Q. 65 - Calculus

Question

Let a, b, and c be positive real numbers. In Exercises 65–68, let T 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.

Step-by-Step Solution

Answer

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

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

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:

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

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