Chapter 8.1, Problem 40E

To determine

Tofind the centre of mass.

Answer to Problem 40E

The centre of the mass is found to be $\left(\frac{3}{4},\frac{12}{5}\right)$ .

Explanation of Solution

Given information:

Suppose we have a finite collection of masses in the co-ordinate plane, the mass $m_k$ located at the point $\left(x_k,y_k\right)$ as shown in the figure.

  

The region bounded by the parabola $y=x^2$ and the line $y=4$ with constant density $\delta$ .

The area is given by

  $\begin{array}{l}A={\displaystyle \underset{0}{\overset{2}{\int }}dx}{\displaystyle \underset{x^2}{\overset{4}{\int }}dy}\\ A={\displaystyle \underset{0}{\overset{2}{\int }}\left(4-x^2\right)}dx={\left[4x-\frac{x^3}{3}\right]}_0^2\\ \Rightarrow 8-\frac{8}{3}=\frac{16}{3}\end{array}$

The weight is $A\cdot \delta$ where $\delta$ is the constant density.

Let $X$ and $Y$ be the co-ordinates of the centre. Then by definition,

  $\begin{array}{l}X=\frac{1}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}xdx}{\displaystyle \underset{x^2}{\overset{4}{\int }}dy}=\frac{1}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}\left(4x-x^3\right)dx}\\ \Rightarrow \frac{1}{A}\cdot \left(8-\frac{16}{4}\right)=\frac{3}{4}\end{array}$

And

  $\begin{array}{l}Y=\frac{1}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}xdx}{\displaystyle \underset{x^2}{\overset{4}{\int }}dy}=\frac{1}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{2}\left(4^2-x^4\right)dx}\\ \Rightarrow \frac{1}{A}\cdot \frac{1}{2}\left(32-\frac{32}{5}\right)=\frac{12}{5}\end{array}$

Therefore, the centre of the mass is found to be $\left(\frac{3}{4},\frac{12}{5}\right)$ .