Q. 57 - Calculus

Question

In Exercises 57–60, let R be the rectangular solid defined by

R = {(x, y, z) | 0 ≤ x ≤ 4, 0 ≤ y ≤ 3, 0 ≤ z ≤ 2}.

Assume that the density of R is uniform throughout.

(a) Without using calculus, explain why the center of mass is (2, 3/2, 1).

(b) Verify that the center of mass is (2, 3/2, 1), using the appropriate integral expressions.

Step-by-Step Solution

Answer

Part (a) The center of mass is $\left(2,\frac{3}{2},1\right)$ because it is given that the density of R is uniform throughout, so the coordinates will be the same as the center of mass coordinates.

Part (b) It is verified that the center of mass is $\left(2,\frac{3}{2},1\right).$

Part (a) Step 1. Given Information.

The given rectangular solid is defined by $\mathfrak{R}=\left\{\right(x,y,z)\mid 0\le x\le 4, 0\le y\le 3, 0\le z\le 2\}.$

Part (a) Step 2. Explanation.

As we know the center of mass is the midpoint of the coordinates, so the center of mass coordinates are:

$$\begin{gathered} (\overline{x},\overline{y},\overline{z})=\left(\frac{0+4}{2},\frac{0+3}{2},\frac{0+2}{2}\right) \\ (\overline{x},\overline{y},\overline{z})=\left(2,\frac{3}{2},1\right) \end{gathered}$$

It is given that the density of R is uniform throughout, so the coordinates will be the same as the center of mass coordinates.

Thus, the coordinates of the center of mass are $\left(2,\frac{3}{2},1\right).$

Part (b) Step 1. Verify the center of mass is 2 , 3 2 , 1 .

We have to use an appropriate integral expression to verify the center of mass. 

Now, the density of R is uniform throughout, so $\rho (x,y,z)=1.$

To find the center of mass let's find the mass of the solid:

$$\begin{gathered} M={\iiint }_R\rho (x,y,z)dzdydx \\ M={\int }_0^4{\int }_0^3{\int }_0^2\left(1\right)dzdydx \\ M={\int }_0^4{\int }_0^3[z{]}_0^{2}dydx \\ M=24 \end{gathered}$$

Now, the center of mass is,

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{T}x\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{x}=\frac{1}{M}{\int }_{x=0}^4{\int }_{y=0}^3{\int }_{z=0}^{2}x dzdydx \\ \overline{x}=\frac{1}{M}{\int }_0^4{\int }_0^3\left[x{\int }_0^{2}dz\right]dydx \\ \overline{x}=\frac{1}{24}\left(48\right) \\ \overline{x}=2 \end{gathered}$$

Part (b) Step 2. Solve.

By proceeding with the calculation further,

$$\begin{gathered} \overline{y}=\frac{M_{xz}}{M}=\frac{{\iiint }_{T}y\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{y}=\frac{1}{M}{\int }_{x=0}^4{\int }_{y=0}^3{\int }_{z=0}^{2}ydzdydx \\ \overline{y}=\frac{1}{24}\left(36\right) \\ \overline{y}=\frac{3}{2} \end{gathered}$$

Now,

$$\begin{gathered} \overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{T}z\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{z}=\frac{1}{M}{\int }_{x=0}^4{\int }_{y=0}^3{\int }_{z=0}^{2}z dzdydx \\ \overline{z}=\frac{1}{24}\left(24\right) \\ \overline{z}=1 \end{gathered}$$

Hence proved, the center of mass is $\left(2,\frac{3}{2},1\right).$