moment of inertia about central axis

Question

A solid D is bounded below by a right circular cone whose generators have length a and make an angle π/6 with the central axis. It is bounded above by a portion of the sphere of radius a centered at the vertex of the cone. Find its moment of inertia about its central axis, assuming the density δ at a point is numerically equal to the distance of the point from a plane through the vertex perpendicular to the central axis. 

Answer 1

Sue,

 

This is a problem in triple integration from Calculus III. There is a formula for finding the moment of inertia about the axis of symmetry.

 

If you think of the cone as having its vertex at the origin in a 3D-coordinate system, you can easily describe the cone and sphere using spherical coordinates. (The cone has equation φ = π/6 and the sphere has equation ρ = a.)

 

You can find the moment of inertia about the z-axis using the appropriate formula, which is:

 

I_z  = ∫∫∫ (x² + y²) δ dV,

 

where you integrate over the solid object as described in your problem. The density can be expressed as z, assuming the plane from which the distance is measured is the xy-plane in xyz-space.

 

If you need more help setting up the integral in spherical coordinates or otherwise, you will need to post a reply to this so I can offer more help. Depending on how the integrand appears, you may wish to re-express the integral in a different coordinate system (spherical or cylindrical).