The region [latex]Q[/latex] is a tetrahedron (Figure 1) meeting the axes at the points [latex](6, 0, 0)[/latex], [latex](0, 3, 0)[/latex], and [latex](0, 0, 2)[/latex]. To find the limits of integration, let [latex]z=0[/latex] in the slanted plane [latex]{z} = {\frac{1}{3}}{({6} - {x} - {{2}{y}})}[/latex]. Then for [latex]x[/latex] and [latex]y[/latex] find the projection of [latex]Q[/latex] onto the [latex]xy[/latex]-plane, which is bounded by the axes and the line [latex]x+2y=6[/latex]. Hence the mass is

We have used this tetrahedron before and know the limits of integration, so we can proceed to the computations right away. First, we need to find the moments about the [latex]xy[/latex]-plane, the [latex]xz[/latex]-plane, and the [latex]yz[/latex]-plane:

[latex]\begin{aligned} {{M}_{xy}} = &{\underset{Q}{\displaystyle\iiint}}{z}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{2}}{{yz}^{2}}{dz} \ {dy} \ {dx} = {\frac{54}{35}} \ {\approx} \ {1.543}, \\ {{M}_{xz}} = & {\underset{Q}{\displaystyle\iiint}}{y}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{2}}{{y}^{2}}{z} \ {dz} \ {dy} \ {dx} = {\frac{81}{35}} \ {\approx} \ {2.314}, \\ {{M}_{yz}} = & {\underset{Q}{\displaystyle\iiint}}{x}{\rho}{({x},{y},{z})}{dV} = {\displaystyle\int^{{x} = {6}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {1/2}{({6} - {x})}}_{{y} = {0}}} \ {\displaystyle\int^{{z} = {1/3}{({6} - {x} - {2y})}}_{{z} = {0}}} \ {{x}^{3}}{yz} \ {dz} \ {dy} \ {dx} = {\frac{243}{35}} \ {\approx} \ {6.943}. \end{aligned}[/latex]

Hence the center of mass is

[latex]\begin{aligned} {\overline{x}} = & {\frac{{M}_{yz}}{m}}, \ {\overline{y}} = {\frac{{M}_{xz}}{m}}, \ {\overline{z}} = {\frac{{M}_{xy}}{m}}, \\ {\overline{x}} = & {\frac{{M}_{yz}}{m}} = {\frac{243/35}{108/35}} = {\frac{243}{108}} = {2.25}, \\ {\overline{y}} = & {\frac{{M}_{xz}}{m}} = {\frac{81/35}{108/35}} = {\frac{81}{108}} = {0.75}, \\ {\overline{z}} = & {\frac{{M}_{xy}}{m}} = {\frac{54/35}{108/35}} = {\frac{54}{108}} = {0.5}. \end{aligned}[/latex]

The center of mass for the tetrahedron [latex]Q[/latex] is the point [latex]{({2.25},{0.75},{0.5})}.[/latex]

Once again, we can almost immediately write the limits of integration and hence we can quickly proceed to evaluating the moments of inertia. Using the formula stated before, the moments of inertia of the tetrahedron [latex]Q[/latex] about the [latex]xy[/latex]-plane, the [latex]xz[/latex]-plane, and the [latex]yz[/latex]-plane are

and

Proceeding with the computations, we have

{{I}_{x}} &= {\underset{Q}{\displaystyle\iiint}}{({{y}^{2}} + {{z}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(y^2+z^2)x^2yz \ dz \ dy \ dx = \frac{117}{35}\approx3.343, \\{{I}_{y}} &= {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{z}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(x^2+z^2)x^2yz \ dz \ dy \ dx = \frac{684}{35}\approx19.543, \\{{I}_{z}} &= {\underset{Q}{\displaystyle\iiint}}{({{x}^{2}} + {{y}^{2}})}x^2yz \ {dV}=\displaystyle\int_{x=0}^{x=6}\displaystyle\int_{y=0}^{y=\frac{1}{2}(6-x)}\displaystyle\int_{z=0}^{z=\frac13(6-x-2y)}(x^2+y^2)x^2yz \ dz \ dy \ dx = \frac{729}{35}\approx20.829.\end{align}[/latex]

Thus, the moments of inertia of the tetrahedron [latex]Q[/latex] about the [latex]yz[/latex]-plane, the [latex]xz[/latex]-plane, and the [latex]xy[/latex]-plane are [latex]{117/35}, {684/35},[/latex] and [latex]{729/35}[/latex], respectively.