Question

Calculo 3!!!
Pergunta e alternativas em anexo..

Answer 1

A região descrita em coordenadas esféricas é

$\left\{ \begin{array}{c} 0\leq \rho\leq 3\\ \\ 0\leq \theta \leq 2\pi\\ \\ 0\leq \varphi\leq \pi \end{array} \right.$


O módulo do Jacobiano para coordenadas esféricas é

$|\mathrm{Jac\,}\phi|=\rho^{2}\,\mathrm{sen\,}\varphi.$


A função integranda escrita em coordenadas esféricas é

$f(x,\;y,\;z)=\sqrt{x^{2}+y^{2}+z^{2}}~\Rightarrow~g(\rho,\;\theta,\;\varphi)=\rho$


Escrevendo a integral iterada já em coordenadas esféricas, temos

$I=\displaystyle\iiint\limits_{R}{g(\rho,\;\theta,\;\varphi)\cdot |\mathrm{Jac\,}\phi|\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{3}{\rho\cdot \rho^{2}\,\mathrm{sen\,}\varphi\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{3}{\rho^{3}\,\mathrm{sen\,}\varphi\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}{\mathrm{sen\,}\varphi\cdot \left.\left(\dfrac{\rho^{4}}{4} \right )\right|_{0}^{3}\,d\varphi\,d\theta}$

$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}{\mathrm{sen\,}\varphi\cdot \left(\dfrac{3^{4}}{4}-\dfrac{0^{4}}{4} \right )\,d\varphi\,d\theta}\\ \\ \\ =\dfrac{81}{4}\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}{\mathrm{sen\,}\varphi\,d\varphi\,d\theta}\\ \\ \\ =\dfrac{81}{4}\int\limits_{0}^{2\pi}{\left(-\cos \varphi \right )|_{0}^{\pi}\,d\theta}\\ \\ \\ =\dfrac{81}{4}\int\limits_{0}^{2\pi}{\left(-\cos \pi+\cos 0 \right )d\theta}\\ \\ \\ =\dfrac{81}{4}\int\limits_{0}^{2\pi}{\left(-(-1)+1 \right )d\theta}\\ \\ \\ =\dfrac{81}{4}\int\limits_{0}^{2\pi}{\left(1+1 \right )d\theta}$

$=\displaystyle\dfrac{81}{4}\int\limits_{0}^{2\pi}{2\,d\theta}\\ \\ \\ =\dfrac{81}{2}\int\limits_{0}^{2\pi}{d\theta}\\ \\ \\ =\dfrac{81}{2}\cdot \theta|_{0}^{2\pi}\\ \\ \\ =\dfrac{81}{2}\cdot (2\pi-0)\\ \\ \\ =81\pi.$


Resposta: alternativa e.