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\documentclass[a4paper,10pt,draft]{article}


\usepackage[latin1]{inputenc}
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\usepackage{amssymb,amsmath}
\usepackage{enumerate}
\usepackage{float}
\usepackage{a4wide}

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\begin{document}


\def\testjob{resteste7}
\def\solutionname{resteste7 }
\edef\thisJobName{\jobname}
\edef\thisJobName{\meaning\thisJobName}
\edef\testjob{\meaning\testjob}
\newif\ifsolution
\ifx\thisJobName\testjob
 \solutiontrue
 \else
 \solutionfalse
\fi


\thispagestyle{empty}
\sffamily
\begin{center} 
{\Large{\bf \sf An\'alise Matem\'atica III} \\
\vspace{.5cm}
\large{\sf 1${}^{\text{\underline{o}}}$ semestre de 2005/2006}}\\
\end{center}

\vspace{1cm} 

\noindent {\large{\bf \sf %
Exerc\'{\i}cio Teste 7}} (a entregar na semana de 7/11/2005)

\vspace{.5cm} 

\begin{enumerate}[1.]
\item Use os teoremas de mudan\c{c}a de coordenadas e da fun\c{c}\~ao
  inversa para calucular
$$\int_U e^{x^2 + y^2 -x^2y^2}\sqrt{1-x^2}$$ 
em que $U = \{(x,y) \in \R^2 \, |\, 0 < x < 1 \text{ e } 0 < y < 1\}$.

\noindent\textit{Sugest\~ao:} Considere a aplica\c{c}\~ao $\Phi(x,y) = (x,y\sqrt{1-x^2})$.

\end{enumerate}

\ifsolution \immediate\write16{running resolution }

\vspace{1cm}
  
\noindent{\large\bf Resolu\c{c}\~ao:} 

\begin{enumerate}[1.]
\item Considere a fun\c{c}\~ao $\Phi: U \to \R^2$ dada por $\Phi(x,y)
  = (x,y\sqrt{1-x^2})$.  Tem-se que $\Phi$ \'e (i) de classe
  $\mathcal{C}^1$, (ii) injectiva, e (iii) $\det D\Phi =
  \sqrt{1-x^2}\neq 0$ para cada $(x,y) \in U$.  Pelo teorema da
  fun\c{c}\~ao inversa, o subconjunto $V = \Phi(U) \subseteq \R^2$ \'e
  aberto e existe uma fun\c{c}\~ao $\Psi : V \to U$ de classe
  $\mathcal{C}^1$ tal que $\Phi\circ \Psi (u,v) = (u,v)$ e
  $\Psi\circ\Phi (x,y) =(x,y)$. Portanto $\Phi : U \to V$ \'e um
  difeomorphismo de classe $\mathcal{C}^1$. Seja
  $$f(x,y) = e^{x^2 + y^2 -x^2y^2}\sqrt{1-x^2}.$$  Pelo teorema de
  mudan\c{c}a de coordenadas obt\'em-se
  $$\int_U f(x,y) = \int_{\Psi(V)}f(x,y) = 
  \int_V f\circ\Psi(u,v) |\det D\Psi(u,v)|.$$ 

  Seja $\Psi(u,v) = \bigl(\psi_1(u,v),\psi_2(u,v)\bigr)$.  Ent\~ao
  $$|\det D\Psi(u,v)| = |\det D\Phi(\Psi(u,v))|^{-1} =
  \bigl(1-\psi_1(u,v)^2\bigr)^{-1/2},$$ e como $\Phi\circ\Psi(u,v) =
  (u,v)$ obt\'em-se $\psi_1(u,v) = u$ e
  $\psi_2(u,v)\sqrt{1-\psi_1(u,v)^2} = v$.  Segue-se
  \begin{align*}
    f\circ\Psi(u,v) |\det D\Psi(u,v)| &=
    f\bigl(\psi_1(u,v),\psi_2(u,v)\bigr)\bigl(1-\psi_1(u,v)^2\bigr)^{-1/2} \\
    &= e^{\psi_1(u,v)^2 + \psi_2(u,v)^2 - \psi_1(u,v)^2\psi_2(u,v)^2} \sqrt{1-\psi_1(u,v)^2}\bigl(1-\psi_1(u,v)^2\bigr)^{-1/2} \\
    &= e^{\psi_1(u,v)^2 + \psi_2(u,v)^2(1-\psi_1(u,v)^2)} \\
    &= e^{u^2 + v^2}.
  \end{align*}

  Para cada $b \in \; ]0,1[$ considere o segmento horizontal $S_b
  =\{(x,b) \in U \, |\, 0 < x < 1\}$.  Tem-se $(u,v) = \Phi(x,b) = (x,
  b\sqrt{1-x^2})$.  Logo $\Phi(S_b)$ pertence ao elipse $u^2 +
  \frac{v^2}{b^2} = 1$, e segue-se que
  $$V= \{(u,v) \in \R^2 \,|\, u^2 + v^2 < 1, 0< u < 1 
  \text{ e } 0 < v < 1\}.$$ Usando o teorema de mudan\c{c}a de
  coordenadas, com $(u,v) = (r \cos \theta, r \sen \theta)$, obt\'em-se
  \begin{align*}
    \int_U e^{x^2 + y^2 -x^2y^2}\sqrt{1-x^2} &= \int_V e^{u^2 + v^2}
    \\
    &= \int_0^1\int_0^{\pi/2} e^{r^2} r \, d\theta dr \\
    &= \frac{\pi}{4}(e- 1).
  \end{align*}
\end{enumerate}


\else 

\immediate\write16{Para fazer uma resolucao deve escrever latex
  -jobname=\solutionname \jobname} \fi

\immediate\write16{testjob:\testjob}
\immediate\write16{thisJobName:\thisJobName}

\end{document}