Question

A integral $\displaystyle\int_0^{2\pi}\int_0^{\begin{array}{l}\!\!\!\scriptsize\frac{\theta}{2\pi}\end{array}}\int_0^{3+24r^2}\,r\,dz\,dr\,d\theta$ vale:

a) 3,9π

b) 2,4π

c) 4,2π

d) 2,8π

e) 3,4π

Answer 1

$I=\displaystyle\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}\int_{0}^{3+24r^{2}}r\,dz\,dr\,d\theta=\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}\bigg[\int_{0}^{3+24r^{2}}r\,dz\bigg]\,dr\,d\theta$

Encontrando a integral destacada:

$I=\displaystyle\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}\bigg[rz\bigg|_{z=0}^{z=3+24r^{2}}\bigg]\,dr\,d\theta\\\\\\I=\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}\big[r(3+24r^{2})-r(0)\big]\,dr\,d\theta\\\\\\I=\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}r(3+24r^{2})\,dr\,d\theta\\\\\\I=\int_{0}^{2\pi}\int_{0}^{\theta/2\pi}(3r+24r^{3})\,dr\,d\theta\\\\\\I=\int_{0}^{2\pi}\bigg[\int_{0}^{\theta/2\pi}(3r+24r^{3})\,dr\bigg]\,d\theta$

Sendo assim, temos

$I=\displaystyle\int_{0}^{2\pi}\bigg[\dfrac{3}{2}r^{2}+6r^{4}\bigg|_{r=0}^{r=\theta/2\pi}\bigg]\,d\theta\\\\\\I=\int_{0}^{2\pi}\bigg[\dfrac{3}{2}\bigg(\dfrac{\theta}{2\pi}\bigg)^{2}+6\bigg(\dfrac{\theta}{2\pi}\bigg)^{4}\bigg]\,d\theta\\\\\\I=\int_{0}^{2\pi}\bigg[\dfrac{3}{2}\cdot\dfrac{\theta^{2}}{4\pi^{2}}+6\dfrac{\theta^{4}}{16\pi^{4}}\bigg]\,d\theta\\\\\\I=\int_{0}^{2\pi}\bigg[\dfrac{3}{8\pi^{2}}\theta^{2}+\dfrac{3}{8\pi^{4}}\theta^{4}\bigg]\,d\theta$

Integrando novamente:

$I=\bigg[\dfrac{3}{8\pi^{2}}\cdot\dfrac{1}{3}\theta^{3}+\dfrac{3}{8\pi^{4}}\cdot\dfrac{1}{5}\theta^{5}\bigg]_{\theta=0}^{\theta=2\pi}\\\\\\I=\bigg[\dfrac{\theta^{3}}{8\pi^{2}}+\dfrac{3\theta^{5}}{40\pi^{4}}\bigg]_{\theta=0}^{\theta=2\pi}\\\\\\I=\bigg[\dfrac{(2\pi)^{3}}{8\pi^{2}}+\dfrac{3(2\pi)^{5}}{40\pi^{4}}\bigg]-0\\\\\\I=\dfrac{8\pi^{3}}{8\pi^{2}}+\dfrac{3\cdot32\pi^{5}}{40\pi^{4}}\\\\\\I=\pi+\dfrac{12\pi}{5}\\\\\\I=\dfrac{5\pi+12\pi}{5}\\\\\\I=\dfrac{17}{5}\pi\\\\\\\boxed{\boxed{I=3,4\pi}}$