Q. 61 - Calculus

Question

In Exercises 61–64, let R be the rectangular solid defined by

$\mathfrak{R}=\left\{(x,y,z)|0\le a_1\le x\le a_2, 0\le b_1\le y\le b_2, 0\le c_2\le z\le c_2\right\}.$

Assume that the density of R is uniform throughout.

(a) Without using calculus, explain why the center of

mass is  $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right).$

(b) Verify that $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right)$ is the center of mass by using the appropriate integral expressions.

Step-by-Step Solution

Answer

Part (a) The center of mass is $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right)$ because it is given that density is uniform so $m_1=m_2=k.$

Part (b) It is verified that $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right)$ is the center of a mass.

Part (a) Step 1. Given Information.

The given rectangular solid is defined by  $\mathfrak{R}=\left\{(x,y,z)|0\le a_1\le x\le a_2, 0\le b_1\le y\le b_2, 0\le c_2\le z\le c_2\right\}.$

Part (a) Step 2. Explanation.

As we know the center of a mass of the rectangular solid is 

$\overline{x}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}, \overline{y}=\frac{m_1{b}_1+m_2{b}_2}{m_1+m_2}, \overline{z}=\frac{m_1{c}_1+m_2{c}_2}{m_1+m_2}$ where the $m_1 \mathrm{and} m_2$ are the masses of the body.

Since the density is uniform $m_1=m_2=k.$

Thus, the center of the masses are $\overline{x}=\frac{a_1+a_2}{2}, \overline{y}=\frac{b_1+b_2}{2}, \overline{z}=\frac{c_1+c_2}{2}.$

Part (b) Step 1. Verification.

It is given that the density of R is uniform throughout, so $\rho (x,y,z)=k.$

To find the center of a mass using the integral expression, let's first find the mass,

$$\begin{gathered} M={\iiint }_R\rho (x,y,z)dxdydz \\ M={\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}kdxdydz \\ M=k(a_2-a_1)(b_2-b_1)(c_2-c_1) \end{gathered}$$

Now, the center of the mass of the rectangular solid is

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{T}x\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{x}=\frac{1}{M}{\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}\left(kx\right)dxdydz \\ \overline{x}=\left(\frac{k}{2M}\right)(a_2^2-a_1^2)(b_2-b_1)(c_2-c_1) \\ \overline{x}=\left(\frac{{\displaystyle k}}{{\displaystyle 2\left(k(a_2-a_1)(b_2-b_1)(c_2-c_1)\right)}}\right)(a_2^2-a_1^2)(b_2-b_1)(c_2-c_1) \\ \overline{x}=\frac{a_1+a_2}{2} \end{gathered}$$

Part (b) Step 2. Verification.

By proceeding with the center of a mass,

$$\begin{gathered} \overline{y}=\frac{M_{xz}}{M}=\frac{{\iiint }_{T}y\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{y}=\frac{1}{M}{\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}\left(ky\right)dxdydz \\ \overline{y}=\left(\frac{k}{2M}\right)(a_2-a_1)({b_2}^2-{b_1}^2)(c_2-c_1) \\ \overline{y}=\left(\frac{{\displaystyle k}}{{\displaystyle 2\left(k(a_2-a_1)(b_2-b_1)(c_2-c_1)\right)}}\right)(a_2-a_1)({b_2}^2-{b_1}^2)(c_2-c_1) \\ \overline{y}=\frac{b_1+b_2}{2} \end{gathered}$$

Now, 

$$\begin{gathered} \overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{T}z\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{z}=\frac{1}{M}{\int }_{x=a_1}^{a_2}{\int }_{y=b_1}^{b_2}{\int }_{z=c_1}^{c_2}\left(kz\right)dxdydz \\ \overline{z}=\left(\frac{k}{2M}\right)(a_2-a_1)(b_2-b_1)({c_2}^2-{c_1}^2) \\ \overline{z}=\left(\frac{{\displaystyle k}}{{\displaystyle 2\left(k(a_2-a_1)(b_2-b_1)(c_2-c_1)\right)}}\right)(a_2-a_1)(b_2-b_1)({c_2}^2-{c_1}^2) \\ \overline{z}=\frac{c_1+c_2}{2} \end{gathered}$$

Hence proved, the center of mass is $\left(\frac{a_1+a_2}{2},\frac{b_1+b_2}{2},\frac{c_1+c_2}{2}\right).$