# Moment of inertia

• March 30th 2009, 11:22 AM
s7b
Moment of inertia
Can someone explain how to set this up;

Find the moment of inertia about the z-axis of a this shell of constant density cut from the cone 4x^2+4y^2-z^2=0, z is greater than 0 by the circular cylinder x^2+y^2=2x

The formula for moment of inertia is:
I=double surface integral of (x^2+y^2)density ds

-How do you know which is the surface?
• March 30th 2009, 11:56 AM
NonCommAlg
Quote:

Originally Posted by s7b
Can someone explain how to set this up;

Find the moment of inertia about the z-axis of a this shell of constant density cut from the cone 4x^2+4y^2-z^2=0, z is greater than 0 by the circular cylinder x^2+y^2=2x

The formula for moment of inertia is:
I=double surface integral of (x^2+y^2)density ds

-How do you know which is the surface?

the problem is telling you that the surface is the cone $z=2 \sqrt{x^2 + y^2}$ and the region of integration is $D=\{(x,y): \ x^2 + y^2 \leq 2x \}.$

you'll eventually need to use polar coordinates. note that then the limits of integration will be: $\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}, \ \ 0 \leq r \leq 2 \cos \theta, \ .$
• March 30th 2009, 12:01 PM
s7b
Do you just use the density for example just as "b"
• March 30th 2009, 12:06 PM
s7b
Also when you find ds,
how the heck do you find (dz/dx)^2