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   \marginpar{#1}
 }{

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\begin{document}
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\exametitle{Calculo Vetorial}{Ficha 2}{\today}
\begin{questao}{}
\noindent Calcule o integral de superfície
$$
I=\int\int_{z=\sqrt{x^2+y^2}, \; 1 \leq \sqrt{x^2+y^2} \leq 2} x^2 z dS.
$$

....................

Pela definição do integral de superfície temos
$$
 I=\int\int_{1 \leq \sqrt{x^2+y^2} \leq 2} x^2 \sqrt{x^2+y^2}
 \sqrt{1+((\sqrt{x^2+y^2})'_x)^2+ ((\sqrt{x^2+y^2})'_y)^2}dxdy
$$
$$
 =\int\int_{1 \leq \sqrt{x^2+y^2} \leq 2} x^2 \sqrt{x^2+y^2}
 \sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy.
$$
$$
 =\int\int_{1 \leq \sqrt{x^2+y^2} \leq 2} x^2 \sqrt{x^2+y^2}
 \sqrt{2}dxdy.
$$

A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (2);
\fill[white] (0,0) circle (1);
\draw[black] (0,0) circle (2);
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\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\sqrt{2}\int_0^{2\pi} \int_{1}^{2} \cos^2(\phi)  r^4  dr d\phi
$$
$$
 =\sqrt{2}\int_0^{2\pi}  \cos^2(\phi)  \left(\frac{2^5}{5}-\frac{1^5}{5}\right) d\phi  = \sqrt{2}\frac{31}{5}  \int_{0}^{2\pi}\left(\frac{1+\cos(2\phi)}{2}\right) d\phi = \sqrt{2}\frac{31}{5}\pi .
$$
\end{questao}
\begin{questao}{}
\noindent Calcule o integral de superfície
$$
I=\int\int_{z=\sqrt{x^2+y^2}, \; 2 \leq \sqrt{x^2+y^2} \leq 3} x^2 z dS.
$$

....................

Pela definição do integral de superfície temos
$$
 I=\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{1+((\sqrt{x^2+y^2})'_x)^2+ ((\sqrt{x^2+y^2})'_y)^2}dxdy
$$
$$
 =\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy.
$$
$$
 =\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{2}dxdy.
$$

A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (3);
\fill[white] (0,0) circle (2);
\draw[black] (0,0) circle (3);
\draw[black] (0,0) circle (2);
\draw[->](-10,0)--(10,0) node[right]{$x$};
\draw[->](0,-10)--(0,10) node[above]{$y$};
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\draw (\x ,-1pt) -- (\x ,1pt);
\foreach \y in {-8,...,8}
\draw (-1pt,\y ) -- (1pt,\y );
\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\sqrt{2}\int_0^{2\pi} \int_{2}^{3} \cos^2(\phi)  r^4  dr d\phi
$$
$$
 =\sqrt{2}\int_0^{2\pi}  \cos^2(\phi)  \left(\frac{3^5}{5}-\frac{2^5}{5}\right) d\phi  = \sqrt{2}\frac{211}{5}  \int_{0}^{2\pi}\left(\frac{1+\cos(2\phi)}{2}\right) d\phi = \sqrt{2}\frac{211}{5}\pi .
$$
\end{questao}
\begin{questao}{}
\noindent Calcule o integral de superfície
$$
I=\int\int_{z=\sqrt{x^2+y^2}, \; 2 \leq \sqrt{x^2+y^2} \leq 3} x^2 z dS.
$$

....................

Pela definição do integral de superfície temos
$$
 I=\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{1+((\sqrt{x^2+y^2})'_x)^2+ ((\sqrt{x^2+y^2})'_y)^2}dxdy
$$
$$
 =\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy.
$$
$$
 =\int\int_{2 \leq \sqrt{x^2+y^2} \leq 3} x^2 \sqrt{x^2+y^2}
 \sqrt{2}dxdy.
$$

A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (3);
\fill[white] (0,0) circle (2);
\draw[black] (0,0) circle (3);
\draw[black] (0,0) circle (2);
\draw[->](-10,0)--(10,0) node[right]{$x$};
\draw[->](0,-10)--(0,10) node[above]{$y$};
\foreach \x/\xtext in { 2/2,  4/4, 6/6, 8/8 }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$\xtext$};
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\draw (\x ,-1pt) -- (\x ,1pt);
\foreach \y in {-8,...,8}
\draw (-1pt,\y ) -- (1pt,\y );
\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\sqrt{2}\int_0^{2\pi} \int_{2}^{3} \cos^2(\phi)  r^4  dr d\phi
$$
$$
 =\sqrt{2}\int_0^{2\pi}  \cos^2(\phi)  \left(\frac{3^5}{5}-\frac{2^5}{5}\right) d\phi  = \sqrt{2}\frac{211}{5}  \int_{0}^{2\pi}\left(\frac{1+\cos(2\phi)}{2}\right) d\phi = \sqrt{2}\frac{211}{5}\pi .
$$
\end{questao}
\begin{questao}{}
\noindent Calcule o integral de superfície
$$
I=\int\int_{z=\sqrt{x^2+y^2}, \; 3 \leq \sqrt{x^2+y^2} \leq 4} x^2 z dS.
$$

....................

