Integral triola I
Enviado: Qui 05 Out, 2017 18:48
\int _0^{2\pi }\int _0^{\frac{\theta }{2\pi }}\int _0^{3+24r^2}\:r\:dzdrd\theta
Observe
Solução:
[tex3]\int\limits_{0}^{2π}\int\limits_{0}^{
\frac{\theta }{2π}}\int\limits_{0}^{3+24r^2}r \ dzdrd\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\int\limits_{0}^{
\frac{\theta }{2π}}[z]_{0}^{3+24r^2}r \ drd\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\int\limits_{0}^{
\frac{\theta }{2π}}(3+24r^2-0).r \ drd\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\int\limits_{0}^{
\frac{\theta }{2π}}(3r+24r^3) \ drd\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\left[\frac{3r^2}{2}+6r^4\right]_{0}^{
\frac{\theta }{2π}} \ d\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\left(\frac{3\theta ^2}{8π^2}+\frac{6\theta ^4}{16π^4}-0-0\right) \ d\theta = [/tex3]
[tex3]\int\limits_{0}^{2π}\left(\frac{3\theta ^2}{8π^2}+\frac{3\theta ^4}{8π^4}\right) \ d\theta = [/tex3]
[tex3]\frac{3}{8}.\int\limits_{0}^{2π}\left(\frac{\theta ^2}{π^2}+\frac{\theta ^4}{π^4}\right) \ d\theta = [/tex3]
[tex3]\frac{3}{8}.\left[\frac{\theta ^3}{3π^2}+\frac{\theta ^5}{5π^4}\right]_{0}^{2π} = [/tex3]
[tex3]\frac{3}{8}.\left(\frac{8π^3}{3π^2}+\frac{32π^5}{5π^4}-0-0\right) = [/tex3]
[tex3]\frac{3}{8}.\left(\frac{8π}{3}+\frac{32π}{5}\right) = π+\frac{12π}{5}=\frac{17π}{5}[/tex3]
Portanto, [tex3]\int\limits_{0}^{2π}\int\limits_{0}^{
\frac{\theta }{2π}}\int\limits_{0}^{3+24r^2}r \ dzdrd\theta = \frac{17π}{5}[/tex3]
Bons estudos!