# 6.6 Moments and centers of mass  (Page 8/14)

 Page 8 / 14

## Key equations

• Mass of a lamina
$m=\rho {\int }_{a}^{b}f\left(x\right)dx$
• Moments of a lamina
${M}_{x}=\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{M}_{y}=\rho {\int }_{a}^{b}xf\left(x\right)dx$
• Center of mass of a lamina
$\stackrel{–}{x}=\frac{{M}_{y}}{m}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}$

For the following exercises, calculate the center of mass for the collection of masses given.

${m}_{1}=2$ at ${x}_{1}=1$ and ${m}_{2}=4$ at ${x}_{2}=2$

${m}_{1}=1$ at ${x}_{1}=-1$ and ${m}_{2}=3$ at ${x}_{2}=2$

$\frac{5}{4}$

$m=3$ at $x=0,1,2,6$

Unit masses at $\left(x,y\right)=\left(1,0\right),\left(0,1\right),\left(1,1\right)$

$\left(\frac{2}{3},\frac{2}{3}\right)$

${m}_{1}=1$ at $\left(1,0\right)$ and ${m}_{2}=4$ at $\left(0,1\right)$

${m}_{1}=1$ at $\left(1,0\right)$ and ${m}_{2}=3$ at $\left(2,2\right)$

$\left(\frac{7}{4},\frac{3}{2}\right)$

For the following exercises, compute the center of mass $\stackrel{–}{x}.$

$\rho =1$ for $x\in \left(-1,3\right)$

$\rho ={x}^{2}$ for $x\in \left(0,L\right)$

$\frac{3L}{4}$

$\rho =1$ for $x\in \left(0,1\right)$ and $\rho =2$ for $x\in \left(1,2\right)$

$\rho =\text{sin}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(0,\pi \right)$

$\frac{\pi }{2}$

$\rho =\text{cos}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(0,\frac{\pi }{2}\right)$

$\rho ={e}^{x}$ for $x\in \left(0,2\right)$

$\frac{{e}^{2}+1}{{e}^{2}-1}$

$\rho ={x}^{3}+x{e}^{\text{−}x}$ for $x\in \left(0,1\right)$

$\rho =x\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(0,\pi \right)$

$\frac{{\pi }^{2}-4}{\pi }$

$\rho =\sqrt{x}$ for $x\in \left(1,4\right)$

$\rho =\text{ln}\phantom{\rule{0.2em}{0ex}}x$ for $x\in \left(1,e\right)$

$\frac{1}{4}\left(1+{e}^{2}\right)$

For the following exercises, compute the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right).$ Use symmetry to help locate the center of mass whenever possible.

$\rho =7$ in the square $0\le x\le 1,$ $0\le y\le 1$

$\rho =3$ in the triangle with vertices $\left(0,0\right),$ $\left(a,0\right),$ and $\left(0,b\right)$

$\left(\frac{a}{3},\frac{b}{3}\right)$

$\rho =2$ for the region bounded by $y=\text{cos}\left(x\right),$ $y=\text{−}\text{cos}\left(x\right),$ $x=-\frac{\pi }{2},$ and $x=\frac{\pi }{2}$

For the following exercises, use a calculator to draw the region, then compute the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right).$ Use symmetry to help locate the center of mass whenever possible.

[T] The region bounded by $y=\text{cos}\left(2x\right),$ $x=-\frac{\pi }{4},$ and $x=\frac{\pi }{4}$

$\left(0,\frac{\pi }{8}\right)$

[T] The region between $y=2{x}^{2},$ $y=0,$ $x=0,$ and $x=1$

[T] The region between $y=\frac{5}{4}{x}^{2}$ and $y=5$

$\left(0,3\right)$

[T] Region between $y=\sqrt{x},$ $y=\text{ln}\left(x\right),$ $x=1,$ and $x=4$

[T] The region bounded by $y=0,$ $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$

$\left(0,\frac{4}{\pi }\right)$

[T] The region bounded by $y=0,$ $x=0,$ and $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$

[T] The region bounded by $y={x}^{2}$ and $y={x}^{4}$ in the first quadrant

$\left(\frac{5}{8},\frac{1}{3}\right)$

For the following exercises, use the theorem of Pappus to determine the volume of the shape.

Rotating $y=mx$ around the $x$ -axis between $x=0$ and $x=1$

Rotating $y=mx$ around the $y$ -axis between $x=0$ and $x=1$

$\frac{m\pi }{3}$

A general cone created by rotating a triangle with vertices $\left(0,0\right),$ $\left(a,0\right),$ and $\left(0,b\right)$ around the $y$ -axis. Does your answer agree with the volume of a cone?

A general cylinder created by rotating a rectangle with vertices $\left(0,0\right),$ $\left(a,0\right),\left(0,b\right),$ and $\left(a,b\right)$ around the $y$ -axis. Does your answer agree with the volume of a cylinder?

$\pi {a}^{2}b$

A sphere created by rotating a semicircle with radius $a$ around the $y$ -axis. Does your answer agree with the volume of a sphere?

For the following exercises, use a calculator to draw the region enclosed by the curve. Find the area $M$ and the centroid $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ for the given shapes. Use symmetry to help locate the center of mass whenever possible.

[T] Quarter-circle: $y=\sqrt{1-{x}^{2}},$ $y=0,$ and $x=0$

$\left(\frac{4}{3\pi },\frac{4}{3\pi }\right)$

[T] Triangle: $y=x,$ $y=2-x,$ and $y=0$

[T] Lens: $y={x}^{2}$ and $y=x$

$\left(\frac{1}{2},\frac{2}{5}\right)$

[T] Ring: ${y}^{2}+{x}^{2}=1$ and ${y}^{2}+{x}^{2}=4$

[T] Half-ring: ${y}^{2}+{x}^{2}=1,$ ${y}^{2}+{x}^{2}=4,$ and $y=0$

$\left(0,\frac{28}{9\pi }\right)$

Find the generalized center of mass in the sliver between $y={x}^{a}$ and $y={x}^{b}$ with $a>b.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.

Find the generalized center of mass between $y={a}^{2}-{x}^{2},$ $x=0,$ and $y=0.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.

Center of mass: $\left(\frac{a}{6},\frac{4{a}^{2}}{5}\right),$ volume: $\frac{2\pi {a}^{4}}{9}$

Find the generalized center of mass between $y=b\phantom{\rule{0.2em}{0ex}}\text{sin}\left(ax\right),$ $x=0,$ and $x=\frac{\pi }{a}.$ Then, use the Pappus theorem to find the volume of the solid generated when revolving around the y -axis.

Use the theorem of Pappus to find the volume of a torus (pictured here). Assume that a disk of radius $a$ is positioned with the left end of the circle at $x=b,$ $b>0,$ and is rotated around the y -axis.

Volume: $2{\pi }^{2}{a}^{2}\left(b+a\right)$

Find the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ for a thin wire along the semicircle $y=\sqrt{1-{x}^{2}}$ with unit mass. ( Hint: Use the theorem of Pappus.)

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
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
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