whilst the polar quadratic moment with respect to O is
I0 = Σ {m (x2 + y2 + z2)}.
(11)
We note that
Iyz = Iy + Iz, Izx = Iz + Ix, Ixy = Ix + Iy,
(12)
and
I0 = Ix + Iy + Iz = 1⁄2 (Iyz + Izx + Ixy).
(13)
In the case of continuous distributions of matter the summations in (9), (10), (11) are of course to be replaced by integrations. For a uniform thin circular plate, we find, taking the origin at its centre, and the axis of z normal to its plane, I0 = 1⁄2Ma2, where M is the mass and a the radius. Since Ix = Iy, Iz = 0, we deduce Izx = 1⁄2Ma2, Ixy = 1⁄2Ma2; hence the value of the squared radius of gyration is for a diameter 1⁄4a2, and for the axis of symmetry 1⁄2a2. Again, for a uniform solid sphere having its centre at the origin we find I0 = 3⁄5Ma2, Ix = Iy = Iz = 1⁄5Ma2, Iyz = Izx = lxy = 3⁄5Ma2; i.e. the square of the radius of gyration with respect to a diameter is 2⁄5a2. The method of homogeneous strain can be applied to deduce the corresponding results for an ellipsoid of semi-axes a, b, c. If the co-ordinate axes coincide with the principal axes, we find Ix = 1⁄5Ma2, Iy = 1⁄5Mb2, Iz = 1⁄5Mc2, whence Iyz = 1⁄5M (b2 + c2), &c.