\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);
\draw[black] (0,0) circle (1);
\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
$$
 I=\sqrt{2}\int_0^{2\pi} \int_{1}^{2} \cos^2(\phi)  r^4  dr d\phi
$$
$$
 =\sqrt{2}\int_0^{2\pi}  \cos^2(\phi)  \left(\frac{2^5}{5}-\frac{1^5}{5}\right) d\phi  = \sqrt{2}\frac{31}{5}  \int_{0}^{2\pi}\left(\frac{1+\cos(2\phi)}{2}\right) d\phi = \sqrt{2}\frac{31}{5}\pi .
$$