The parallel-axis theorem is a theorem of physics that makes it easier to find some moments of inertia. Note that this writeup should be more helpful if you understand what moment of inertia is, otherwise it's basically meaningless. Sorry, Charlie (a brief analogy: mass:translational motion::moment of inertia:rotational motion)
You have a rigid object (mass is m) whose axis of rotation passes through its center of mass. Let's say you know the moment of inertia of the object around said axis. Now you have another axis, parallel to this axis, and you wanna know the moment of inertia of the object about this particular axis. Let's call distance between these two axes is d. The moment of inertia I of the object through an axis parallel to that axis that passes through its center of mass is:
I = ICM + m d2
Where ICM is the moment of inertia about an axis that passes through the center of mass. N.B., this axis must be PARALLEL.
Okay, so it may be easier to understand this with an example. Let's take the case of a uniform rod of length L, mass m, and negligible radius. If the axis of rotation were perpendicular to the rod and passed through its center of mass (which is, in this case, the geometric center), the moment of inertia would be m L2 / 12. (yes it is, and if you want proof, you can do the integral yourself). However, let's say we wanted to rotate the rod about a pivot, fixed on the end of the rod, so that the rod swung, instead of spun. This new axis of rotation would be parallel to the one passing through its center of mass, and would be L / 2 away from the center of mass (distance between center of mass and end of rod). With this in mind, we use the parallel axis theorem:
I = ICM + m d2 = m L2 / 12 + m(L/2)2 = m L2 / 12 + m L2 / 4 = m L2 / 3
So the moment of inertia about an axis that passes through the end of the uniform rod is m L2 / 3, which is exactly what other sources confirm. This other source is Unperson's writeup in the moment of inertia node.
So, the moral of this theorem is: why integrate for the moment of inertia all over again? For the low, low price of $999.99, you can use the Parallel-Axis TheoremTM!!!
my physics textbook, specific details to come later
http://en.wikipedia.org/w/index.php?title=Parallel_axes_rule&oldid=32913513 (27 December, 2005)