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

Figure 1. Finding the mass of a three-dimensional solid [latex]Q[/latex].

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]