# 6.6 Moments and centers of mass  (Page 4/14)

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Next, we need to find the total mass of the rectangle. Let $\rho$ represent the density of the lamina (note that $\rho$ is a constant). In this case, $\rho$ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by $\rho .$ Then, the mass of the rectangle is given by $\rho f\left({x}_{i}^{*}\right)\text{Δ}x.$

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

$m\approx \sum _{i=1}^{n}\rho f\left({x}_{i}^{*}\right)\text{Δ}x.$

This is a Riemann sum. Taking the limit as $n\to \infty$ gives the exact mass of the lamina:

$m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho f\left({x}_{i}^{*}\right)\text{Δ}x=\rho {\int }_{a}^{b}f\left(x\right)dx.$

Next, we calculate the moment of the lamina with respect to the x -axis. Returning to the representative rectangle, recall its center of mass is $\left({x}_{i}^{*},\left(f\left({x}_{i}^{*}\right)\right)\text{/}2\right).$ Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the x -axis is given by the mass of the rectangle, $\rho f\left({x}_{i}^{*}\right)\text{Δ}x,$ multiplied by the distance from the center of mass to the x -axis: $\left(f\left({x}_{i}^{*}\right)\right)\text{/}2.$ Therefore, the moment with respect to the x -axis of the rectangle is $\rho \left({\left[f\left({x}_{i}^{*}\right)\right]}^{2}\text{/}2\right)\text{Δ}x.$ Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the x -axis is

${M}_{x}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho \frac{{\left[f\left({x}_{i}^{*}\right)\right]}^{2}}{2}\text{Δ}x=\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx.$

We derive the moment with respect to the y -axis similarly, noting that the distance from the center of mass of the rectangle to the y -axis is ${x}_{i}^{*}.$ Then the moment of the lamina with respect to the y -axis is given by

${M}_{y}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho {x}_{i}^{*}f\left({x}_{i}^{*}\right)\text{Δ}x=\rho {\int }_{a}^{b}xf\left(x\right)dx.$

We find the coordinates of the center of mass by dividing the moments by the total mass to give $\stackrel{–}{x}={M}_{y}\text{/}m\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}={M}_{x}\text{/}m.$ If we look closely at the expressions for ${M}_{x},{M}_{y},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}m,$ we notice that the constant $\rho$ cancels out when $\stackrel{–}{x}$ and $\stackrel{–}{y}$ are calculated.

We summarize these findings in the following theorem.

## Center of mass of a thin plate in the xy -plane

Let R denote a region bounded above by the graph of a continuous function $f\left(x\right),$ below by the x -axis, and on the left and right by the lines $x=a$ and $x=b,$ respectively. Let $\rho$ denote the density of the associated lamina. Then we can make the following statements:

1. The mass of the lamina is
$m=\rho {\int }_{a}^{b}f\left(x\right)dx.$
2. The moments ${M}_{x}$ and ${M}_{y}$ of the lamina with respect to the x - and y -axes, respectively, are
${M}_{x}=\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{M}_{y}=\rho {\int }_{a}^{b}xf\left(x\right)dx.$
3. The coordinates of the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ are
$\stackrel{–}{x}=\frac{{M}_{y}}{m}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}.$

In the next example, we use this theorem to find the center of mass of a lamina.

## Finding the center of mass of a lamina

Let R be the region bounded above by the graph of the function $f\left(x\right)=\sqrt{x}$ and below by the x -axis over the interval $\left[0,4\right].$ Find the centroid of the region.

The region is depicted in the following figure.

Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant $\rho$ cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume $\rho =1.$

First, we need to calculate the total mass:

$\begin{array}{cc}\hfill m& =\rho {\int }_{a}^{b}f\left(x\right)dx={\int }_{0}^{4}\sqrt{x}\phantom{\rule{0.2em}{0ex}}dx\hfill \\ & ={\frac{2}{3}{x}^{3\text{/}2}|}_{0}^{4}=\frac{2}{3}\left[8-0\right]=\frac{16}{3}.\hfill \end{array}$

Next, we compute the moments:

$\begin{array}{cc}\hfill {M}_{x}& =\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\hfill \\ & ={\int }_{0}^{4}\frac{x}{2}dx={\frac{1}{4}{x}^{2}|}_{0}^{4}=4\hfill \end{array}$

and

$\begin{array}{cc}\hfill {M}_{y}& =\rho {\int }_{a}^{b}xf\left(x\right)dx\hfill \\ & ={\int }_{0}^{4}x\sqrt{x}dx={\int }_{0}^{4}{x}^{3\text{/}2}dx\hfill \\ & ={\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{2}{5}\left[32-0\right]=\frac{64}{5}.\hfill \end{array}$

Thus, we have

$\stackrel{–}{x}=\frac{{M}_{y}}{m}=\frac{64\text{/}5}{16\text{/}3}=\frac{64}{5}·\frac{3}{16}=\frac{12}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{y}=\frac{4}{16\text{/}3}=4·\frac{3}{16}=\frac{3}{4}.$

The centroid of the region is $\left(12\text{/}5,3\text{/}4\right).$

why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg