# Moments,Center of Mass, & Centroid

1. Jan 26, 2012

### chapsticks

1. The problem statement, all variables and given/known data
Find Mx,My, & (x bar, y bar) for the laminas of uniform density ρ bounded by the graphs of the equations. (Use rho for ρ as necessary.)
x=-y
x=5y-y2

2. Relevant equations

m= ∫f(x)-g(x) dx
my= ∫x(f(x)-g(x)) dx =>x bar my/m
mx= 1/2 ∫ (f(x))2-g(x))2dx => y bar=mx/m

3. The attempt at a solution
So this is my work

x=-y <-- g(y)
x=5y-y^2 <----f(y)

a=0
b=6

*note I don't know how to put 0 to 6 on the integral

m=p ∫ [(5y-y^2)-(y)]dy
=p [3y^2 -(y^3/3)]= 36 p

My= p∫[(5y-y^2)+((-y)/2)][(5y-y^2)-(-y)]
=p/2∫ (4y-y^2)(6y-y^2)dy
=p/2∫ (y^4-10y^3+24y^2) dy
= p/2 [(y^5/5)-(5y^4/2)+8y^3]
=216/5 p is wrong I don't know why :?:

2. Jan 27, 2012

### vela

Staff Emeritus
The general form of the relevant equations is
\begin{align*}
m &= \iint \rho\,dx\,dy \\
M_x &= \iint \rho y\,dx\,dy \\
M_y &= \iint \rho x\,dx\,dy
\end{align*}
When the region of interest is between x=a and x=b and is bounded on the top by f(x) and on the bottom by g(x), you get the equations you cited. For instance, for the moment about the x-axis, you get
$$M_x = \rho \int_a^b \int_{g(x)}^{f(x)} y\,dy\,dx = \rho \int_a^b \left.\frac{y^2}{2}\right|_{g(x)}^{f(x)} \,dx = \frac{1}{2}\rho \int_a^b [f(x)^2-g(x)^2]\,dx$$
If you sketch the region for this particular problem, however, you'll see the roles of x and y appear to be reversed, so the formulas you were trying to use don't work. You'll need to derive the correct ones or adapt the ones you have for this particular case.