To determine the centre of mass of a system with a hole, the hole should be treated as negative mass. If the geometry of the system is **symmetrical**, the centre of mass is also symmetrical in the x and y dimensions.
For each mass, its surface density $\sigma$ is equal to:
$$
\sigma = \frac{m}{A} \\
m = \sigma A
$$
Holes have negative mass, i.e., $m = -\sigma A$.
For a **one-dimensional** hole, the linear mass density uses a similar formula:
$$
\lambda =\frac{m}{L} \\
\lambda = \frac{dm}{dx}
$$
This means that a hole in a rod can use a different formula: