I need to calculate the area moment of inertia of a cross for a homework assignment. The cross is a symmetric plus sign:

```
    _
   | |
|==   ==|
   | |
    -

```

where the length/width of each arm is 1" (so the total area is 5 square in.).

My understanding is that I can just separate the cross into three sections (left, middle, right), so the area moment of inertia would be

$$ I = \sum \frac{bh^3}{12} = \frac{1\times 1^3 + 3\times 1^3 + 1\times 1^3}{12} = \frac{5}{12}$$

This is surprising to me, because a 2.2" x 2.2" square would have the same area but a much larger moment of inertia:

$$ I = \frac{bh^3}{12} = \frac{(\sqrt{5})^4}{12} = 2.08$$

Since the arms of the cross span 3 inches, I would expect there to be more material further from the center, whereas the square only has a small amount of area in the corners that the cross doesn't have.

Are these formulas correct or am I doing something wrong? Thanks in advance.