Integral duplo em coordenadas 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: Integral duplo 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 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[1]

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