Coordenadas polares

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CONTEXTO : Primeiro ciclo universitário
AREA: Matemática
DISCIPLINA: Calculo diferencial e integral 2
ANO: 1
LINGUA: pt
AUTOR: Equipa Calculo diferencial e integral 2
MATERIA PRINCIPAL:
DESCRICAO:
DIFICULDADE: easy
TEMPO MEDIO DE RESOLUCAO: 15 mn
TEMPO MAXIMO DE RESOLUCAO: 30 mn
PALAVRAS CHAVE:

Sendo f uma função positiva e integrável, a seguinte soma de integrais em coordenadas polares $\begin{array}{c}\text{}\int_0^1\int_0^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_1^{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_{\sqrt{2}}^2\int_{-\arccos\left(-\frac{\sqrt{2}}{r}\right)}^{-\frac{3\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\\end{array}$ pode também ser dada por:

A)$\fbox{$\begin{array}{c}\text{}\int_0^1\int_0^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_1^{\sqrt{2}}\int_{\arccos\left(\frac{1}{r}\right)}^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_{\sqrt{2}}^2\int_{-\arccos\left(-\frac{\sqrt{2}}{r}\right)}^{-\frac{3\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\\end{array}$}$

B)$\fbox{$\begin{array}{c}\text{}\int_0^1\int_0^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_1^{\sqrt{2}}\int_{\arccos\left(\frac{1}{r}\right)}^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_{\sqrt{2}}^2\int_{-\arccos\left(-\frac{\sqrt{2}}{r}\right)}^{-\frac{3\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\\end{array}$}$

C)$\fbox{$\begin{array}{c}\text{}\int_0^1\int_0^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_1^{\sqrt{2}}\int_{\arccos\left(\frac{1}{r}\right)}^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\+\int_{\sqrt{2}}^2\int_{-\arccos\left(-\frac{\sqrt{2}}{r}\right)}^{-\frac{3\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\thetadr\\\end{array}$}$

D)Nenhuma das anteriores

Para obter o zip que contém as instâncias deste exercício clique aqui(coordPolaCartes)

Se deseja obter o código fonte que gera os exercícios contacte miguel.dziergwa@ist.utl.pt

Coordenadas polares

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