Double Integration - Moment of Inertia problem?

Question

A lamina occupies the part of the disk x^2 + y^2 < or = 1 in the first quadrant. The density at any point is proportional to the square of its distance from the origin.

Find the moment of inertia about the origin, x-axis, and y-axis. It is ok if you stop at the integral. I just wanted to know how to get everything inside the integral.

Thanks!

Answer 1

Since the density D is proportional to r^2, it&#039;s convenient to do this problem in polar coordinates.  Think of your disk as being made up of many concentric rings, each with radius r and width dr.  The mass of each ring is then:

m=D*A=(r^2)*(2pi*r)=2pi*r^3

Moment of inertia L is defined as the integral of mass times distance from the center, (i.e. mass times r), or the integral from zero to one of 2pi*r^3*r=2/5*pi*1^5=2pi/5

I notice that the problem simply says the density is PROPORTIONAL to the square of distance from the origin, so strictly speaking, D=k*r^2 where k is some constant I can&#039;t ascertain from the information you&#039;ve given.  If k is some number other than one, it just flows right through all the calculations and you wind up with k*2pi/5 at the end.