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 T is uniform throughout.

(a) Find the x-coordinate of the center of mass of T.

(b) Explain how to use your answer from part (a) to find the y- and z-coordinates of the center of mass without doing any other computations.

Short answer

Part (a) The x-coordinate of the center of mass of Tis$\frac{a}{4}.$

Part (b) The y- and z-coordinates of the center of mass are$\left(\frac{b}{4},\frac{c}{4}\right).$

Step-by-step solution

Part (a) Step 1. Given Information. 

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

Part (a) Step 2. Find the x-coordinate of the center of mass of T. 

It is given that the density at each point in T is uniform throughput, so$\rho \left(x,y,z\right)=k.$

The x-coordinate of the center of a mass of T is $\overline{x}=\frac{M_{yz}}{M}.$

To find the x-coordinate of the center of a mass, let's first find the mass:

$$\begin{gathered} M={\iiint }_{\tau }\rho (x,y,z)dv \\ M={\int }_0^a{\int }_0^b{\int }_c^{c}kdzdydx \\ M=k{\int }_0^a{\int }_0^b\left[{\int }_0^{c}dz\right]dydx \\ M=kabc \end{gathered}$$

Now,

$$\begin{gathered} M_{yz}={\iiint }_{\tau }x\rho (x,y,z)dv \\ {M}_{yz}={\int }_0^a{\int }_0^b{\int }_0^{c}x\left(k\right)dzdydx \\ {M}_{yz}=k{\int }_0^a{\int }_0^b\left[x{\int }_0^{c}dz\right]dydx \\ {M}_{yz}=\frac{k{a}^{2}bc}{2} \end{gathered}$$

Part (a) Step 3. Solve.

By proceeding with the calculation further,

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M} \\ \overline{x}=\frac{1}{kabc}\left(\frac{k{a}^{2}bc}{2}\right) \\ \overline{x}=\frac{a}{4} \end{gathered}$$

Part (b) Step 1. Find the y- and z-coordinates of the center of mass.

We have to find y- and z-coordinates of the center of mass without doing any other computations, so as we know the center of a mass is the midpoint of the coordinates, $0\le x\le a,0\le y\le b,0\le z\le c.$

Now, the center of a mass of coordinates is:

role="math" localid="1650368135775" $$\begin{gathered} (\overline{x},\overline{y},\overline{z})=\left(\frac{0+a+0+0}{4},\frac{0+0+b+0}{4},\frac{0+0+0+c}{4}\right) \\ (\overline{x},\overline{y},\overline{z})=\left(\frac{a}{4},\frac{b}{4},\frac{c}{4}\right) \end{gathered}$$

Hence the y-and z-coordinates of the center of mass are$\left(\frac{b}{4},\frac{c}{4}\right).$