\noindent Calcule o integral de superfície
$$
I=\int\int_{z=2(x^2+y^2)^{-2}, \; x^2+y^2\leq 4} (1+64 (x^2+y^2)^{-5})^{-1} (x^2+y^2)^{-6}dS.
$$

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

Pela definição do integral de superfície temos
$$
 I=\int\int_{x^2+y^2\leq 4} (1+64 (x^2+y^2)^{-5})^{-1} (x^2+y^2)^{-6}
 \sqrt{1+((2 (x^2+y^2)^{-2})'_x)^2+ ((2 (x^2+y^2)^{-2})'_y)^2}dxdy
$$
$$
 =\int\int_{x^2+y^2\leq 4} (1+64 (x^2+y^2)^{-5})^{-1} (x^2+y^2)^{-6} \sqrt{1+64 (x^2+y^2)^{-5}}dxdy.
$$
A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (2);
\draw[black] (0,0) circle (2);
\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
$$
 I=\int_0^{2\pi}d\phi \int_0^{2} (1+64 r^{-10})^{-1}r^{-12}\sqrt{1+64 r^{-10}}rdr.
$$
Fazendo $u=1+64 r^{-10}$, obtemos
$$
 I=\frac{2\pi}{-640} \int_1^{{{17}\over{16}}} u^{-1+\frac{1}{2}}du = {{4-\sqrt{17}}\over{640}} \pi .
$$