91Ó°ÊÓ

The given vertices of the tetrahedron are$(0,0,0),(a,0,0),(0,b,0),\mathrm{and}(0,0,c).$

It is given that the density at each point in Tis proportional to the distance of the point from the xz-plane, so $\mathrm{ÒÏ}(x,y,z)=ky.$

To find the center of a mass of tetrahedron, let's first find the mass:

$$\begin{gathered} M=∭\mathrm{ÒÏ}(x,y,z)dxdydz \\ M={∫}_{x=0}^a{∫}_{y=0}^{b\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{∫}_{z=0}^{c\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}−\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kydxdydz \end{gathered}$$

Now, the center of a mass of Tis,

$$\begin{gathered} \stackrel{¯}{x}=\frac{M_{yz}}{M}=\frac{{∭}_{T}x\mathrm{ÒÏ}(x,y,z)dxdydz}{∭\mathrm{ÒÏ}(x,y,z)dxdydz} \\ \stackrel{¯}{x}=\frac{1}{M}{∫}_{x=0}^a{∫}_{y=0}^{b\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{∫}_{z=0}^{c\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}−\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kxydxdydz \end{gathered}$$

By proceeding with the calculation further,

$$\begin{gathered} \stackrel{¯}{y}=\frac{M_{xz}}{M}=\frac{{∭}_{T}y\mathrm{ÒÏ}(x,y,z)dxdydz}{∭\mathrm{ÒÏ}(x,y,z)dxdydz} \\ \stackrel{¯}{y}=\frac{1}{M}{∫}_{x=0}^a{∫}_{y=0}^{b\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{∫}_{z=0}^{c\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}−\frac{{\displaystyle y}}{{\displaystyle b}}\right)}k{y}^{2}dxdydz \end{gathered}$$

Now,

$$\begin{gathered} \stackrel{¯}{z}=\frac{M_{xy}}{M}=\frac{{∭}_{T}z\mathrm{ÒÏ}(x,y,z)dxdydz}{∭\mathrm{ÒÏ}(x,y,z)dxdydz} \\ \stackrel{¯}{z}=\frac{1}{M}{∫}_{x=0}^a{∫}_{y=0}^{b\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{∫}_{z=0}^{c\left(1−\frac{{\displaystyle x}}{{\displaystyle a}}−\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kyzdxdydz \end{gathered}$$