Fortunately, we do not need to use calculus to solve this problem. Notice that all the mass is the same distance R from the axis of rotation. Thus we do not need to integrate over a range, but can calculate the total moment of inertia. Each small element dm has a rotational inertia of R ² dm , where r is constant. Summing over all elements, we see that I = R ² dm = R ² M . The sum of all the small elements of mass is simply the total mass. This value for I of MR ² agrees with experiment, and is the accepted value for a hoop.

To solve this problem we split the cylinder into small hoops of mass dm , and width dr :

This small element of mass has a volume of (2Πr)(L)(dr) , where dr is the width of the hoop. Thus the mass of this element can be expressed in terms of volume and density:

We also know that the total volume of the entire cylinder is given by: V = AL = ΠR ² L . In addition, our density is given by the total mass of the cylinder divided by the total volume of the cylinder. Thus:

Substituting this into our equation for dm ,

Now that we have dm in terms of r , we simply have to integrate over all possible values of r to get our rotational inertia:

Thus the rotational inertia of a cylinder is simply . Once again, it has the form of kMR ² , where k is some constant less than one.