Moment of Inertia of a Cube About an Edge
\[a\]
of variable density \[\rho = k(x+y+z)\]
with one vertex at the origin at edges parallel to the axes is allowed to rotate about an axis. What is the moment of inertia of the cube about that axis?A volume element
\[dV=dxdydz\]
at point \[(x,y,z)\]
has mass \[dm= \rho dV= k(x+y+z) dxdydz\]
and the moment of inertia of the volume element about the \[z\]
axis is \[dI= (x^2+y^2) dm=(x^2+y^2)k(x+y+z)dxdydz\]
.The moment of inertia of the cuboid is
\[\begin{equation} \begin{aligned}I &= \int^{z=a}_{z=0} \int^{y=a}_{y=0} \int^{x=a}_{x=0} (x^2+y^2)k(x+y+z)dxdydz \\ &=k \int^{z=a}_{z=0} \int^{y=a}_{y=0} \int^{x=a}_{x=0} x^3+y^3+x^2y+x^2z+xy^2+y^2zdxdydz \\ &=k \int^{z=a}_{z=0} \int^{y=a}_{y=0} [\frac{x^4}{4}+xy^3+ \frac{x^3y}{3}+ \frac{x^3z}{3}+ \frac{x^2y^2}{2}+xy^2z]^{x=a}_{x=0} ]dydz \\ &=k \int^{z=a}_{z=0} \int^{y=a}_{y=0}\frac{a^4}{4}+ay^3 + \frac{a^3y}{3}+ \frac{a^3z}{3}+ \frac{a^2y^2}{2}+ay^2z \\ &=k \int^{z=a}_{z=0} [ \frac{a^4y}{4}+ \frac{ay^4}{4}+ \frac{a^3y^2}{6}+ \frac{a^3yz}{3}+ \frac{a^2y^3}{6}+ \frac{ay^3z}{3} ]^{y=a}_{y=0} dz \\ &=k \int^{z=a}_{z=0} \frac{a^5}{4}+ \frac{a^5}{4}+ \frac{a^5}{6}+ \frac{a^4z}{3}+ \frac{a^5}{6}+ \frac{a^4z}{3} dz \\ &=k \int^{z=a}_{z=0} \frac{17a^5}{6}+ \frac{2a^4z}{3} dz \\ &= k[ \frac{17a^5z}{6}+ \frac{a^4z^2}{3} ]^{z=a}_{z=0} \\ &= \frac{19ka^6}{6}\end{aligned} \end{equation}\]
