt més senzill. En efecte: 
%\begin{displaymath}
%\det \frac{\partial (u,v,w)}{\partial (x,y,z)}=\begin{vmatrix}[2.25]
%-\dfrac{y^{2}}{x^{2}} & 2\dfrac{y}{x} & 0\\
%0 & -\dfrac{z^{2}}{y^{2}} & 2\dfrac{z}{y}\\
%2\dfrac{x}{z} & 0 & -\dfrac{x^{2}}{z^{2}}
%\end{vmatrix} = -7,
%\end{displaymath}
\begin{displaymath}
	\det J_{T} (x,y,z) =
	\begin{vmatrix}[2.25] -\dfrac{y^{2}}{x^{2}} & 2\dfrac{y}{x} & 0\\
0 & -\dfrac{z^{2}}{y^{2}} & 2\dfrac{z}{y}\\
2\dfrac{x}{z} & 0 & -\dfrac{x^{2}}{z^{2}}
\end{vmatrix} = -7,
\end{displaymath}
d'on:
%\begin{displaymath}
%\left|\det \frac{\partial (x,y,z)}
%{\partial (u,v,w)}\right| = 
%\frac{1}{\left|\det \dfrac{\partial (u,v,w)}{\partial
%      (x,y,z)}\right|} 
%= \frac{1}{7}.
%\end{displaymath}
\begin{displaymath}
	\left|\det J_{T^{-1}}(u,v,w)\right| =
	\frac{1}%
	{\left|\det J_{T}\left(x(u,v,w),y(u,v,w),z(u,v,w)\right)\right|}
= \frac{1}{7}.
\end{displaymath}
Així doncs, aplicant el canvi~\eqref{eq:xyx} resulta el volum: 
\begin{displaymath}
\begin{split}
V=\iiint_{\mathcal{D}}\rmd x\:\rmd y\:\rmd z
&= \iiint_{\mathcal{D'}} 
\left|\det J_{T} (u,v,w)\right|
\rmd u\:\rmd v\:\rmd w\\
&= \frac{1}{7}\int_{1}^{a} \rmd u\int_{1}^{a} \rmd v
\int_{1}^{a}\rmd w = 
\frac{1}{7}\left(\int_{1}^{a} \rmd u\right)^{3}=
\frac{(a-1)^{3}}{7}.
\end{split}
\end{displaymath}
\begin{figure}[!h] 
\subfigure[][$\mathcal{D}$: Domini definit per les superfícies,

\centering
$\begin{matrix} 
x^{2}=z,& y^{2}=x,& z^{2}=y,\\
x^{2}=az,& y^{2}=ax,& z^{2}=ay.
\end{matrix}$

\label{fig:A}]%
{\includegraphics[scale=0.56]{fig_b_prob_23f_tema2.png}}
\subfigure[][$\mathcal{D':}\; 1\le u\le a,\ 1\le v\le a,\ 
1\le w\le a.$  
\label{fig:B}]%
  {\includegraphics[scale=0.56]{fig_a_prob_23f_tema2.png}}
\caption{Transformació del domini $\mathcal{D}$ en 
$\mathcal{D'}$ pel canvi~\eqref{eq:xyx}.\label{fig:1}}
\end{figure}

\end{document}
