Expert: Scotto - 2/16/2009

Question
A lamina with uniform density p= 18 is bounded by the curve 9x-x2 and the x axis.
Give moments and center of mass. (in exact, simplified, non calculator answers.)  
   Mx=
   My=
   Center of Mass=


Answer
I found in libraryofmath.com/moments-and-center-of-mass.html
that
⌠
⌡yf(x) dA = Mx and
⌠
⌡xf(y) dA = My where the integral is over the region and
dA is for the variable used to find the area,
once we have computed the appropriate f(x) and f(y).
They are suppose to be probability distributions with
an integral over the region of 1.

To get a better understanding, read the paper yourself.
It is only 2 ro 3 pages long.

What it has at the top is that m(x bar) = My and m(y bar) = My.

I believe when you have found My and Mx, the answer needs to be
multiplied by 18 to get the mass.

From what I see, to calculate My, I would take
⌠9
⌡0 x(9x-x²)dx =
⌠9
⌡0 9x²-x³ dx = 9x³/3 - x^4/4 = 3x³ - x^4/4 evaluated(0 to 9).
We can see that putting in 0 gives us 0, so putting in 9 gives us
3*729 - 6561/4 = 2,187 - 1640.25 = 546.75.

If I understand it correctly, to find Mx, I would take
⌠9
⌡0 y(9x-x²) dy.  It would seem that 9x-x² would need to be
inverted into a function on y.

To find the center of mass, first we obtain the mass.
⌠9⌠9x-x²
⌡0⌡0        1 dy dx =
⌠9
⌡0 9x-x² dx =
9x²/2 - x³/3(0 to 9) =
9*729/2 - 729/3 = 6561/2 - 243 = (6561-486)/2 = 3037.5.

I hope this is useful, but read the paper to
really get a grasp on it.

I wish I could offer more assistance, but I'm still getting a grasp
on howt Mx and My are found.