Let us now solve a few examples.

Example1: Let us start with a simple example of a square of side a with its center of the origin (see figure 2).

Figure 3

To calculate this, we choose the elemental area as shown in figure 4 and integrate. Then

dA = ady

so that

Similarly for calculating I_YY we choose a vertical elemental area and calculate

Let us also calculate the product of inertia. Choose on elemental area dxdy and calculate (see figure 5)

As noted earlier, I_YX is equal to I_XY and therefore it also vanishes.

A related problem is that of a rectangular area of size a x b. Its length of side a is parallel to the x-axis and the other side of length b is parallel to the y-axis. I leave this as an exercise for you to show that in this case , . Notice that due to the area being symmetrically distributed about the x- and y-axes, the product of the area comes out to be zero.

Example 2 : Next let us consider a quarter of an ellipse as shown in figure 6 and calculate the moment and product of area for this area.

Equation of the ellipse whose quarter is shown in figure 6 is: . For calculating choose an area element parallel to the x-axis to calculate dA=xdy and perform the integral

Using the equation for ellipse, we get

which gives

This integral can easily be performed by substituting y = b sin θ and gives

Similarly by taking a vertical strip to perform the integral, we calculate

and get

Next we calculate the product of area I_XY . To calculate I_XY , we take a small element ( ) as shown in figure 7, multiply it by x and y and integrate to get

For a given x , the value of y changes from 0 to so the integral is

This integral is easily performed to get

Thus for a quarter of an ellipse, the moments and products of area are

If we put a = b , these formulas give the moments and products of area for a quarter of a circle of radius a . I will leave it for you to work out what will be for the full ellipse about its centre.

Using the second moment of an area, we define the concept of the radii of gyration. This is the point which will give the same moment of inertia as the area under consideration if the entire area was concentrated there. Thus

define the radii of gyration k_X and k_Y about the x- and the y-axes, respectively. In the example of a rectangular area of size a x b with side a parallel to the x-axis, we had , . So for this rectangle, the radii of gyration are and .

Having defined the moments and products of area, we now describe a relationship between the second moment of an area about a set of axes passing through the centroid of that area and another set of (x-y) axes which are parallel to those passing through the centroid. This is known a transfer theorem.