Find the static moment of a homogeneous regular hexagon
Find the static moment of a homogeneous regular hexagon with side $a$ ($\varrho=1$) with respect to one of its sides.
Explanation of Method
With one side on the $x$-axis, the hexagon's centroid coincides with its geometric center at height equal to the apothem, so $M=m\bar y$ gives the answer immediately from two standard hexagon facts. A direct decomposition into two trapezoidal strips (matching the hexagon's linearly-varying width) confirms the same result by explicit integration.
- Recall the hexagon's area and centroid height.
- Compute $M=m\bar y$.
- Verify by direct integration over two trapezoidal strips.
Step 1 · Hexagon geometry
With one side on the $x$-axis: apothem (centroid height) $\bar y=\dfrac{a\sqrt3}2$, area $m=\dfrac{3\sqrt3}2a^2$.
Step 2 · Static moment via centroid
$$ M = m\bar y = \frac{3\sqrt3}2a^2\cdot\frac{a\sqrt3}2 = \frac{9a^3}4. $$
Step 3 · Direct verification by strips
The width varies linearly: $w(y)=a+\dfrac{2y}{\sqrt3}$ for $0\le y\le\frac{a\sqrt3}2$, and $w(y)=3a-\dfrac{2y}{\sqrt3}$ for $\frac{a\sqrt3}2\le y\le a\sqrt3$. Direct integration of $\int yw(y)\,dy$ over both strips gives $I_1=\dfrac{5a^3}8$ and $I_2=\dfrac{13a^3}8$, summing to $\dfrac{18a^3}8=\dfrac{9a^3}4$, matching Step 2.
Additional Notes
- This derivation matches the source exactly, and its own internal cross-check (centroid shortcut vs. direct strip integration) agrees.
Result
The static moment is: