Use coordenadas esféricas para calcular a integral tripla de f(x, y, z) sobre a região dada.
(a) f(x, y, z)=y ; \quad x^{2}+y^{2}+z^{2} \leq 1, \quad x, y, z \leq 0
(b) f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}} ; \quad x^{2}+y^{2}+z^{2} \leq 2 z
(c) f(x, y, z)=z\left(x^{2}+y^{2}+z^{2}\right)^{-3 / 2} ; \quad x^{2}+y^{2}+z^{2} \leq 16, \quad z \geq 2
(d) f(x, y, z)=\frac{z}{x^{2}+y^{2}+z^{2}} ; \quad região delimitada pelas superfícies z=\sqrt{\frac{x^{2}+y^{2}}{3}}, z= \sqrt{3 x^{2}+3 y^{2}} e x^{2}+y^{2}+z^{2}=4

Passo 1

O enunciado pede para calcular quatro integrais triplas usando coordenadas esféricas. Em todas elas vamos usar
– as relações
x = \rho \sin\varphi \cos\theta,\quad y = \rho \sin\varphi \sin\theta,\quad z = \rho \cos\varphi,
– o elemento de volume
dV = \rho^{2}\sin\varphi\,d\rho\,d\varphi\,d\theta,
– e determinar cuidadosamente os limites de \rho, \varphi e \theta conforme cada região.

Passo 2

(a) f(x,y,z)=y na bola de raio 1, com x,y,z\le0.
  1. Em esféricas y = \rho\sin\varphi\sin\theta.
  2. A região é 0\le\rho\le1, z\le0\Rightarrow\varphi\in[\tfrac\pi2,\pi], x\le0,y\le0\Rightarrow\theta\in[\pi,\,3\pi/2].
  3. A integral fica
\int_{\theta=\pi}^{3\pi/2} \int_{\varphi=\pi/2}^{\pi} \int_{\rho=0}^{1} \bigl(\rho\sin\varphi\sin\theta\bigr) \bigl(\rho^{2}\sin\varphi\bigr) \,d\rho\,d\varphi\,d\theta.
  1. Reúna fatores: \rho^{3}\sin^{2}\varphi\sin\theta.
  2. Calcule
    \displaystyle\int_{0}^{1}\rho^{3}d\rho=\tfrac14,
    \displaystyle\int_{\pi/2}^{\pi}\sin^{2}\varphi\,d\varphi=\tfrac\pi4,
    \displaystyle\int_{\pi}^{3\pi/2}\sin\theta\,d\theta=-1.
  3. Multiplicando dá \tfrac14\cdot\tfrac\pi4\cdot(-1)=-\tfrac\pi{16}.

Passo 3

(b) f=\sqrt{x^{2}+y^{2}+z^{2}}=\rho na região x^{2}+y^{2}+z^{2}\le2z.
  1. A desigualdade vira \rho^{2}\le2\rho\cos\varphi\Rightarrow0\le\rho\le2\cos\varphi, e é preciso \cos\varphi\ge0\Rightarrow\varphi\in[0,\pi/2].
  2. \theta\in[0,2\pi].
  3. A integral é
\int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{2\cos\varphi} \rho\;\bigl(\rho^{2}\sin\varphi\bigr) \,d\rho\,d\varphi\,d\theta =\int_{0}^{2\pi}\!d\theta\int_{0}^{\pi/2}\!\sin\varphi\!\int_{0}^{2\cos\varphi}\rho^{3}d\rho\,d\varphi.
  1. \displaystyle\int_{0}^{2\cos\varphi}\rho^{3}d\rho=4\cos^{4}\varphi.
  2. Logo dá
2\pi\cdot4\int_{0}^{\pi/2}\cos^{4}\varphi\sin\varphi\,d\varphi =8\pi\int_{0}^{1}u^{4}du\quad(u=\cos\varphi) =8\pi\cdot\tfrac15 =\tfrac{8\pi}{5}.

Passo 4

(c) f=z(\,x^{2}+y^{2}+z^{2})^{-3/2} no interior da esfera de raio 4 com z\ge2.
  1. Em esféricas f=\rho\cos\varphi\;\rho^{-3}=\cos\varphi\,\rho^{-2}.
  2. A condição z\ge2 é \rho\cos\varphi\ge2\Rightarrow\rho\ge2/\cos\varphi. Como \rho\le4 pela esfera, precisamos \cos\varphi\ge\tfrac12\Rightarrow\varphi\in[0,\pi/3].
  3. \theta\in[0,2\pi], \rho\in[2/\cos\varphi,\,4].
  4. A integral é
\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{2/\cos\varphi}^{4} \cos\varphi\,\rho^{-2}\;(\rho^{2}\sin\varphi) \,d\rho\,d\varphi\,d\theta =2\pi\int_{0}^{\pi/3}\cos\varphi\sin\varphi\int_{2/\cos\varphi}^{4}1\,d\rho\,d\varphi.
  1. \displaystyle\int_{2/\cos\varphi}^{4}d\rho=4-2\operatorname{sec}\varphi.
  2. Logo
2\pi\int_{0}^{\pi/3}\bigl(4\cos\varphi\sin\varphi-2\sin\varphi\bigr)d\varphi =2\pi\bigl[2\sin^{2}\varphi+2\cos\varphi\bigr]_{0}^{\pi/3} = \pi.

Passo 5

(d) f=\frac{z}{x^{2}+y^{2}+z^{2}} na região entre
\;z=\sqrt{\tfrac{x^{2}+y^{2}}{3}} (cone de semiângulo \pi/3),
\;z=\sqrt{3(x^{2}+y^{2})} (cone de semiângulo \pi/6),
e a esfera de raio 2.
  1. Em esféricas f=\rho\cos\varphi\;\rho^{-2}=\cos\varphi/\rho.
  2. A esfera dá \rho\le2. Os cones fixam \varphi\in[\pi/6,\pi/3]. \theta\in[0,2\pi].
  3. A integral é
\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2} \frac{\cos\varphi}{\rho}\;(\rho^{2}\sin\varphi) \,d\rho\,d\varphi\,d\theta =2\pi\int_{\pi/6}^{\pi/3}\cos\varphi\sin\varphi\int_{0}^{2}\rho\,d\rho\,d\varphi.
  1. \displaystyle\int_{0}^{2}\rho\,d\rho=2.
  2. Então
2\pi\cdot2\int_{\pi/6}^{\pi/3}\cos\varphi\sin\varphi\,d\varphi =4\pi\cdot\tfrac12\bigl[\sin^{2}\varphi\bigr]_{\pi/6}^{\pi/3} =\pi.

Resposta

(a) -\dfrac{\pi}{16}
(b) \dfrac{8\pi}{5}
(c) \pi
(d) \pi

Exercícios relacionados