Question

Calcule usando Coordenadas cilindricas

Answer 1

Como o enunciado pede para calcular usando coordenadas cilíndricas, está implícito que trata-se de uma aplicação da integral tripla.


O volume do sólido $D$ é dado por

$\text{Volume}(D)=\displaystyle\iiint_{D}{1\,d\mathbf{V}}$


sendo que neste caso, podemos descrever $D$ assim:

$\begin{array}{cc} D=\left\{~(x,\;y,\;z)\in \mathbb{R}^{3}:\right.&-2\le x\le 2\,;\\\\ &-\,\sqrt{4-x^{2}}\le y\le \sqrt{4-x^{2}}\,;\\\\ &\left.-\,\sqrt{9-x^{2}-y^{2}}\le z\le \sqrt{9-x^{2}-y^{2}}~\right \} \end{array}$

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Mudança para coordenadas cilíndricas:

$\begin{array}{cc} \left\{ \begin{array}{l} x=r\,\cos \theta\\\\ y=r\,\mathrm{sen\,}\theta\\\\ z=z \end{array} \right.&~~~\begin{array}{c} 0\le \theta\le 2\pi\\\\ 0\le r\le 2\\\\ -\,\sqrt{9-r^{2}}\le z\le \sqrt{9-r^{2}} \end{array} \end{array}$


O módulo do Jacobiano desta transformação é $|\mathrm{Jac\,}\phi|=r.$
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Escrevendo as integrais iteradas na ordem $dz\,dr\,d\theta,$ temos

$\text{Volume}(D)=\displaystyle\iiint_{D}{1\,d\mathbf{V}}\\\\\\ =\iiint_{D_{r,\,\theta,\,z}}{1\cdot |\mathrm{Jac\,}\phi|\,dz\,dr\,d\theta}\\\\\\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{-\sqrt{9-r^{2}}}^{\sqrt{9-r^{2}}}{1\cdot r\,dz\,dr\,d\theta}\\\\\\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{r\cdot (z)\left|_{-\sqrt{9-r^{2}}}^{~\;\sqrt{9-r^{2}}}\right.\,dr\,d\theta}$

$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{r\cdot \left[\sqrt{9-r^{2}}-(-\sqrt{9-r^{2}}) \right ]\,dr\,d\theta}\\\\\\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{r\cdot \left[\sqrt{9-r^{2}}+\sqrt{9-r^{2}} \right ]\,dr\,d\theta}\\\\\\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{r\cdot 2\sqrt{9-r^{2}}\,dr\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}\int\limits_{0}^{2}{\sqrt{9-r^{2}}\cdot (-2r)\,dr\,d\theta}~~~~~~\mathbf{(i)}$


Fazendo a seguinte mudança de variável:

$9-r^{2}=u~~\Rightarrow~~-2r\,dr=du$


Mudando os extremos de integração em $r:$

$\text{Quando }~r=0~~\Rightarrow~~u=9\\\\ \text{Quando }~r=2~~\Rightarrow~~u=5$


Substituindo em $\mathbf{(i)},$ a integral do volume fica

$=-\displaystyle\int\limits_{0}^{2\pi}\int\limits_{9}^{5}{\sqrt{u}\,du\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}\int\limits_{9}^{5}{u^{1/2}\,du\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}{\left.\left(\dfrac{u^{1/2+1}}{\frac{1}{2}+1} \right )\right|_{9}^{5}\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}{\left.\left(\dfrac{u^{3/2}}{\frac{3}{2}} \right )\right|_{9}^{5}\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}{\left.\left(\dfrac{2}{3}\,u^{3/2} \right )\right|_{9}^{5}\,d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}{\left.\left(\dfrac{2}{3}\,u^{3/2} \right )\right|_{9}^{5}\,d\theta}$

$=-\displaystyle\int\limits_{0}^{2\pi}{\dfrac{2}{3}\cdot \left(5^{3/2}-9^{3/2} \right )d\theta}\\\\\\ =-\int\limits_{0}^{2\pi}{\dfrac{2}{3}\cdot \left(5\sqrt{5}-27 \right )d\theta}\\\\\\ =-\,\dfrac{2}{3}\cdot \left(5\sqrt{5}-27 \right )\int\limits_{0}^{2\pi}{d\theta}\\\\\\ =\dfrac{2}{3}\cdot \left(27-5\sqrt{5} \right )\cdot \left(\theta \right )|_{0}^{2\pi}\\\\\\ =\dfrac{2}{3}\cdot \left(27-5\sqrt{5} \right )\cdot (2\pi-0)\\\\\\ \therefore~~\boxed{\begin{array}{c}\text{Volume}(D)=\dfrac{4\pi}{3}\left(27-5\sqrt{5} \right )~~\mathrm{u.v.} \end{array}}$