Application for Center of Mass and Moment of Inertia?

Question

Evaluate the intreagl for the moment of inertai about the x-axes for the solid region region lying between the hemisphere z=sqrt(4 -x^2 -y^2) on the xy plane


∫ (from -2 to2) ∫ (from -√(4-x^2 ) to √(4-x^2)) ∫ ( from 0 to √(4 - x^2 - y^2)

for the equation kz(x^2+y^2) dz dy dx

Answer 1

You seem to want to do it the hard way. I can't tell if that is required here because I can't figure out what the geometry actually looks like, but surely it would be easier to recognize either disks or half disks for what they are and just do one simple integral?

For example, the center of gravity of a hemisphere of radius 1 would be:

integral from r = 0 to r = 1 of rV(r)
where V(r) is the volumetric element of the disk at distance r from the center.

Thus V(r) = A(r)dr = (pi)(R(r)^2)dr = (pi)(1 - r^2)dr

(Note that the disk at distance r has a radius R(r) such that r^2 + R(r)^2 = 1)

So the integral is a simple integral of a polynomial in r and can be easily computed.

Once you have the center of gravity for a hemisphere of radius 1, you can scale it to any size you need.

The integral for the moment of inertia of a hemisphere around around an axis perpendicular to its axis of symmetry is similarly that of a simple polynomial.

In any case, remember the Parallel Axis Theorem for moments of inertia as it makes computations much simpler:
http://en.wikipedia.org/wiki/Moment_of_inertia

HTH.

Answer 2

For the moment of inertia, the formula for the function in the integral is k*w^2 where w is the distance from a point in the body to the axis of rotation (here it's x).

So in this case,
w= [y^2 + z^2] ^1/2 (draw it to see it) and you need to integrate

k[y^2 + z^2] = k(r^2*sin^2(theta) +z^2)
The limits are ok, but if you switch to cylindrical coordinates rdrd(theta)dz your integration will be possibly easier:

∫ (from 0 to 2pi∫ (from -2 to2) ∫ ( from 0 to √(4 - r^2) of (kw^2) dz rdrdtheta

Answer is 2/5* m*4^2

Answer 3

1 day has passed and no response!! work the integral in cylindrical coords. [0, 2 pi] [0, 2] [0, 4 - r^2] are the new limits....64 pi ????