Question

Assinale a alternativa correta, correspondente aos momentos de Inércia lx ly e lz, considerando que a integral (em anexo) representa um cubo no 1º octante, cuja função densidade é p(x,y,z) = x + 2.

Escolha uma:

a = Ix = 7/4, Iy = 7/4, Iz =7/4
b = Ix = 5/3, Iy = 7/4, Iz =7/4
c = Ix = 11/6, Iy = 11/6, Iz = 11/6
d = Ix = 7/4, Iy = 7/4, Iz = 11/6
e = Ix = 11/6, Iy = 11/6, Iz 7/4

Answer 1

$\displaystyle\\ \\ R=[0,1]\times [0,1]\times [0,1]\\ \\ I_x=\iiint\limits_R(y^2+z^2)\rho(x,y,z)dV\\ \\ \\ I_x=\iiint\limits_R(y^2+z^2)(x+2)dV\\ \\ \\ I_x=\iiint\limits_Rxy^2+xz^2+2y^2+2z^2\; dV\\ \\ \\ I_x=\int\limits_{0}^{1}\int\limits_{0}^{1}\int\limits_{0}^{1}xy^2+xz^2+2y^2+2z^2\;dx\, dy\,dz\\ \\ \\ \boxed{I_x=\dfrac{5}{3}}$


$\displaystyle I_y=\iiint\limits_R(x^2+z^2)\rho(x,y,z)dV\\ \\ \\ I_y=\iiint\limits_R(x^2+z^2)(x+2)dV\\ \\ \\ I_y=\iiint\limits_Rx^3+xz^2+2x^2+2z^2\; dV\\ \\ \\ I_y=\int\limits_{0}^{1}\int\limits_{0}^{1}\int\limits_{0}^{1}x^3+xz^2+2x^2+2z^2\;dx\,dy\,dz\\ \\ \\ \boxed{I_y=\dfrac{7}{4}}$


$\displaystyle I_z=\iiint\limits_R(x^2+y^2)\rho(x,y,z)dV\\ \\ \\ I_z=\iiint\limits_R(x^2+y^2)(x+2)dV\\ \\ \\ I_z=\iiint\limits_Rx^3+xy^2+2x^2+2y^2\;dV\\ \\ \\ I_z=\int\limits_{0}^{1}\int\limits_{0}^{1}\int\limits_{0}^{1}x^3+xy^2+2x^2+2y^2\;dx\,dy\,dz\\ \\ \\ \boxed{I_z=\dfrac{7}{4}}$