Q. 59 - 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 at each point in R is proportional to the distance of the point from the xy-plane.

(a) Without using calculus, explain why the x- and y-coordinates of the center of mass are $\overline{x}=2 \mathrm{and} \overline{y}=\frac{3}{2},$ respectively.

(b) Use an appropriate integral expression to find the z-coordinate of the center of mass.

Step-by-Step Solution

Answer

Part (a) The x- and y-coordinates of the center of mass are $\overline{x}=2 \mathrm{and} \overline{y}=\frac{3}{2},$ respectively because it is given that density at each point in R is proportional to the distance of the point from the xy-plane, so x- and y-coordinates will be the same as the center of mass coordinates and the z-coordinate of the center of the mass should be zero.

Part (b) The z-coordinate of the center of mass is $\overline{z}=\frac{4}{3}.$

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 density at each point in R is proportional to the distance of the point from the xy-plane, so x- and y-coordinates will be the same as the center of mass coordinates and the z-coordinate of the center of the mass should be zero.

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

Part (b) Step 1. Find the z- coordinate of the center of mass.

We have to use an appropriate integral expression to find the z-coordinate of the center of mass. 

Now,  the density at each point in R is proportional to the distance of the point from the xy-plane, so $\rho (x,y,z)=kz.$

The z-coordinate of the center of mass is $\overline{z}=\frac{M_{xy}}{M}.$

Let's find the mass of the solid:

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

Now,

$$\begin{gathered} M_{xy}={\iiint }_{R}z\rho (x,y,z)dzdydx \\ {M}_{xy}={\int }_0^4{\int }_0^3{\int }_0^{2}z\left(kz\right)dzdydx \\ {M}_{xy}=k\left(4\right)\left(3\right)\left(\frac{8}{3}\right) \\ {M}_{xy}=32k \end{gathered}$$

Thus, the z-coordinate of the center of mass is,

$$\begin{gathered} \overline{z}=\frac{M_{xy}}{M} \\ \overline{z}=\frac{32k}{24k} \\ \overline{z}=\frac{4}{3} \end{gathered}$$