Mudança da ordem de integracao polares
Metadata
- 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 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