Diferenças entre edições de "Coordenadas polares"
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− | 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\ | + | 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\ | + | 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\ | + | 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\ | + | 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 | D)Nenhuma das anteriores |
Revisão das 10h13min de 30 de agosto de 2016
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 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