Problem of the Week #132 - October 6th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Find the center of gravity of the square lamina with vertices $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$ if the density is proportional to the square of the distance from the origin.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by MarkFL. You can find his solution below.

[sp][box=blue]We define the mass of the lamina by the double integral:

$$m=\iint\limits_{R}\rho(x,y)\,dA\tag 1$$​

We define the coordinates of the center of mass (or gravity) of the lamina by:

$$\overline{x}=\frac{M_y}{m}\quad\overline{y}=\frac{M_x}{m}\tag 2$$​

where:

$$M_y=\iint\limits_{R}x\rho(x,y)\,dA\quad\text{ and }\quad M_x=\iint\limits_{R}y\rho(x,y)\,dA\tag 3$$​

are the moments of the lamina about the $y$- and $x$-axes respectively.[/box]

For this problem, we are told:

$$\rho(x,y)=k\left(x^2+y^2\right)$$

And so, using (1) we find that the mass $m$ of the given lamina is:

$$m=k\iint\limits_{R} x^2+y^2\,dA=k\int_0^1\int_0^1 x^2+y^2\,dx\,dy=k\int_0^1\left[\frac{x^3}{3}+xy^2\right]_0^1\,dy=k\int_0^1\frac{1}{3}+y^2\,dy=k\left[\frac{1}{3}y+\frac{1}{3}y^3\right]_0^1=\frac{2k}{3}$$

Now, using (3), we find:

$$M_y=k\iint\limits_{R} x^3+xy^2\,dA=k\int_0^1\int_0^1 x^3+xy^2\,dx\,dy=k\int_0^1\left[\frac{x^4}{4}+\frac{x^2}{2}y^2\right]_0^1\,dy=\frac{k}{4}\int_0^1 2y^2+1\,dy=\frac{k}{4}\left[\frac{2}{3}y^2+y\right]_0^1=\frac{5k}{12}$$

$$M_x=k\iint\limits_{R} yx^2+y^3\,dA=k\int_0^1\int_0^1 yx^2+y^3\,dx\,dy=k\int_0^1\left[y\frac{x^3}{3}+xy^3\right]_0^1\,dy=\frac{k}{3}\int_0^1 3y^3+y\,dy=\frac{k}{3}\left[\frac{3}{4}y^4+\frac{y^2}{2}\right]_0^1=\frac{5k}{12}$$

As we would expect, the moments of the lamina are the same, given the symmetry of the lamina about the line $y=x$, and so we would expect that the center of gravity would lie on this line.

Hence, by (2), we find:

$$\overline{x}=\overline{y}=\frac{\dfrac{5k}{12}}{\dfrac{2k}{3}}=\frac{5}{8}$$

Thus, we find that the center of gravity of the given lamina is:

$$(\overline{x},\overline{y})=\left(\frac{5}{8},\frac{5}{8}\right)$$[/sp]