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Cálculo 2 - Integral Dupla
Enviado: Dom 10 Fev, 2019 22:14
por Leleco075
Integral dupla
[tex3]\int\limits_{0}^{2}\int\limits_{0}^{\pi }r\sen^2 \theta \,d\theta dr[/tex3]
Re: Cálculo 2 - Integral Dupla
Enviado: Dom 10 Fev, 2019 23:33
por Cardoso1979
Observe
Solução:
[tex3]\int\limits_{0}^{2}\int\limits_{0}^{\pi }r\sen^2 \theta \ d\theta dr=[/tex3]
[tex3]r.\int\limits_{0}^{2}\int\limits_{0}^{\pi }\sen^2 \theta \,d\theta dr=[/tex3]
Obs. sen ² θ = [tex3]\frac{1}{2}-\frac{1}{2}cos (2\theta )[/tex3]
.
[tex3]r.\int\limits_{0}^{2}\int\limits_{0}^{\pi }[\frac{1}{2}-\frac{1}{2}\cos ( \ 2\theta ) ]\ d\theta dr=[/tex3]
[tex3]r.\int\limits_{0}^{2}[\int\limits_{0}^{\pi }\frac{1}{2}-\frac{1}{2}\cos (2\theta ) \ d\theta ] \ dr=[/tex3]
[tex3]r.\int\limits_{0}^{2}[\frac{\theta }{2}-\frac{1}{4}\sen ( 2\theta )]_{0}^{π} \ dr=[/tex3]
[tex3]r.\int\limits_{0}^{2}[\frac{π}{2}-\frac{1}{4}\sen ( 2π )-\frac{0}{2}+\frac{1}{4}.sen (2.0)] \ dr=[/tex3]
[tex3]r.\int\limits_{0}^{2}(\frac{π}{2}-\frac{1}{4}.0-\frac{0}{2}+\frac{1}{4}.0) \ dr=[/tex3]
[tex3]\int\limits_{0}^{2}\frac{r.π}{2}\ dr=[/tex3]
[tex3][\frac{r^2π}{2.2}]_{0}^{2}=[\frac{r^2π}{4}]_{0}^{2}=\frac{2^2π}{4}-\frac{0^2.π}{4}=π[/tex3]
Portanto, [tex3]\int\limits_{0}^{2}\int\limits_{0}^{\pi }r\sen^2 \theta \,d\theta dr=π[/tex3]
.
Bons estudos!