Question

Determine o volume do solido abaixo do paraboloide z = ln(x2 +y2) entre os discos 1≤ x2 +y2 ≤4 e y e x nao negativos.

Answer 1

Resposta:

O volume do sólido é:

$\displaystyle\boxed{\iiint\limits_\textrm{s\'{o}lido} \textrm{ d}V = \pi\int\limits_1^2\rho\ln\rho\textrm{ d}\rho = \pi \left(2\ln 2-\dfrac{3}{4}\right)}.$

Explicação passo-a-passo:

Devido à simetria do problema, utilizamos coordenadas cilíndricas:

$\displaystyle\begin{cases}x=\rho\cos\theta \\ y=\rho\sin\theta \\ z=z\end{cases}.$

Note que $\rho^2 = x^2 + y^2$, logo a região de interesse corresponde a $1 \leq x^2+y^2 \leq 4 \iff 1 \leq \rho^2 \leq 4 \iff 1 \leq \rho \leq 2.$

Por outro lado, como $x$ e $y$ são não-negativos, estamos no 1.º octante, pelo que $\theta\in\left[0, \dfrac{\pi}{2}\right[$.

Finalmente, a coordenada $z$ varia entre $0$ e $\ln(x^2+y^2) = \ln(\rho^2).$

Como o jacobiano das coordenadas cilíndricas é $\rho$, o volume pretendido é dado por:

$\displaystyle\iiint\limits_\textrm{s\'{o}lido}\textrm{ d}V = \int\limits_1^2 \int\limits_0^{\pi/2}\int\limits_0^{\ln(\rho^2)} \rho \textrm{ d}z\textrm{ d}\theta \textrm{ d}\rho = \underbrace{\int\limits_0^{\pi/2}\textrm{ d}\theta}_{=\pi/2}\int\limits_1^2 \rho \underbrace{\int\limits_0^{\ln(\rho^2)}\textrm{ d}z}_{=\ln(\rho^2)}\textrm{ d}\rho =\\= \dfrac{\pi}{2}\int\limits_1^2 \rho\ln(\rho^2)\textrm{ d}\rho = \dfrac{\pi}{2}\int\limits_1^2 2\rho\ln\rho\textrm{ d}\rho = \pi\int\limits_1^2\rho\ln\rho\textrm{ d}\rho.$

Para calcular este integral, utilizamos a técnica de integração por partes. Definindo:

$u = \ln \rho \implies u' = \dfrac{1}{\rho}; \\\\v' = \rho \implies v = \dfrac{\rho^2}{2},$

vem:

$\displaystyle\int\rho\ln\rho\textrm{ d}\rho = \int u v' = uv-\int u'v = \dfrac{\rho^2}{2}\ln\rho - \int \dfrac{1}{\rho}\dfrac{\rho^2}{2} \textrm{ d}\rho = \\=\dfrac{\rho^2}{2}\ln\rho - \dfrac{1}{2}\int\rho\textrm{ d}\rho =\dfrac{\rho^2}{2}\ln\rho - \dfrac{1}{2}\dfrac{\rho^2}{2} +C = \dfrac{\rho^2}{2}\ln\rho -\dfrac{\rho^2}{4}+C.$

Vem então:

$\displaystyle\int\limits_1^2\rho\ln\rho\textrm{ d}\rho = \left[\dfrac{\rho^2}{2}\ln\rho -\dfrac{\rho^2}{4}\right]_{\rho=1}^{\rho=2} = \left(\dfrac{2^2}{2}\ln 2 - \dfrac{2^2}{4}\right) -\left(\dfrac{1^2}{2}\ln 1 - \dfrac{1^2}{4}\right) =\\= 2\ln 2-1+\dfrac{1}{4} = 2\ln 2-\dfrac{3}{4}.$

Finalmente, obtemos o volume pretendido:

$\displaystyle\iiint\limits_\textrm{s\'{o}lido} \textrm{ d}V = \pi\int\limits_1^2\rho\ln\rho\textrm{ d}\rho = \pi \left(2\ln 2-\dfrac{3}{4}\right).$

Answer 2

A resposta correta é: 28π