Question

Calculate the center of mass of the area between f(x)=x and f(x)=x^2 mass=12x

Answer 1

i dont have time to answer but others may find the diagram useful

Answer 2

i know the diagram, im just confused of calculating the Center of Mass Coordinates and the moments of y and x.

Answer 3

isn't the center of mass found by this sum:
\[\int\limits_{0}^{1}x*f(x) dx\]
where f(x) is height of region
f(x) = x-x^2

Answer 4

i haven't actually done this kind before but drawing from what i know of moments

would it be:

\[\frac{\int\limits{12x(x-x^2)x}dx}{\int\limits{12x(x-x^2)}dx}\]

Answer 5

aah i have to go

Answer 6

the mass dm=(12x)(x-x^2)dx

Answer 7

aw ok

Answer 8

moment of y=∫1/2(x^2)(12x)(x-x^2)dx from 0 to 1 right?

Answer 9

that would be moment of x, my bad. so the solution would give me y bar. am i correct?

Answer 10

let (x',y') be the centroid. then
\[x'=\ M_y/Area\]
\[y'=\ M_x/2/Area\]
where M_y is moment about y axis and M_x is moment about x axis
find first Area bounded by the graphs:
\[A=\int\limits_{0}^{1}x-x^2dx=(x^2/2 -x^3/3)|_{0}^{1}=1/6\]
Moment about the x axis:
\[1/2\int\limits_{0}^{1}x^2-x^4dx=1/2(x^3/3-x^5/5)|_{0}^{1}=1/2(2/15)=1/15\]
Moment about y-axis:
\[\int\limits_{0}^{1}(x-x^2)xdx=\int\limits_{0}^{1}x^2-x^3dx=(x^3/3-x^4/4)|_{0}^{1}=1/12\]
x'=1/12/1/6=1/2
y'=1/15/1/6=6/15
so our centroid is (1/2,6/15)

Answer 11

here is a good reference
http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

Answer 12

@anonymoustwo44 but you didnt even use the rate of mass. it wasnt constant it was 12x

Answer 13

@Zarkon how would you factor in the rate of density?

Answer 14

wouldn't the mass be

\[m = \int\limits{(x - x^2) \times 12x}dx\]