Question

Utilizando coordenadas esféricas Calcule ;


a) ∫∫∫H (9-x²-y²) dV, onde H é o hemisferio sólido x²+y²+z² ≤ 9, z≥0


b) ∫∫∫E (Z) dv, onde E está entre as esferas x²+y² +z² =1 e x²+ y²+z²= 4 acima do plano xy e abaixo do plano z= x+2

Answer 1

a) $\displaystyle\iiint\limits_{H}{(9-x^{2}-y^{2})\,dx\,dy\,dz}$

$\bullet\;\;$ Mudança para coordenadas esféricas:

$\begin{array}{cc} \left\{ \begin{array}{l} x=\rho\,\mathrm {sen\,}\varphi\cos \theta\\ \\ y=\rho\,\mathrm{sen\,}\varphi\, \mathrm{sen\,}\theta\\ \\ z=\rho\cos \varphi \end{array} \right. ~~&~~\begin{array}{c} 0\leq \theta \leq 2\pi\\ \\ 0\leq \varphi \leq \dfrac{\pi}{2}\\ \\ 0\leq \rho \leq 3 \end{array} \end {array}$


O módulo do Jacobiano desta transformação é

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


Efetuando a mudança de coordenadas, temos

$\displaystyle\iiint\limits_{H}{(9-x^{2}-y^{2})\,dx\,dy\,dz}\\ \\ \\ =\iiint\limits_{H}{(9-(x^{2}+y^{2}))\,dx\,dy\,dz}\\ \\ \\ =\iiint\limits_{H_{\rho\theta\varphi}}{(9-(\rho^{2}\,\mathrm{sen^{2}\,\varphi}\cos^{2}\theta+\rho^{2}\,\mathrm{sen^{2}\,\varphi}\,\mathrm{sen^{2}\,}\theta))\cdot|\mathrm{Jac\,\phi}|\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\iiint\limits_{H_{\rho\theta\varphi}}{(9-\rho^{2}\,\mathrm{sen^{2}\,\varphi}(\cos^{2}\theta+\mathrm{sen^{2}\,}\theta))\cdot\rho^{2}\,\mathrm{sen\,\varphi}\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\iiint\limits_{H_{\rho\theta\varphi}}{(9-\rho^{2}\,\mathrm{sen^{2}\,\varphi})\cdot\rho^{2}\,\mathrm{sen\,\varphi}\,d\rho\,d\varphi\,d\theta}$


Escrevendo os extremos de integração, temos:

$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}\int\limits_{0}^{3}{(9-\rho^{2}\,\mathrm{sen^{2}\,\varphi})\cdot\rho^{2}\,\mathrm{sen\,\varphi}\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}\int\limits_{0}^{3}{(9\rho^{2}\,\mathrm{sen\,\varphi}-\rho^{4}\,\mathrm{sen^{3}\,}\varphi)\,d\rho\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left.\left(3\rho^{3}\,\mathrm{sen\,\varphi}-\dfrac{\rho^{5}}{5}\,\mathrm{sen^{3}\,}\varphi\right)\right|_{0}^{3}\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left(3\cdot 3^{3}\,\mathrm{sen\,\varphi}-\dfrac{3^{5}}{5}\,\mathrm{sen^{3}\,}\varphi\right)\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left(81\,\mathrm{sen\,\varphi}-\dfrac{243}{5}\,\mathrm{sen^{3}\,}\varphi\right)\,d\varphi\,d\theta}$

$=\displaystyle\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left(81\,\mathrm{sen\,\varphi}-\dfrac{243}{5}\,\mathrm{sen^{2}\,}\varphi\cdot \mathrm{sen\,}\varphi\right)\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left[81\,\mathrm{sen\,\varphi}-\dfrac{243}{5}\,(1-\cos^{2}\varphi)\cdot \mathrm{sen\,}\varphi\right]\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left[81\,\mathrm{sen\,\varphi}-\dfrac{243}{5}\,(\mathrm{sen\,}\varphi-\cos^{2}\varphi\cdot\mathrm{sen\,}\varphi) \right]\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left(81\,\mathrm{sen\,\varphi}-\dfrac{243}{5}\,\mathrm{sen\,}\varphi+\dfrac{243}{5}\cos^{2}\varphi\cdot\mathrm{sen\,}\varphi \right)\,d\varphi\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/2}{\left(\dfrac{162}{5}\,\mathrm{sen\,}\varphi+\dfrac{243}{5}\cos^{2}\varphi\cdot\mathrm{sen\,}\varphi \right)\,d\varphi\,d\theta}$

$=\displaystyle\int\limits_{0}^{2\pi}{\left.\left(-\dfrac{162}{5}\,\cos\varphi-\dfrac{81}{5}\cos^{3}\varphi \right)\right|_{0}^{\pi/2}\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left(\dfrac{162}{5}\,\cos 0+\dfrac{81}{5}\cos^{3}0 \right)\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left(\dfrac{162}{5}+\dfrac{81}{5} \right)\,d\theta}\\ \\ \\ =\int\limits_{0}^{2\pi}{\dfrac{243}{5}\,d\theta}\\ \\ \\ =\dfrac{243}{5}\cdot (2\pi-0)\\ \\ \\ =\dfrac{486\pi}{5}$