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:

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 most recent figure (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