A lamina occupies the region inside the circle x 2 y 2 4y

A lamina occupies the region inside the circle x ² + y ² = 4y but outside the circle x ² + y ² = 4. Find the center of mass if the density at any point is inversely proportional to its distance from the origin.

Solution

In polar coordinates,
x^2 + y^2 = 4y ==> r^2 = 4r sin ==> r = 4 sin
x^2 + y^2 = 4 ==> r = 2.

Points of intersection:
4 sin = 2
==> sin = 1/2
==> = /6, 5/6.

Moreover, = k/(x^2 + y^2) = k/r for some constant k.

So, m = dA
..........= ( = /6 to 5/6) (r = 2 to 4 sin ) (k/r) * (r dr d)
..........= ( = /6 to 5/6) k(4 sin - 2) d
..........= k(43 - 4/3).

My = x dA
......= ( = /6 to 5/6) (r = 2 to 4 sin ) (r cos ) * (k/r) * (r dr d)
......= ( = /6 to 5/6) (k/2)r^2 cos {for r = 2 to 4 sin } d
......= ( = /6 to 5/6) k(8 sin^2() - 2) cos d
......= k((8/3) sin^3() - 2 sin ) {for = /6 to 5/6}
......= 0.

Mx = y dA
......= ( = /6 to 5/6) (r = 2 to 4 sin ) (r sin ) * (k/r) * (r dr d)
......= ( = /6 to 5/6) (k/2)r^2 sin {for r = 2 to 4 sin } d
......= ( = /6 to 5/6) k(8 sin^2() - 2) sin d
......= ( = /6 to 5/6) k(8(1 - cos^2()) - 2) sin d
......= ( = /6 to 5/6) k(6 sin - 8 cos^2() sin ) d
......= k [-6 cos + (8/3) cos^3()] {for = /6 to 5/6}
......= 4k3.

Hence, the center of mass (My/m, Mx/m) equals

(0, (4k3)/k(43 - 4/3) = (0,4sqrt(3)/(4sqrt(3)-4pi/3)