• Matéria: Matemática
  • Autor: 00029713
  • Perguntado 9 anos atrás

Segue integral montada e respondida.

Anexos:

Respostas

respondido por: andresccp
0
\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\sqrt{5}} \rho^2*sen(\phi)\;d\rho\; d\theta \;d\phi\\\\ = \int_{0}^{\pi}sen(\phi) d\phi *\int_{0}^{2\pi}d\theta* \int_{0}^{\sqrt{5}} \rho^2 d\rho\\\\= \left[-cos(\phi)  \right]_0^{\pi} * \left[ \theta \right]_0^{2\pi} * \left[ \frac{\rho^3}{3}   \right]_0^{\sqrt{5}}\\\\ = [-cos(\pi)-cos(0)] * [2\pi-0]* [ \frac{(\sqrt{5})^3}{3}- \frac{0^3}{3}  ]\\\\ =[-(-1)-1]*[2\pi]*[ \frac{(\sqrt{5})^3}{3} ]\\\\ = 2*2\pi * \frac{5\sqrt{5}}{3} \\\\ = \frac{20\pi\sqrt{5}}{3}
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