Mudança da ordem de integracao polares

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(trocaOrdemPolares)

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