Question

Let a, b, and cbe positive real numbers. In Exercises 65–68, letT be the tetrahedron with vertices(0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0,c).

Assume that the density at each point in Tis proportional to the distance of the point from the xz-plane. Set up the integral expressions required to find the center of mass of T.

Short answer

The integral expression to find the center of a mass ofTis$$\begin{gathered} \overline{x}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kxydxdydz \\ \overline{y}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}k{y}^{2}dxdydz \\ \overline{z}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kyzdxdydz \end{gathered}$$.

Step-by-step solution

Step 1. Given Information. 

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

Step 2. Find the integral expression of the center of a mass of T.

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

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

$$\begin{gathered} M=\iiint \rho (x,y,z)dxdydz \\ M={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kydxdydz \end{gathered}$$

Now, the center of a mass of Tis,

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{T}x\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{x}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kxydxdydz \end{gathered}$$

Step 3. Solve.

By proceeding with the calculation further,

$$\begin{gathered} \overline{y}=\frac{M_{xz}}{M}=\frac{{\iiint }_{T}y\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{y}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}k{y}^{2}dxdydz \end{gathered}$$

Now,

$$\begin{gathered} \overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{T}z\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{z}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kyzdxdydz \end{gathered}$$