00:01
Hello, to find the center of mass of the lunae, we need to find the mass and the momentum of the lunae with respect to the x and y axis.
00:08
The mass of the lunae is given with the double integral of the intensity over the region.
00:12
So, m is equal to double integral over r, k by d of p, o and da.
00:20
So, k is a constant of proportionality, d of p, o is the distance between the point p of x, y and the origin o, 0, 0 and r is the region of the lunae.
00:29
We can write the density as k of, k by d of p, o is equal to k by x square plus y square, the whole power 1 by 2.
00:42
So, the mass becomes m is equal to double integral over r, k by x square plus y square, the whole power 1 by 2, da.
00:52
We can switch to polar coordinates to evaluate this integral.
00:56
So, m is equal to integral theta is equal to 0 to 2 pi and integral r is equal to 3 power 1 by 2 to r is equal to 2 into 3 power 1 by 2.
01:11
So, k by r, r dr d theta and which is equal to we get k into 2 pi, i n of 2.
01:20
So, where we have used the fact that i n of, here we are using i n of 3 power 1 by 2 minus i n of 3 is equal to i n of 2.
01:36
So, the moment of the lunae with respect to the x axis is given by the double integral of the density multiplied by the distance from the point to the x axis.
01:48
So, mx is equal to double integral over r y and k by d of p, x axis da.
02:00
So, now we can write y is equal to r sin theta and d of p, x axis is equal to r sin theta...