The part of the disk x2 y2 a2 in the first quadrant lx ly

The part of the disk x² + y² a² in the first quadrant

l_{x = ?}

l_y = ?

Solution

First of all,
m = dA = * (1/4)a^2 = (1/4)a^2.

Moments of inertia:
Ix = y^2 dA
....= ( = 0 to /2) (r = 0 to a) (r sin )^2 * (r dr d)
....= ( = 0 to /2) sin^2() d * (r = 0 to a) r^3 dr
....= ( = 0 to /2) (1/2)(1 - cos(2)) d * (r = 0 to a) r^3 dr
....= (1/2)( - sin(2)/2) {for = 0 to /2} * (1/4)r^4 {for r = 0 to a}
....= (/4) * (1/4)a^4
....= (1/16)a^4.

Iy = x^2 dA
....= ( = 0 to /2) (r = 0 to a) (r cos )^2 * (r dr d)
....= ( = 0 to /2) cos^2() d * (r = 0 to a) r^3 dr
....= ( = 0 to /2) (1/2)(1 + cos(2)) d * (r = 0 to a) r^3 dr
....= (1/2)( + sin(2)/2) {for = 0 to /2} * (1/4)r^4 {for r = 0 to a}
....= (/4) * (1/4)a^4
....= (1/16)a^4.

Hence, the radii of gyration are given by
x^ = Ix / M = (1/4)a^2 and y^ = Iy / M = (1/4)a^2.

I hope this helps!