• Matéria: Matemática
  • Autor: Cavassini275
  • Perguntado 8 anos atrás

calcule o valor da integral ∫pi/2 0 ∫2 0 r^4 cos(2theta) dr dtheta

Respostas

respondido por: Lukyo
34
Calcular o valor da integral dupla em coordenadas polares:

\displaystyle\int_0^{\pi/2}\int_0^2 r^4\cos(2\theta)\,dr\,d\theta\\\\\\ =\int_0^{\pi/2} \cos(2\theta)\cdot \frac{r^5}{5}\bigg|_{r=0}^{r=2}\,d\theta\\\\\\ =\int_0^{\pi/2} \cos(2\theta)\cdot\left(\frac{2^5}{5}-\frac{0^5}{5}\right)d\theta\\\\\\ =\int_0^{\pi/2} \cos(2\theta)\cdot \frac{32}{5}\,d\theta\\\\\\ =\frac{32}{5}\int_0^{\pi/2} \cos(2\theta)\,d\theta

\displaystyle=\frac{32}{5}\cdot \frac{1}{2}\,\mathrm{sen}(2\theta)\bigg|_{\theta=0}^{\theta=\pi/2}\\\\\\ =\frac{32}{5}\cdot \left[\frac{1}{2}\,\mathrm{sen}\!\left(2\cdot\frac{\pi}{2}\right)-\frac{1}{2}\,\mathrm{sen}(2\cdot 0)\right]\\\\\\ =\frac{32}{5}\cdot \left[\frac{1}{2}\,\mathrm{sen}(\pi)-\frac{1}{2}\,\mathrm{sen}(0)\right]

=0\quad\longleftarrow\quad\textsf{resposta.}

Bons estudos! :-)
respondido por: DemolidorPK
23

aonde foi q eu errei?

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