Coordenadas polares

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Revisão em 10h13min de 30 de agosto de 2016 por Ist178052 (Discussão | contribs)

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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 \(\begin{array}{c}\text{}\int_0^1\int_0^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\theta dr\\+\int_1^{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\theta dr\\+\int_{\sqrt{2}}^2\int_{-\arccos\left(-\frac{\sqrt{2}}{r}\right)}^{-\frac{3\pi}{4}}r\text{f}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)d\theta dr\\\end{array}\) pode também ser dada por:

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

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

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

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