Questão 3. Utilizando coordenadas esféricas, calcule a integral
\iiint_{E} \frac{\sqrt{x^{2}+y^{2}}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \mathrm{~d} V
em que E é o sólido exterior ao paraboloide z=x^{2}+y^{2} e interior à esfera x^{2}+y^{2}+z^{2}=2 z.
Res.: O paraboloide tem equação \rho \cos \varphi=(\rho \sin \varphi)^{2}, ou seja, \rho=\frac{\cos \varphi}{\sin ^{2} \varphi}. A esfera tem equação \rho=2 \cos \varphi. Logo, a interseção das duas superfícies é dada pela equação
2 \cos \varphi=\frac{\cos \varphi}{\sin ^{2} \varphi}
ou seja, ela ocorre quando \varphi=\frac{\pi}{4}\left(\sin \varphi=\frac{\sqrt{2}}{2}\right).
Logo,
\begin{aligned} \iiint_{E} \frac{\sqrt{x^{2}+y^{2}}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \mathrm{~d} V & =\int_{0}^{2 \pi} \int_{\pi / 4}^{\pi / 2} \int_{\frac{\cos \varphi}{\sin ^{2} \varphi}}^{2 \cos \varphi} \frac{\rho \sin \varphi}{\rho^{3}} \rho^{2} \sin \varphi d \rho d \varphi d \theta \\\ & =\int_{0}^{2 \pi} \int_{\pi / 4}^{\pi / 2}\left(2 \cos \varphi-\frac{\cos \varphi}{\sin ^{2} \varphi}\right) \sin ^{2} \varphi d \varphi d \theta \\\ & =\int_{0}^{2 \pi} \int_{\pi / 4}^{\pi / 2}\left(2 \cos \varphi \sin ^{2} \varphi-\cos \varphi\right) d \varphi d \theta \\\ & =2 \pi\left[2 \frac{\sin ^{3} \varphi}{3}-\sin \varphi\right]_{\pi / 4}^{\pi / 2}=\frac{2 \pi}{3}(\sqrt{2}-1) \end{aligned}

Passo 1

O enunciado pede o cálculo da integral tripe
\iiint_{E}\frac{\sqrt{x^{2}+y^{2}}}{\bigl(x^{2}+y^{2}+z^{2}\bigr)^{3/2}}\,dV
na região E que fica exterior ao paraboloide z=x^{2}+y^{2} e interior à esfera x^{2}+y^{2}+z^{2}=2z. Vamos usar coordenadas esféricas para resolver.

Passo 2

Em coordenadas esféricas temos
x=\rho\sin\varphi\cos\theta,\quad y=\rho\sin\varphi\sin\theta,\quad z=\rho\cos\varphi,
logo
\sqrt{x^{2}+y^{2}}=\rho\sin\varphi,\quad \bigl(x^{2}+y^{2}+z^{2}\bigr)^{3/2}=\rho^{3},\quad dV=\rho^{2}\sin\varphi\,d\rho\,d\varphi\,d\theta.
Assim o integrando vira
\frac{\rho\sin\varphi}{\rho^{3}}\times\rho^{2}\sin\varphi =\sin^{2}\varphi.

Passo 3

As superfícies em esféricas são
esfera: \rho^{2}=2\rho\cos\varphi\implies\rho=2\cos\varphi,
paraboloide: \rho\cos\varphi=(\rho\sin\varphi)^{2}\implies\rho=\frac{\cos\varphi}{\sin^{2}\varphi}.
Elas se intersectam quando
2\cos\varphi=\frac{\cos\varphi}{\sin^{2}\varphi} \quad\Longrightarrow\quad \sin^{2}\varphi=\tfrac12 \quad\Longrightarrow\quad \varphi=\frac{\pi}{4}.
Para \varphi\in[\tfrac{\pi}{4},\tfrac{\pi}{2}] temos \rho variando de \frac{\cos\varphi}{\sin^{2}\varphi} a 2\cos\varphi e \theta\in[0,2\pi].

Passo 4

A integral torna-se
\int_{0}^{2\pi}\!\int_{\pi/4}^{\pi/2}\!\int_{\frac{\cos\varphi}{\sin^{2}\varphi}}^{2\cos\varphi} \sin^{2}\varphi\,d\rho\,d\varphi\,d\theta =2\pi\int_{\pi/4}^{\pi/2}\Bigl(2\cos\varphi-\frac{\cos\varphi}{\sin^{2}\varphi}\Bigr)\sin^{2}\varphi\,d\varphi
ou seja
2\pi\int_{\pi/4}^{\pi/2}\bigl(2\cos\varphi\sin^{2}\varphi-\cos\varphi\bigr)\,d\varphi.

Passo 5

Calculemos cada termo:
\int2\cos\varphi\sin^{2}\varphi\,d\varphi =2\int \sin^{2}\varphi\,d(\sin\varphi) =2\frac{\sin^{3}\varphi}{3}, \quad \int\cos\varphi\,d\varphi=\sin\varphi.
Logo
2\pi\Bigl[2\frac{\sin^{3}\varphi}{3}-\sin\varphi\Bigr]_{\pi/4}^{\pi/2} =2\pi\frac{-1+ \sqrt{2}}{3} =\frac{2\pi}{3}\bigl(\sqrt{2}-1\bigr).

Resposta

\displaystyle \frac{2\pi}{3}\bigl(\sqrt{2}-1\bigr)

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