Question

EXAMPLE 2 Find the center of mass of a thin plate covering the region bounded
above by the parabola y = 4 - x? and below by the x-axis (Figure 6.51). Assume the den-
sity of the plate at the point (x, y) is 8 = 2x?, which is twice the square of the distance
from the point to the y-axis,​

Answer 1

Correct Question:-

Find the center of mass of a thin plate covering the region bounded above by the parabola $\sf{y=4-x^2}$ and below by the x-axis. Assume the density of the plate at the point $\sf{(x,\ y)}$ is $\sf{\rho=2x^2,}$ which is twice the square of the distance

from the point to the y-axis.

Solution:-

Let the position of the center of mass be $\sf{(\bar x,\ \bar y).}$

The plate is bounded above, by the parabola $\sf{y=4-x^2}$ which is an even function. So the parabola, or the plate itself, is symmetrical along y axis and hence the center of mass of the plate lies along y axis, i.e., it's x coordinate is zero.

$\sf{\longrightarrow \bar x=0}$

As in the figure we consider a rectangular element of length $\sf{2x}$ and breadth $\sf{dy}$ of mass $\sf{dm ,}$ from the plate.

The area of this element,

$\sf{\longrightarrow dA=2x\ dy}$

But,

$\sf{\longrightarrow y=4-x^2}$

$\sf{\longrightarrow dy=-2x\ dx }$

Then,

$\sf{\longrightarrow dA=2x\ (-2x\ dx)}$

$\sf{\longrightarrow dA=-4x^2\ dx}$

Given that the density of the plate is,

$\sf{\longrightarrow\rho=2x^2}$

in case of the element,

$\sf{\longrightarrow\dfrac{dm}{dA}=2x^2}$

$\sf{\longrightarrow dm=2x^2\ dA}$

$\sf{\longrightarrow dm=2x^2\ (-4x^2\ dx)}$

$\sf{\longrightarrow dm=-8x^4\ dx}$

Now, the y coordinate of the center of mass of the plate is given by,

$\sf{\longrightarrow \bar y=\dfrac{\displaystyle\int\limits_0^My\ dm}{\displaystyle\int\limits_0^Mdm}}$

$\sf{\longrightarrow \bar y=\dfrac{\displaystyle\int\limits_{-2}^2\left(4-x^2\right)\left(-8x^4\ dx\right)}{\displaystyle\int\limits_{-2}^2-8x^4\ dx}}$

$\sf{\longrightarrow \bar y=\dfrac{-8\displaystyle\int\limits_{-2}^2x^4\left(4-x^2\right)\ dx}{\displaystyle-8\int\limits_{-2}^2x^4\ dx}}$

$\sf{\longrightarrow \bar y=\dfrac{\displaystyle\int\limits_{-2}^2\left(4x^4-x^6\right)\ dx}{\displaystyle\int\limits_{-2}^2x^4\ dx}}$

$\sf{\longrightarrow \bar y=\dfrac{\dfrac{4}{5}\left[x^5\right]_{-2}^2-\dfrac{1}{7}\left[x^7\right]_{-2}^2}{\dfrac{1}{5}\left[x^5\right]_{-2}^2}}$

$\sf{\longrightarrow \bar y=\dfrac{\dfrac{4}{5}\times64-\dfrac{1}{7}\times256}{\dfrac{1}{5}\times64}}$

$\sf{\longrightarrow \bar y=\dfrac{8}{7}}$

Hence the position of center of mass of the plate is $\bf{\left(0,\ \dfrac{8}{7}\right).}$