Section 4.8 Mass

4.8 Mass

We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function

over a bounded region. In this section we develop computational techniques for finding the center of mass. 

Definition: Center of Mass in Two Dimensions

The center of mass is the center of gravity if the object is in a uniform gravitational field. To find the coordinates of the center

of mass $P(\bar{x},\bar{y})$ of a lamina, we need to find the moment $M_x$ of the lamina about the $x$-axis and the moment $M_y$ about the $y$-axis. We also need to find the mass $m$ of the lamina. Then $\bar{x}=\frac{M_y}{m}$ and $\bar{y}=\frac{M_x}{m}$. 

Theorem: The mass of the lamina with density function $\rho (x,y)$ is 

$$m=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l{m}_{ij}=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l\rho (x_{ij}^{\ast },y_{ij}^{\ast })\mathrm{△}A=\int {\int }_R\rho (x,y)dA$$

Example 1: Consider a triangular lamina $R$ with vertices $(0,0)$, $(0,2)$ and $(2,0)$ and with density $\rho (x,y)=xy$ $\text{kg}/{\text{m}}^2$. Find the total mass. 

Exercise 1: Consider a triangular lamina $R$ with vertices $(0,0)$, $(0,1)$ and $(1,1)$ and with density $\rho (x,y)=x+y$ $\text{kg}/{\text{m}}^2$. Find the total mass. 

Example 2: Electric charge is distributed over the disk $x^2+y^2\le 4$ so the charge density is $\rho (x,y)=e^{x^2+y^2}$ at the point $(x,y)$. Find the total charge. 

Exercise 2: Electric charge is distributed over the disk $x^2+y^2\le 1$ so the charge density is $\rho (x,y)=e^{x^2+y^2}$ at the point $(x,y)$. Find the total charge. 

Theorem:

The moment $M_x$ about the $x$-axis for $R$ is the limit of the sums of moments of the regions $R_{ij}$ about the $x$-axis. Hence

$$M_x=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l{y}_{ij}^{\ast }{m}_{ij}=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l{y}_{ij}^{\ast }\rho (x_{ij}^{\ast },y_{ij}^{\ast })\mathrm{△}A=\int {\int }_{R}y\rho (x,y)dA$$

The moment $M_y$ about the $y$-axis for $R$ is the limit of the sums of moments of the regions $R_{ij}$ about the $y$-axis. Hence

$$M_y=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l{x}_{ij}^{\ast }{m}_{ij}=\underset{k,l\to \mathrm{\infty }}{lim}\sum _{i=1}^k\sum _{j=1}^l{x}_{ij}^{\ast }\rho (x_{ij}^{\ast },y_{ij}^{\ast })\mathrm{△}A=\int {\int }_{R}x\rho (x,y)dA$$

Example 3: Consider a triangular lamina $R$ with vertices $(0,0)$, $(0,2)$ and $(2,0)$ and with density $\rho (x,y)=xy$ $\text{kg}/{\text{m}}^2$. Find the moments $M_x$ and $M_y$ and center of the mass. 

Exercise 3: Consider a triangular lamina $R$ with vertices $(0,0)$, $(0,1)$ and $(1,1)$ and with density $\rho (x,y)=x+y$ $\text{kg}/{\text{m}}^2$. Find the moments $M_x$ and $M_y$ and center of the mass. 

Example 4: A lamina occupies the part of the disk $x^2+y^2\le 4$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the $x$-axis. 

Exercise 4: A lamina occupies the part of the disk $x^2+y^2\le 1$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the $y$-axis.

Example 5: The boundary of a lamina consists of the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{4-x^2}$together with the portions of $x$-axis that join them. Find the center of the mass of the lamina if the density at any point is proportional to its distance from the origin. 

Group work:

1. The boundary of a lamina consists of the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{4-x^2}$ together with the portions of $x$-axis that join them. Find the center of the mass of the lamina if the density at any point is inversely proportional to its distance from the origin.

2. Find the mass, moments, and the center of mass of the lamina of density $\rho (x,y)=x+y$ occupying the region $R$ under the curve $y=x^2$ in the interval $0\le x\le 2$.