Mudança da ordem de integração polares

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Metadata

  • CONTEXTO : Primeiro ciclo universitário
  • AREA: Matemática
  • DISCIPLINA: Calculo Diferencial e Integral 2
  • ANO: 1
  • LINGUA: pt
  • AUTOR: Ana Moura Santos e Miguel Dziergwa
  • MATERIA PRINCIPAL: Teorema de mudança de variáveis
  • DESCRICAO: Mudança da ordem de integração em coordenadas polares
  • DIFICULDADE: ***
  • TEMPO MEDIO DE RESOLUCAO: 15 mn
  • TEMPO MAXIMO DE RESOLUCAO: 30 mn
  • PALAVRAS CHAVE: integral duplo, ordem de integração, extremos de integração, coordenadas polares

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

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

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

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

D)\(\fbox{$\begin{array}{c}\int_0^{\frac{\pi}{4}}\int_0^{\frac{1}{\cos(\theta)}}\text{rf}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)\text{d}r\text{d}\theta+\int_{\frac{\pi}{4}}^{\pi}\int_0^{\sqrt{2}}\text{rf}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)\text{d}r\text{d}\theta\\+\int_{\pi}^{\frac{5\pi}{4}}\int_0^{-\frac{\sqrt{2}}{\cos(\theta)}}\text{rf}\left(\begin{array}{c}r\cos(\theta)\\r\sin(\theta)\\\end{array}\right)\text{d}r\text{d}\theta\\\end{array}$}\)


Para obter o zip que contém as instâncias deste exercício clique aqui[1]

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