The polar moment of inertia with respect to an axis through O perpendicular to the plane of the figure is defined by the integral:
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To illustrate this integral, let's look at an arbitrary figure with a differential area dA:

We can see from the above figure that the integral of polar moment of inertia is similar to those for moments of inertia Ix and Iy. In fact:
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This is a important relationship because it shows that 'the polar moment of inertia with respect to an axis perpendicular to the plane of the figure at any point O is equal to the sum of the moments of inertia with respect to any two perpendicular axes x and y passing through that same point and lying in the plane of the figure' (Gere 2004, p.841).
The above theory can be furthered to include the parallel-axis theorem for polar moments of inertia:

Figure 1
Let's designate the polar moments of inertia at points O and C by Ipo and Ipc, respectively. Then, by using the equation from polar moment of inertia derived above, we can write:
Ipo = Ix + Iy and Ipc = Ixc + Iyc
Now we use the parallel-axis theorem to get:
Ix + Iy = Ixc + Iyc + A(a2 + b2)
Ipo = Ipc + A(a2 + b2)
In fact a2 + b2 is the squared distance from point O to C.
Integral for the product of inertia with respect to the x and y axes is given by:
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The result may be positive, negative or zero depending on where the x and y axes are located. Also, if an axis acts through the line of symmetry of the object, its product of inertia will be zero.
If we refer back to figure 1 (above), we can derive the following expression for the parallel-axis theorem for products of inertia:

Gere J.M. 2004, Mechanics of Materials, 6th edn, Thomson, USA