For a planar body—a body whose mass can be treated as lying in one plane—the moments of inertia about three mutually perpendicular axes through the same point are related by the perpendicular-axis theorem.

Let the body lie in the $xy$-plane. The $z$-axis is perpendicular to the plane. Then

$$\boxed{I_z=I_x+I_y}.$$

Why the theorem works

For a mass element at coordinates $(x,y)$, its squared distance from the $z$-axis is

$$r_z^2=x^2+y^2.$$

Its squared distance from the $x$-axis is $y^2$, while its squared distance from the $y$-axis is $x^2$. Therefore

$$I_z=\sum_i m_i(x_i^2+y_i^2),$$

$$I_x=\sum_i m_iy_i^2,$$

and

$$I_y=\sum_i m_ix_i^2.$$

Adding the last two equations gives

$$I_x+I_y=I_z.$$

The same argument holds for a continuous planar mass distribution by replacing the sums with integrals.

Example: a thin circular ring

A thin ring of mass $M$ and radius $R$ has every mass element a distance $R$ from the perpendicular symmetry axis, so

$$I_z=MR^2.$$

By rotational symmetry in the plane,

$$I_x=I_y.$$

The perpendicular-axis theorem therefore gives

$$MR^2=2I_x,$$

so

$$\boxed{I_x=I_y=\frac12MR^2}.$$

Thus the theorem can turn one easily known moment of inertia into moments about axes lying in the plane.

Scope of the theorem

The theorem requires the mass distribution to be planar and all three axes to pass through the same point. It does not apply directly to a thick three-dimensional body, because a mass element can then have a nonzero coordinate perpendicular to the chosen plane.

The perpendicular-axis theorem changes the orientation of axes through one point. This is different from the parallel-axis theorem, which shifts an axis to a parallel one without changing its orientation.