How to find moments and center of mass for a given shape using integration?

tnutty
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Not really a homework problem but related. In one of my homework problem, it says ,
Calculate the moments Mx and My and the center of mass of a lamina with density ρ = 4 and the given shape.
Mx = 2
My = 12
(x, y) = (2,1/3 )

I got the answer. YOUR WELCOME.


But, I only got it because I looked up the formula to find Mx,My and (x,y).
the formula is given in this site near the intro http://www.math.utep.edu/Faculty/javila/Calculus%20II/Math%201312%20problems%20for%20section%207.6.pdf

But I was hoping one of you could prove this, since my book didn't.
 
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What is the given shape?
 
it is a function of f(x) = 1/3x.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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