let P be the portion of the disk x^2 + y^2 <= 1, lying in the first quadrant. having mass density of p=x, if the mass of P is 1/3 , find the y coordinate of the center of mass of P.

the answer is 3/8

It lies on the 45 degree line by symmetry so all we need to do is find the y coordinate, the mass does not matter, only uniform density call it 1

mass = 1/4 pi (1)^ 2 = pi/4
divide moment by this in the end

moment about x axis of strip at y of width dy = y (x dy)
integrate from y = 0 to y = 1

y (1 - y^2)^.5 dy from 0 to 1

let y = sin t
dy = cos t dt

sin t cos t cos t dt from 0 to pi/2
sin t cos^2 t dt
so
(-1/3)cos^3 t at t = pi/2 -at t = 0
= 1/3

now divide by pi/4
(1/3)/pi/4 = 4/(3pi)