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

nt (0,2π) int (0,t/2π)int (0,3+24r^2) rdzdrdt\
int (0,2π) int (0,t/2π) rz |(0,3+24r^2)  dzdt\
int (0,2π) int (0,t/2π) 3r+24 r^3 -0   drdt\
int (0,2π) int (0,t/2π) (3r^2)/2+(24 r^4)/4 -0 |(0,t/2π)  drdt\
int (0,2π) (3r^2)/2+6.r^4 |(0,t/2π)    drdt \
int (0,2π) 3t^2/8π^2+6t^4/16π^4 dt\= t^3/8π^2+6t^5/80π^4)| 2π\
= (2π)^3/8π^2+6(2π)^5/80π^4)=8π^3/8π^2+ 192π^5/80π^4= π+2,4π=3,4π\