Question

Calcule usando integral m ́ultipla o volume da bola do R 3 centrada na origem de raio 2.

Answer 1

A bola centrada na origem com raio 2 é descrita pela desigualdade

     $x^2+y^2+z^2\le 2^2\\\\ x^2+y^2+z^2\le 4$


Vamos usar integrais triplas. O volume do sólido é dado por

     $\displaystyle\iiint_S 1\,d\mathbf{V}\\\\\\ =\iiint_{x^2+y^2+z^2\le 4}~1\,dz\,dy\,dx$


Dado a natureza do sólido, é mais viável fazer uma mudança de coordenadas e trabalhar com coordenadas esféricas.

     $\left\{\! \begin{array}{l} x=r\cos\theta\\ y=r\,\mathrm{sen\,}\theta\\ z=\rho\cos\phi \end{array} \right.$


Mas r = ρ sen θ. Então, a relação de transformação de coordenadas cartesianas para esféricas é

     $\left\{\! \begin{array}{l} x=\rho\,\mathrm{sen\,}\phi\cos\theta\\\\ y=\rho\,\mathrm{sen\,}\phi\,\mathrm{sen\,}\theta\\\\ z=\rho\cos\phi \end{array} \right.\qquad\quad \begin{array}{c}0\le\rho\le 2\\\\0\le \phi\le \pi\\\\ 0\le \theta\le 2\pi\end{array}$


O módulo do Jacobiano desta mudança é

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


Então, a integral fica

     $\displaystyle\iiint_{\rho \le 2}~|\mathrm{Jac}|\,d\rho\,d\phi\,d\theta\\\\\\ =\displaystyle\iiint_{\rho \le 2}~\rho^2\,\mathrm{sen\,}\phi\,d\rho\,d\phi\,d\theta\\\\\\ =\displaystyle\int_0^{2\pi}\int_0^\pi\int_0^2 ~\rho^2\,\mathrm{sen\,}\phi\,d\rho\,d\phi\,d\theta\\\\\\ =\displaystyle\int_0^{2\pi}\int_0^\pi ~\mathrm{sen\,}\phi\cdot \frac{\rho^3}{3}\bigg|_{\rho=0}^{\rho=2}~d\phi\,d\theta\\\\\\ =\displaystyle\int_0^{2\pi}\int_0^\pi ~\mathrm{sen\,}\phi\cdot \left(\frac{2^3}{3}-\frac{0^3}{3}\right)d\phi\,d\theta$

    $=\displaystyle\frac{8}{3}\int_0^{2\pi}\int_0^\pi \mathrm{sen\,}\phi\,d\phi\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\int_0^{2\pi}(-\cos\phi)\Big|_0^{\pi}\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\int_0^{2\pi}\big[(-\cos\pi)-(-\cos 0)\big]\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\int_0^{2\pi}\big[\!-\cos\pi+\cos\,0\big]\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\int_0^{2\pi}\big[-(-1)+1\big]\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\int_0^{2\pi}\big[1+1\big]\,d\theta\\\\\\ =\displaystyle\frac{8}{3}\cdot 2\int_0^{2\pi}d\theta$

     $=\displaystyle\frac{16}{3}\cdot \theta\Big|_0^{2\pi}\\\\\\ =\displaystyle\frac{16}{3}\cdot (2\pi-0)$

     $=\displaystyle\frac{32\pi}{3}~\mathrm{u.v.}\quad\longleftarrow\quad \mathsf{esta~\acute{e}~a~resposta.}$


Dúvidas? Comente.


Bons estudos! :-)