Moment of Intertia?

Question

I am having trouble setting up this problem, if you could help me I would really appreciate it!

Thank you in advance.

A lamina occupies the part of the disk x^2 + y^2 ≤ 25 in the first quadrant and the density at each point is given by the function p(x,y) = 3(x^2+y^2).

What is the moment of inertia about the origin, ∫(x^2+y^2) ρ(x,y)dxdy ?

Answer 1

it would probably be simple to change the integral from cartesian to polar.

it becomes...
∫(0.. π/2) ∫(0.. 5) 3r^5 dr dΘ , right?

Answer 2

Convert everthing to polar coordinates to start:

x = r cos (a)
y = r sin(a) where a = arctan(y/x) and r =sqrt(x^2+y^2)

then p(x,y) = p(r, a) = 3r^2

Now the integral
S(x^2+y^2)p(x,y)dxdy = Sr^2*(3*r^2)*(rdrda) = S(3*r^5)drda

the integral over a varies from 0 to pi/2, the integral on r varies from 0 to 5, so
S(3*r^5)drda = 1/2r^6*pi/2 = pi/4*(5^6) = 1.23e4 in whatever units you are given.