Find the moment of inertia about the $z$-axis of the homogeneous solid of mass density $\rho$ in the first octant enclosed by the surface $z=4xy$ and the planes $z=0$, $x=3$, and $y=2$.

Figure 47

Solution The moment of inertia about the $z$-axis is $$I_{z}=\iiint\limits_{\kern-2ptE}r^{2}\rho \,{\it dV}$$

where $r=\sqrt{x^{2}+y^{2}}$ is the distance of the point $(x, y, z)$ from the $z$-axis.

Figure 47(a) shows the solid $E$, which is $xy$-simple. The upper surface is $z=z_{2}(x, y)=4xy$ and the lower surface is the plane $z=z_{1}(x, y)=0$. The region $R$ in the $xy$-plane is enclosed by the $x$-axis, the $y$-axis, the line $x=3$, and the line $y=2$, as shown in Figure 47(b). The region $R$ is both $x$-simple and $y$-simple. $$\begin{eqnarray*} I_{z} &=&\iiint\limits_{\kern-2ptE}(x^{2}+y^{2})\rho \,{\it dV}=\rho \int_{0}^{3}\int_{0}^{2}\int_{0}^{4xy}(x^{2}+y^{2})\,{\it dz}\,{\it dy}\,{\it dx} \notag \\[4pt] &=&\rho \int_{0}^{3}\int_{0}^{2}\big[ z(x^{2}+y^{2})\big] _{0}^{4xy}\,{\it dy}\,{\it dx}=\rho \int_{0}^{3}\int_{0}^{2}( 4x^{3}y+4xy^{3}) \,{\it dy}\,{\it dx}\\[4pt] &=&\rho \int_{0}^{3}\big[ 2x^{3}y^{2}+xy^{4}\big] _{0}^{2}\,\,{\it dx} \notag \\[4pt] &=&\rho \int_{0}^{3}( 8x^{3}+16x) \,{\it dx}=\rho \big[ 2x^{4}+8x^{2}\big] _{0}^{3}=\rho ( 162+72) =234\rho \end{eqnarray*}$$