# 6.6 Moments and centers of mass  (Page 5/14)

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Let R be the region bounded above by the graph of the function $f\left(x\right)={x}^{2}$ and below by the x -axis over the interval $\left[0,2\right].$ Find the centroid of the region.

The centroid of the region is $\left(3\text{/}2,6\text{/}5\right).$

We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function $f\left(x\right),$ as before, but now, instead of having the lower bound for the region be the x -axis, suppose the region is bounded below by the graph of a second continuous function, $g\left(x\right),$ as shown in the following figure.

Again, we partition the interval $\left[a,b\right]$ and construct rectangles. A representative rectangle is shown in the following figure.

Note that the centroid of this rectangle is $\left({x}_{i}^{*},\left(f\left({x}_{i}^{*}\right)+g\left({x}_{i}^{*}\right)\right)\text{/}2\right).$ We won’t go through all the details of the Riemann sum development, but let’s look at some of the key steps. In the development of the formulas for the mass of the lamina and the moment with respect to the y -axis, the height of each rectangle is given by $f\left({x}_{i}^{*}\right)-g\left({x}_{i}^{*}\right),$ which leads to the expression $f\left(x\right)-g\left(x\right)$ in the integrands.

In the development of the formula for the moment with respect to the x -axis, the moment of each rectangle is found by multiplying the area of the rectangle, $\rho \left[f\left({x}_{i}^{*}\right)-g\left({x}_{i}^{*}\right)\right]\text{Δ}x,$ by the distance of the centroid from the x -axis, $\left(f\left({x}_{i}^{*}\right)+g\left({x}_{i}^{*}\right)\right)\text{/}2,$ which gives $\rho \left(1\text{/}2\right)\left\{{\left[f\left({x}_{i}^{*}\right)\right]}^{2}-{\left[g\left({x}_{i}^{*}\right)\right]}^{2}\right\}\text{Δ}x.$ Summarizing these findings, we arrive at the following theorem.

## Center of mass of a lamina bounded by two functions

Let R denote a region bounded above by the graph of a continuous function $f\left(x\right),$ below by the graph of the continuous function $g\left(x\right),$ 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
$m=\rho {\int }_{a}^{b}\left[f\left(x\right)-g\left(x\right)\right]dx.$
2. The moments ${M}_{x}$ and ${M}_{y}$ of the lamina with respect to the x - and y -axes, respectively, are
${M}_{x}=\rho {\int }_{a}^{b}\frac{1}{2}\left({\left[f\left(x\right)\right]}^{2}-{\left[g\left(x\right)\right]}^{2}\right)dx\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{M}_{y}=\rho {\int }_{a}^{b}x\left[f\left(x\right)-g\left(x\right)\right]dx.$
3. The coordinates of the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ are
$\stackrel{–}{x}=\frac{{M}_{y}}{m}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}.$

We illustrate this theorem in the following example.

## Finding the centroid of a region bounded by two functions

Let R be the region bounded above by the graph of the function $f\left(x\right)=1-{x}^{2}$ and below by the graph of the function $g\left(x\right)=x-1.$ Find the centroid of the region.

The region is depicted in the following figure.

The graphs of the functions intersect at $\left(-2,-3\right)$ and $\left(1,0\right),$ so we integrate from −2 to 1. Once again, for the sake of convenience, assume $\rho =1.$

First, we need to calculate the total mass:

$\begin{array}{cc}\hfill m& =\rho {\int }_{a}^{b}\left[f\left(x\right)-g\left(x\right)\right]dx\hfill \\ & ={\int }_{-2}^{1}\left[1-{x}^{2}-\left(x-1\right)\right]dx={\int }_{-2}^{1}\left(2-{x}^{2}-x\right)dx\hfill \\ & ={\left[2x-\frac{1}{3}{x}^{3}-\frac{1}{2}{x}^{2}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{1}=\left[2-\frac{1}{3}-\frac{1}{2}\right]-\left[-4+\frac{8}{3}-2\right]=\frac{9}{2}.\hfill \end{array}$

Next, we compute the moments:

$\begin{array}{cc}\hfill {M}_{x}& =\rho {\int }_{a}^{b}\frac{1}{2}\left({\left[f\left(x\right)\right]}^{2}-{\left[g\left(x\right)\right]}^{2}\right)dx\hfill \\ & =\frac{1}{2}{\int }_{-2}^{1}\left({\left(1-{x}^{2}\right)}^{2}-{\left(x-1\right)}^{2}\right)dx=\frac{1}{2}{\int }_{-2}^{1}\left({x}^{4}-3{x}^{2}+2x\right)dx\hfill \\ & =\frac{1}{2}{\left[\frac{{x}^{5}}{5}-{x}^{3}+{x}^{2}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{1}=-\frac{27}{10}\hfill \end{array}$

and

$\begin{array}{cc}\hfill {M}_{y}& =\rho {\int }_{a}^{b}x\left[f\left(x\right)-g\left(x\right)\right]dx\hfill \\ & ={\int }_{-2}^{1}x\left[\left(1-{x}^{2}\right)-\left(x-1\right)\right]dx={\int }_{-2}^{1}x\left[2-{x}^{2}-x\right]dx={\int }_{-2}^{1}\left(2x-{x}^{4}-{x}^{2}\right)dx\hfill \\ & ={\left[{x}^{2}-\frac{{x}^{5}}{5}-\frac{{x}^{3}}{3}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{1}=-\frac{9}{4}.\hfill \end{array}$

Therefore, we have

$\stackrel{–}{x}=\frac{{M}_{y}}{m}=-\frac{9}{4}·\frac{2}{9}=-\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{y}=-\frac{27}{10}·\frac{2}{9}=-\frac{3}{5}.$

The centroid of the region is $\left(\text{−}\left(1\text{/}2\right),\text{−}\left(3\text{/}5\right)\right).$

questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
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
2x^3+6xy-4y^2)^2 solve this
femi
moe
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