Question

Coordenadas cilíndricas, acima de cone e abaixo de esfera. tem a questão na imagem
Preciso do passo a passoa a resposta é 10$\pi$

Answer 1

O volume do sólido $D$ descrito em coordenadas esféricas é dado por

$\mathrm{Volume}(D)=\displaystyle\iiint_{D} \rho^2\,\mathrm{sen\,}\phi\,d\rho\,d\phi\,d\theta$

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O sólido $D$ é descrito pelas seguintes desigualdades:

$0\le \rho \le 4\cos \phi\\\\ 0\le \phi \le \dfrac{\pi}{3}\\\\\\ 0\le \theta \le 2\pi$


Escrevendo a integral com os limites de integração, obtemos

$\mathrm{Volume}(D)=\displaystyle\int_{0}^{2\pi}\int_{0}^{\pi/3}\int_{0}^{4 \cos \phi}\rho^2\,\mathrm{sen\,}\phi\,d\rho\,d\phi\,d\theta\\\\\\ =\int_{0}^{2\pi}\int_{0}^{\pi/3}\mathrm{sen\,}\phi\cdot \left.\left(\dfrac{\rho^3}{3} \right )\right|_{0}^{4 \cos \phi}\,d\phi\,d\theta\\\\\\ =\int_{0}^{2\pi}\int_{0}^{\pi/3}\mathrm{sen\,}\phi\cdot \left(\dfrac{(4 \cos \phi)^3-0^3}{3} \right )d\phi\,d\theta\\\\\\ =\int_{0}^{2\pi}\int_{0}^{\pi/3}\mathrm{sen\,}\phi\cdot \dfrac{64\cos^3\phi}{3}\,d\phi\,d\theta\\\\\\ =\dfrac{64}{3}\int_{0}^{2\pi}\int_{0}^{\pi/3}\mathrm{sen\,}\phi\cdot \cos^3\phi\,d\phi\,d\theta$

$=\displaystyle-\,\dfrac{64}{3}\int_{0}^{2\pi}\int_{0}^{\pi/3}\cos^3\phi\cdot (-\mathrm{sen\,}\phi)\,d\phi\,d\theta\\\\\\ =-\,\dfrac{64}{3}\int_{0}^{2\pi}\left.\left(\dfrac{\cos^4 \phi}{4} \right )\right|_{0}^{\pi/3}\,d\theta\\\\\\ =-\,\dfrac{64}{3}\int_{0}^{2\pi}\dfrac{\cos^4 \frac{\pi}{3}-\cos^4 0}{4}\,d\theta\\\\\\ =-\,\dfrac{64}{3}\int_{0}^{2\pi}\dfrac{\left(\frac{1}{2} \right )^{\!\!4}-1^4}{4}\,d\theta$

$=\displaystyle-\,\dfrac{64}{3}\int_{0}^{2\pi}\dfrac{\frac{1}{16}-1}{4}\,d\theta\\\\\\ =-\,\dfrac{16}{3}\int_{0}^{2\pi}\left(\dfrac{1}{16}-1\right)d\theta\\\\\\ =-\,\dfrac{16}{3}\int_{0}^{2\pi}\left(\dfrac{1}{16}-\dfrac{16}{16}\right)d\theta\\\\\\ =-\,\dfrac{16}{3}\int_{0}^{2\pi}\left(-\,\dfrac{15}{16}\right)d\theta\\\\\\ =5\int_{0}^{2\pi}d\theta\\\\\\ =5\cdot \theta\big|_0^{2\pi}$

$=5\cdot (2\pi-0)\\\\ =10\pi \mathrm{~u.v.}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathrm{Volume}(D)=10\pi \mathrm{~u.v.} \end{array}}$