Coordenadas Esféricas - Cálculo por Integral Tripla

[raimundojr]

(Livro: Cálculo - Autor: James Stewart - Volume 2 - 7ª Edição - Q. 23 - Pág.: 931)
Utilize coordenadas esféricas.
Calcule \(\iiint_E (x^2+y^2)dV\), onde E está entre as esferas x²+y²+z²=4 e x²+y²+z²=9.


Resposta: \(1,688\pi/15\)

[Davi Constant]

Vamos lá...

Em coordenadas esféricas temos o seguinte:

\(\Large x=\rho\sin(\varphi )\cos(\theta );y=\rho\sin(\varphi )\sin(\theta );z=\rho\cos(\varphi );dxdydz=\rho^2\sin(\varphi)d\rho d\varphi d\theta\)

Região de integração \(\Large E:4\leq x^2+y^2+z^2\leq 9 \rightarrow 2\leq \rho\leq 3\). Sabemos também que \(\Large 0\leq\varphi\leq\pi;0\leq \theta\leq 2\pi\)

Do integrando vem que \(\Large x^2+y^2=\rho^2\sin^2(\varphi)\cos^2(\theta)+\rho^2\sin^2(\varphi)\sin^2(\theta)=\rho^2\sin^2(\varphi)\).

Então \(\Large x^2+y^2dV=\rho^2\sin^2(\varphi)\rho^2\sin(\varphi)d\rho d\varphi d\theta=\rho^4\sin^3(\varphi)d\rho d\varphi d\theta\).

Podemos reescrever \(\Large \iiint_{E}x^2+y^2dV\) como \(\Large \int_{0}^{2\pi}\int_{0}^{\pi}\int_{2}^{3}\rho^4\sin^3(\varphi)d\rho d\varphi d\theta\).

\(\Large\int_{0}^{2\pi}\int_{0}^{\pi}\int_{2}^{3}\rho^4\sin^3(\varphi)d\rho d\varphi d\theta=\int_{0}^{2\pi}\int_{0}^{\pi}\left [\frac{\rho^5}{5}\right]_{2}^{3}\sin^3(\varphi)d\varphi d\theta=\frac{211}{5}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3(\varphi)d\varphi d\theta\)

Usando \(\Large \sin^3(\varphi)=\sin^2(\varphi)\sin(\varphi)=(1-\cos^2(\varphi))\sin(\varphi)\) faço a substituição \(\Large u=\cos(\varphi)\rightarrow du=-\sin(\varphi)d\varphi\)...

\(\Large \frac{211}{5}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3(\varphi)d\varphi d\theta=\frac{211}{5}\int_{0}^{2\pi}\left[ \frac{1}{3}\cos^3(\varphi)-\cos(\varphi)\right]_{0}^{\pi}d\theta=\frac{211}{5}\times\frac{4}{3}\int_{0}^{2\pi}d\theta=\frac{211}{5}\times\frac{4}{3}\times 2\pi=\frac{1688}{15}\pi\)

Espero ter ajudado,
Qualquer dúvida sinalize.