Question

Coordenadas Cilíndricas, ajuda ai galera.
Calcule a integral usando coordenadas cilíndricas:

Answer 1

Observando os limites de integração, temos que o sólido de integração $D$ é descrito em coordenadas retangulares como

$D=\left\{(x,\;y,\;z)\in\mathbb{R}^{3}\;|\right.\\ \\~~~~~~~~~\left 0\leq y\leq 2\,;\;0\leq x\leq \sqrt{4-y^{2}}\,;\;\sqrt{x^{2}+y^{2}}\leq z\leq \sqrt{8-x^{2}-y^{2}} \right \}$


Geometricamente,

a projeção de $D$ sobre o plano $xy$ é apenas o primeiro quadrante do disco descrito pela seguinte desigualdade:

$x^{2}+y^{2}\leq 4$


$z$ é positivo, e varia entre o cone de equação

$z^{2}=x^{2}+y^{2}$


e a esfera de equação

$x^{2}+y^{2}+z^{2}=8$

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Transformando para coordenadas cilíndricas:

$\begin{array}{cc} \left\{ \begin{array}{l} x=r\cos \theta\\ \\ y=r\,\mathrm{sen\,}\theta\\ \\ z=z \end{array} \right.~~&~~\begin{array}{c} r\leq z\leq \sqrt{8-r^{2}}\\ \\ 0\leq r\leq 2\\ \\ 0\leq \theta\leq \dfrac{\pi}{2} \end{array} \end{array}$


O módulo do Jacobiano desta transformação é $|\mathrm{Jac~\phi}|=r.$

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Escrevendo a integral iterada em coordenadas cilíndricas:

$\displaystyle\iiint\limits_{D}{z^{2}\,dz\,dx\,dy}\\ \\ \\ =\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}\int\limits_{r}^{\,\sqrt{8-r^{2}}}{z^{2}\cdot |\mathrm{Jac~\phi}|\,dz\,dr\,d\theta}\\ \\ \\ =\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}\int\limits_{r}^{\,\sqrt{8-r^{2}}}{z^{2}\cdot r\,dz\,dr\,d\theta}\\ \\ \\ =\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{r\cdot \left.\left(\dfrac{z^{3}}{3} \right )\right|_{r}^{\sqrt{8-r^{2}}}\,dr\,d\theta}\\ \\ \\ =\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{r\cdot \left(\dfrac{(\sqrt{8-r^{2}})^{3}}{3}-\dfrac{r^{3}}{3} \right )\,dr\,d\theta}$
$=\dfrac{1}{3}\displaystyle\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{r\cdot \left[(8-r^{2})^{3/2}-r^{3}\right]\,dr\,d\theta}\\ \\ \\ =\dfrac{1}{3}\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{\left[r(8-r^{2})^{3/2}-r^{4}\right]\,dr\,d\theta}\\ \\ \\ =\dfrac{1}{3}\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{r(8-r^{2})^{3/2}\,dr\,d\theta}-\dfrac{1}{3}\int\limits_{0}^{\,\pi/2\,}\int\limits_{0}^{2}{r^{4}\,dr\,d\theta}~~~~~~\mathbf{(i)}$

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Calculando a primeira integral:

$\displaystyle\int\limits_{0}^{2}{r(8-r^{2})^{3/2}\,dr}\\ \\ \\ =-\dfrac{1}{2}\cdot \int\limits_{0}^{2}{(-2)r\,(8-r^{2})^{3/2}\,dr}\\ \\ \\ =-\dfrac{1}{2}\cdot \int\limits_{0}^{2}{(8-r^{2})^{3/2}\,\cdot (-2r)\,dr}~~~~~~\mathbf{(ii)}$


Mudança de variável:

$8-r^{2}=u~\Rightarrow~-2r\,dr=du$


Mudando os extremos de integração:

$\text{Quando }r=0~\Rightarrow~u=8\\ \\ \text{Quando }r=2~\Rightarrow~u=4$


Substituindo na integral $\mathbf{(ii)},$ temos

$=-\dfrac{1}{2}\displaystyle\int\limits_{8}^{4}{u^{3/2}\,du}\\ \\ \\ =-\dfrac{1}{2}\cdot \left.\dfrac{2}{5}\,u^{5/2}\right|_{8}^{4}\\ \\ \\ =-\dfrac{1}{5}\cdot (4^{5/2}-8^{5/2})\\ \\ \\ =-\dfrac{1}{5}\cdot (32-128\sqrt{2})\\ \\ \\ =\dfrac{128\sqrt{2}-32}{5}$

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Voltando à integral $\mathbf{(i)},$ temos

$=\displaystyle\dfrac{1}{3}\int\limits_{0}^{\,\pi/2\,}{\dfrac{128\sqrt{2}-32}{5}\,d\theta}-\dfrac{1}{3}\int\limits_{0}^{\,\pi/2\,}{\left.\dfrac{r^{5}}{5}\right|_{0}^{2}\,d\theta}\\ \\ \\ =\dfrac{128\sqrt{2}-32}{15}\int\limits_{0}^{\,\pi/2\,}{d\theta}-\dfrac{1}{3}\cdot \dfrac{2^{5}}{5}\int\limits_{0}^{\,\pi/2\,}{d\theta}\\ \\ \\ =\dfrac{128\sqrt{2}-32}{15}\int\limits_{0}^{\,\pi/2\,}{d\theta}-\dfrac{32}{15}\int\limits_{0}^{\,\pi/2\,}{d\theta}\\ \\ \\ =\dfrac{128\sqrt{2}-64}{15}\int\limits_{0}^{\,\pi/2\,}{d\theta}\\ \\ \\ =\dfrac{128\sqrt{2}-64}{15}\cdot \left(\dfrac{\pi}{2}-0 \right )\\ \\ \\ =\dfrac{64\pi\sqrt{2}-32\pi}{15}\\ \\ \\ =\dfrac{32\pi}{15}\!\left(2\sqrt{2}-1 \right ).$