Pela definição do integral de superfície temos
$$
 I=\int\int_{3 \leq \sqrt{x^2+y^2} \leq 4} x^2 \sqrt{x^2+y^2}
 \sqrt{1+((\sqrt{x^2+y^2})'_x)^2+ ((\sqrt{x^2+y^2})'_y)^2}dxdy
$$
$$
 =\int\int_{3 \leq \sqrt{x^2+y^2} \leq 4} x^2 \sqrt{x^2+y^2}
 \sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy.
$$
$$
 =\int\int_{3 \leq \sqrt{x^2+y^2} \leq 4} x^2 \sqrt{x^2+y^2}
 \sqrt{2}dxdy.
$$

A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (4);
\fill[white] (0,0) circle (3);
\draw[black] (0,0) circle (4);
\draw[black] (0,0) circle (3);
\draw[->](-10,0)--(10,0) node[right]{$x$};
\draw[->](0,-10)--(0,10) node[above]{$y$};
\foreach \x/\xtext in { 2/2,  4/4, 6/6, 8/8 }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$\xtext$};
\foreach \x/\xtext in {-8/8, -6/6, -4/4, -2/2 }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$-\xtext$};
\foreach \y/\ytext in { 2/2,  4/4, 6/6, 8/8}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$\ytext$};
\foreach \y/\ytext in {-8/8, -6/6, -4/4, -2/2}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$-\ytext$};
\foreach \x in {-8,...,8}
\draw (\x ,-1pt) -- (\x ,1pt);
\foreach \y in {-8,...,8}
\draw (-1pt,\y ) -- (1pt,\y );
\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\sqrt{2}\int_0^{2\pi} \int_{3}^{4} \cos^2(\phi)  r^4  dr d\phi
$$
$$
 =\sqrt{2}\int_0^{2\pi}  \cos^2(\phi)  \left(\frac{4^5}{5}-\frac{3^5}{5}\right) d\phi  = \sqrt{2}\frac{781}{5}  \int_{0}^{2\pi}\left(\frac{1+\cos(2\phi)}{2}\right) d\phi = \sqrt{2}\frac{781}{5}\pi .
$$
\end{questao}
\begin{questao}{}
\noindent Calcule o integral de superfície
$$
I=\int\int_{z=3(x^2+y^2)^{-2}, \; x^2+y^2\leq 9} (1+144 (x^2+y^2)^{-5})^{-2} (x^2+y^2)^{-6}dS.
$$

....................

Pela definição do integral de superfície temos
$$
 I=\int\int_{x^2+y^2\leq 9} (1+144 (x^2+y^2)^{-5})^{-2} (x^2+y^2)^{-6}
 \sqrt{1+((3 (x^2+y^2)^{-2})'_x)^2+ ((3 (x^2+y^2)^{-2})'_y)^2}dxdy
$$
$$
 =\int\int_{x^2+y^2\leq 9} (1+144 (x^2+y^2)^{-5})^{-2} (x^2+y^2)^{-6} \sqrt{1+144 (x^2+y^2)^{-5}}dxdy.
$$
A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (3);
\draw[black] (0,0) circle (3);
\draw[->](-7,0)--(7,0) node[right]{$x$};
\draw[->](0,-7)--(0,7) node[above]{$y$};
\foreach \x/\xtext in { 2/2,  4/4,  6/6, }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$\xtext$};
\foreach \x/\xtext in {-6/6, -4/4, -2/2 }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$-\xtext$};
\foreach \y/\ytext in { 2/2, 4/4, 6/6}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$\ytext$};
\foreach \y/\ytext in {-6/6, -4/4, -2/2}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$-\ytext$};
\foreach \x in {-7,...,7}
\draw (\x ,-1pt) -- (\x ,1pt);
\foreach \y in {-7,...,7}
\draw (-1pt,\y ) -- (1pt,\y );
\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\int_0^{2\pi}d\phi \int_0^{3} (1+144 r^{-10})^{-2}r^{-12}\sqrt{1+144 r^{-10}}rdr.
$$
Fazendo $u=1+144 r^{-10}$, obtemos
$$
 I=\frac{2\pi}{-1440} \int_1^{{{6577}\over{6561}}} u^{-2+\frac{1}{2}}du = {{81-\sqrt{6577}}\over{360\,\sqrt{6577}}} \pi .
$$
\end{questao}


\end{document}
